Mathematics Tricks & Tips And Key Problem Solving Concepts of Complex Number for NDA

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Tricks & Tips And Key Problem Solving Concepts of Complex Number for NDA

Solving powers of iota

`i^(4n) = 1, i^(4n +1) =i , i^(4n +2) = -1, i^(4n +3) = - i , \ \ n in I`

The sum of four consecutive powers of `i` is zero. i.e `i^n + i^(n +1) + i^(n +2) + i^(n +3) = 0, n in I`

Q 2723680541

The value of `i^(2n) + i^(2n + 1) + i^(2n + 2) + i^(2n + 3)`, where `i = sqrt(-1)`, is
NDA Paper 1 2017

(A)

`0`

(B)

`1`

(C)

`i`

(D)

`-i`

Solution:

`i^(2n) (1+i +i^2 +i^3)`

`= i^(2n)( 1+i -1 -i)=0`

Correct Answer is `=>` (A) `0`

Q 1659591414

What `( (1 + i)^( 4n+ 5))/((1 - i)^( 4n+ 3))` equal to, where `n` is a natural

number and `i = sqrt(-1)`?
NDA Paper 1 2014

(A)

`2`

(B)

`2i`

(C)

`-2i`

(D)

`i`

Solution:

We have, ` ((1 + i)^( 4n+ 5))/((1 - i)^( 4n+ 3))`

` = ((1 + i)^( 4n+ 3) . ( 1 + i)^2) /( (1 - i)^( 4n+ 3))`

` = ((1 + i)/(1 - i))^( 4n+ 3) . ( 1 + i)^2`

` = [ ((1 + i)(1 + i))/ ((1 - i)(1 - i))] ^( 4n+ 3) . ( 1 + i^2 + 2i) `

` = [ (1 + i^2 + 2i)/( 1 + 1) ]^( 4n+ 3) . 2i`

` = (i)^( 4n+ 3) . 2i = 2 (i) ^( 4n+ 4) = 2`

Correct Answer is `=>` (A) `2`

Q 2386734677

The value of the sum `sum _(n = 1)^(13) (i^n + i^(n+ 1) )`, where `i = sqrt(-1)` ,
is
NDA Paper 1 2012

(A)

`i`

(B)

`-i`

(C)

`0`

(D)

`i - 1`

Solution:

Given, ` sum_(n = 1)^(13) (i^n + i^(n + 1)) = ( 1 + i) sum_(n = 1)^(13) (i^n)`

` = (1 + i) (i + i^2 + i^3 + i^4 + ... + i^(13))`

`= (1 + i ) (i^(13))`

(since, sum of four consecutive power of i is zero)

` = (1 + i) (i^4)^3 .i`

`= (i + i^2) ( ∵ i^4 = 1)`

`= ( i - 1)`

Correct Answer is `=>` (D) `i - 1`

Q 2326545471

The smallest positive integral value of `n` for which
`( (1 - i)/(1 + i))^n` is purely imaginary with positive
imaginary part is
NDA Paper 1 2011

(A)

`1`

(B)

`3`

(C)

`4`

(D)

`5`

Solution:

Let `z = ( (1 - i)/(1 + i))^n = {((1- i) (1 - i))/((1 + i)(1 - i)) }^n`

` = { (1 - i)^2/(1 - i) }^n = { (1 + i^2 - 2i)/(1 + 1) }^n`

` = { ( 1- 1 - 2i)/2 }^n = (-i)^n`

Here, the smallest positive integral value of `n` tor which '`z`' is

purely imaginary with positive imaginary part should be `3`.

`=> (-i)^3 = -i^3 = -i^2 - i = - (-1) - i = i`

Correct Answer is `=>` (B) `3`

Q 2356856774

What is the value of `1 + i^2 + i^4 + i^6 + ... + i^(100)` ,
where `i = sqrt(-1)`?
NDA Paper 1 2010

(A)

`0`

(B)

`1`

(C)

`-1`

(D)

None of these

Solution:

`1 + i^2 + i^ 4 + i^6 + ... + i^( 100)`

`= 1 - 1 + 1 - 1 + ... + 1 = 1`

Correct Answer is `=>` (B) `1`

Q 2386778677

For positive whole number of `n`, what is the value
of `i^(4n+1)`?
NDA Paper 1 2008

(A)

`1`

(B)

`-1`

(C)

`i`

(D)

`-i`

Solution:

`i^(4n +1) = (i^4 )^n xx i`

`= (1)^n xx i = 1 xx i = i`

Correct Answer is `=>` (C) `i`

Q 1628523401

What is the square root of i, where `i = sqrt(-1)`?
NDA Paper 1 2015

(A)

` (1 +i)/2`

(B)

` (1 - i)/2`

(C)

` (1 +i)/sqrt(2)`

(D)

None of these

Solution:

Given, `i = sqrt(-1)`

`:. sqrt(i) = sqrt(e^(i pi/2)) = e^(i pi/4) = cos (pi/4) + i sin ( pi/4) = 1/ sqrt(2) + i/sqrt(2) = (1 + i)/sqrt(2)`

Correct Answer is `=>` (C) ` (1 +i)/sqrt(2)`

Q 1785423367

For a positive integer `n`, find the value of `(1 - i)^n ( 1 - 1/i)^n`.
NCERT Exemplar

Solution:

Given expression ` (1 - i)^n ( 1 - 1/i)^n`

`= (1- i)^n (i -1)^n . i^(-n) = (1 - i)^n (1- i)^n (-1)^n .i^(-n)`

` = [(1- i)^2 ]^n (-1)^n . i^(-n) = (1 + i^2 -2i)^n (-1)^n i^(-n)`

` = (1-1 - 2i)^n (-1)^n i^(-n) = (-2)^n . i^n(-1)^n i^(-n)`

` = (-1)^(2n) .2^n = 2^n`

Q 2356645574

What is the value of `(1 + i)^5 + (1 - i)^5` , where
`i = sqrt(-1)`?
NDA Paper 1 2011

(A)

`-8`

(B)

`8`

(C)

`8i`

(D)

`-8i`

Solution:

Given, `(1 + i)^5 + (1 - i)^5`

`={ text()^5C_0 + text()^5C_1 i + text()^5C_2 i^2 + text()^5C_3 i^3+ text()^5C_4 i^4 + text()^5C_5 i^5}`

`- { text()^5C_0 - text()^5C_1 i + text()^5C_2 i^2 - text()^5C_3 i^3 + text()^5C_4 i^4 - text()^5C_5 i^5 }`

(`∵` by Binomial theorem)

`= 2{ text()^5C_1 i + text()^5C_3 i^3 + text()^5C_5 i^5}`

`= 2 { 5i - 10i + i) = 2(-4i)} = - 8i`

Correct Answer is `=>` (D) `-8i`

Q 1735523462

Evaluate ` sum_(n=1)^(13) (i^n + i^(n+1))`, where `n in N`.
NCERT Exemplar

(A)

i+1

(B)

i-1

(C)

-i+1

(D)

None of these

Solution:

` sum_(n = 1)^(13) (i^n+ i^(n + 1)) . n in N = sum_(n = 1)^(13) i^n (1 + i)`

` = (1 + i) [i + i^2 + i^3 + i^4 + i^5 + i^6 + i^7 + i^8 + i^9 + i^(10) + i^(11) + i^(12) + i^(13)]`

` = (1 + i) [i^(13) ] [∵ i^n + i^(n + 1) + i^(n+ 2) + i^(n+ 3) = 0`, where `n in N` i.e., `sum_(n = 1)^(13) i^n = 0`

` = (1 + i) i`

` = (i^ 2 + i) = i - 1`

Alternate Method
Given that, ` sum_(n=1)^(13) (i^n + i^(n+1)) . n in N`

` = (i + i^2 + i^3 + i^ 4 + i^5 + i^6 + i^7 + i^8 + i^9 + i^(10) + i^(11) + i^(12) + i^(13))`

` + (i^2 + i^3 + i ^4 + i ^5 + i^6 + i ^7 + i^8 + i^9 + i^(10) + i^911) + i^(12) + i^(13) + i^(14))`

`= (i + 2i^2 + 2i^3 + 2i^ 4 + 2i^ 5 + 2i^ 6 + 2i^ 7 + 2i^ 8 + 2i^ 9 + 2i^(10) + 2i^(11) + 2i^( 12) + 2i^(13) + i^(14))`

` = i -2 - 2i + 2 + 2i + 2(i^ 4)i^ 2 + 2(i)^4 i^3 + 2(i^ 2)^4 + 2(i^2)^4 i + 2(i^ 2)^5`

` +2(i^2)^5. i + 2(i^2)^6 + 2(i^2)^6. i +(i^2)^7`

`= i - 2- 2i + 2 + 2i - 2 - 2i + 2 + 2i - 2- 2i + 2 + 2i - 1 - 1 + i = i-1`

Correct Answer is `=>` (B) i-1

Q 2346767673

What is the least positive integer `n` for which `((1 + i)/(1 - i))^n =1 ` ?
NDA Paper 1 2010

(A)

`16`

(B)

`12`

(C)

`8`

(D)

`4`

Solution:

`(1 + i)/(1 - i) = (1 + i)^2/(1 + 1) = ( 1 + i^2 + 2i)/2 = i`

`:. ((1 + i)/(1 - i))^n = i^n = 1`

which is possible for `n = 4`

Correct Answer is `=>` (D) `4`

Q 2549780613

The least positive integral value of `n` for which `( sqrt3 + i)^n = ( sqrt3 - i)^n`, is
BCECE Mains 2015

(A)

`3`

(B)

`4`

(C)

`6`

(D)

None of these

Solution:

We have,

`( sqrt3 + i)^n = ( sqrt3 - i)^n`

` => [ ( sqrt3 + i)/( sqrt3 - i) ]^n = 1`

` => [ ( -1 + i sqrt3)/( 1 + i sqrt3) ]^n = 1`

` => [ (2 omega)/(- 2 omega^2) ]^n = 1`

` [ ∵ (-1)/2 + i sqrt(3/2) = omega ` and ` - 1/2 - i sqrt3/2 = omega^2 ]`

` => (-1)^n omega^(2n) = 1`

So, `n` is an even multiple of `3`.

`:. n = 6`

Correct Answer is `=>` (C) `6`

Q 1163256145

`((1+i)/(1-i))^4+((1-i)/(1+i))^4=`
EAMCET

(A)

`0 `

(B)

`1``

(C)

`2`

(D)

`4`

Solution:

`(1+i)/(1-i)=((1+i)(1+i))/((1-i)(1+i))=(1+i^2+2i)/(1-i^2)=(1-1+2i)/(1+1)`

`=(2i)/2=i`

`And, (1-i)/(1+i)=1/((1+i)/(1-i))=1/i=i/i^2=-i`

`i^4+(-i)^4=2`

Correct Answer is `=>` (C) `2`

Finding modulus and argument of complex number

The unique value of e such that `- pi < theta le pi` is called principal value of argument.

Mathematically to find Principal Argument of `z`

Let `z = a + ib`

Find `alpha= tan^(-1) ((|b|)/(|a|))`

Case I : If `z` lies in `I` quadrant i.e. `a, b > 0` then `amp (z) = theta = alpha`.

Case II : lf `z` lies in `II` quadrant i.e. `a < 0, b > 0` then `amp(z) = theta = (pi - alpha)`

Case III : If `z` lies in Ill quadrant i.e. `a < 0, b < 0` then `amp(z) = theta =- (pi - alpha)`

Case IV : lf `z` lies in IV quadrant i.e `a > 0, b < 0` then `amp(z) = theta =- alpha`.

Q 2723880741

The modulus and principal argument of the complex number ` (1 + 2i)/(1 - (1 - i)^2)` are respectively
NDA Paper 1 2017

(A)

`1, 0`

(B)

`1, 1`

(C)

`2, 0`

(D)

`2, 1`

Solution:

`z= (1+2i)/(1-(1-i)^2) =(1+ 2i)/(1+ 2i) =1`

`|z|=1`

`arg z = tan^(-1) | 0/1| =0`

Correct Answer is `=>` (A) `1, 0`

Q 2670356216

Find the principal argument for following points `A(1+ sqrt3 i) , B(-1 + sqrt 3 i), C(-1 -sqrt 3 i), D(1- sqrt 3 i)`

Solution:

`A (1+ sqrt 3 i)` Here `, a>0 , b > 0` `A` lies in `I` quadrant

`amp (A)= theta_A =tan^(-1) ((|b|)/(|a|))`

`=tan^(-1) |(sqrt 3)/1|= pi/3`

For ` B(-1 + sqrt 3 i) ,` `a < 0, b > 0` then B lies in `II` quadrant

`amp (B)= theta_B= pi - tan^(-1) ((| sqrt 3|)/(|-1|))`

`=pi - pi/3 =(2 pi)/3`

For `C(-1 -sqrt 3 i)`, `a < 0, b < 0` ,`C` lies in Ill quadrant

`amp (C)= theta_C= -pi + tan^(-1) ((|sqrt 3|)/(|-1|))= -pi + pi/3`

`=-(2 pi)/3`

For `D(1- sqrt 3 i)` `a > 0, b < 0` `D` lies in IV quadrant

`amp (D) = theta_D =-tan^(-1) ((|sqrt 3|)/(|-1|))`

`=- pi/3`

Q 2116223170

Let `z` be a complex number satisfying

` | (z-4)/(z-8) |= 1 ` and `|z/(z-2) | = 3/2`

What is `| z |` equal to?
NDA Paper 1 2016

(A)

`6`

(B)

`12`

(C)

`18`

(D)

`36`

Solution:

Let ` z= x + iy`

`:. |(z- 4)/(z-8)| = 1 => | (x+ iy - 4)/(x + iy -8) | =1`

` => (| x + iy- 4|)/ (| x + iy- 8|)= 1`

` => sqrt((x-4)^(2) + y^(2))/ sqrt((x-8)^(2) + y^(2)) = 1`

` => (x- 4)^(2) + y^(2) = (x- 8)^(2) + y^(2)`

` => (x - 4)^(2) = (x - 8 )^(2)`

` => (x - 4) = pm (x - 8)`

` :. x=6`

Also,` |z/(z-2)| = 3/2 => 2| z |= 3| z- 2|`

` => 2 sqrt(x^(2) + y ^(2)) = 3 xx sqrt((x- 2)^(2) + y^( 2))`

`=> 4(x^(2) + y^(2)) = 9 [(x- 2)^(2) + y^(2)]`

For `x = 6, 4(36 + y ^(2)) = 9(16 + y^( 2))`

`=> 144 + 4y^(2) = 144 + 9y^(2) => y = 0`

Hence, `z = 6 + 0i`

` |z| = sqrt((6)^(2) +( 0)^(2)) = 6`

Correct Answer is `=>` (A) `6`

Q 2136323272

Let `z` be a complex number satisfying

` | (z-4)/(z-8) |= 1 ` and `|z/(z-2) | = 3/2`

What is `| (z-6)/(z+6) |` equal to?

NDA Paper 1 2016

(A)

`3`

(B)

`2`

(C)

`1`

(D)

`0`

Correct Answer is `=>` (D) `0`

Q 2270367216

If `z_( 1)` and `z_( 2)` are complex numbers with `| z_(1) |=| z _(2 )| `, then
which of the following is/are correct?
1. `z_(1) =z_(2)`
2. Real part of `z_( 1) =` Real part of `z_( 2)`
3. Imaginary part of `z_( 1) =` Imaginary part of `z _(2)`

Select the correct answer using the code given below
NDA Paper 1 2015

(A)

Only `1`

(B)

Only `2`

(C)

Only `3`

(D)

None of these

Solution:

We have, `z_(1) =z_(2)`

Let `z_(1) = x_(1) + iy_(1)` and `z_(2) = x_(2) + iy_(2)`

` x_(1)^(2) +y_(1)^(2) =x_(2)^(2) +y_(2)^(2) `

`=> (x_(1)^(2) - x_(2)^(2) ) + (y_(1)^(2) -y_(2)^(2) ) = 0`

`=> x_(1)^(2) - x_(2)^(2) = 0` or `y_(1) - y_(2)^(2) = 0`

`=> x_(1) = pm x_(2)` or `y_(1) = pm y_(2)`

e.g., Let `z_(1) = 1 + i` and `z_(2) = - 1 - i`

`| z | = sqrt(2)` and `| z_(2) | = sqrt(2)`

` Re(z_(1) ) != Re(z_(2) )` and `Im (z_(1) ) != Im (z_(2) )` and `z_(1) != z_(2)`

Correct Answer is `=>` (D) None of these

Q 1688423307

If `z = (-2(1 + 2i))/(3 + i)`,where `i = sqrt(-1)`, then the argument

`theta ( - pi < theta <= pi )` of `z` is
NDA Paper 1 2015

(A)

` (3pi)/4`

(B)

`pi/4`

(C)

` (5pi)/6`

(D)

` - (3pi)/4`

Solution:

We have, `z = (-2(1 + 2i))/(3 + i) = (-2(1 + 2i) (3-i))/((3 + i)(3-i))`

` = (-2(3-i + 6i - 2i^2))/(9 - i^2)`

` = (-2)/(10) ( 3 + 5i +2) = (-10)/(10) (i+1) = -i -1`

`:.` Argument `theta = - pi + theta`

`= - pi + tan^(-1) (1/1) =- pi + pi/4 = (-4 pi + pi) /4 = (-3pi)/4`

Correct Answer is `=>` (D) ` - (3pi)/4`

Q 1732380232

If `P` and `Q` are two complex numbers, then the modulus
of the quotient of `P` and `Q` is
NDA Paper 1 2014

(A)

greater than the quotient of their moduli

(B)

less than the quotient of their moduli

(C)

less than or equal to the quotient of their moduli

(D)

equal to the quotient of their moduli

Solution:

Let `P = x + iy` and `Q = a + i beta` are two complex

numbers.

Then, its quotient `= P/Q = (x + iy)/(alpha + ibeta)`

Now ` |P/Q| = | (x + iy)/(alpha + ibeta) | = | (x + iy)/(alpha + ibeta) |`

` = sqrt(x^2 + y^2)/sqrt(alpha^2 + beta^2) = sqrt((x^2 + y^2)/(alpha^2 + beta^2))`

Hence, the modulus of the quotient of `P` and `Q` is equal to

the quotient of their moduli i.e., `|(z_1)/(z_2)| =(|z_1|)/(|z_1|)`

Correct Answer is `=>` (D) equal to the quotient of their moduli

Q 2731545422

If `Re ((z -1 )/(z +1) ) = 0` , where `z = x + iy` is a complex number , then which one of the following is correct ?
NDA Paper 1 2016

(A)

`z = 1 + i`

(B)

`|z | = 2`

(C)

`z =1 - i`

(D)

`|z | = 1`

Solution:

If `Re ((z-1)/(z+1)) =0`

So, `(z-1)/(z+1) + bar((z-1)/(z+1)) =0`

`((z-1)/(z+1)) + ((bar z -1))/((bar z+1))=0`

`(z-1) ( bar z +1) +(bar z -1) (z +1) =0`

`z bar z- bar z + z -1 + z bar z + bar z - z -1 =0`

`2 ( z bar z - 1)=0`

`|z|=1`

Correct Answer is `=>` (D) `|z | = 1`

Q 1715267169

If `f(z) = (7 - z)/(1 - z^2) ` ,where `z = 1 + 2i`, then `| f(z) |` is equal to
NCERT Exemplar

(A)

` (|z|)/2`

(B)

`|z|`

(C)

`2 |z|`

(D)

None of these

Solution:

Let `z = 1 + 2i`

` => | z | = sqrt(1+4) = sqrt(5)`

Now `f(z) = (7 - z)/(1 - z^2) = (7 -1 - 2i)/(1 - (1 + 2i)^2)`

` = (6 -2i)/( 1-1 - 4i^2 -4i) =( 6 - 2i)/( 4 - 4i)`

` = ((3- i) (2 + 2i))/ ((2 - 2i ) (2 + 2i))`

` = (6 - 2i + 6i -2i^2)/(4 - 4i^2) = (6 + 4i + 2)/(4 + 4)`

` = (8+ 4i)/8 = 1+ 1/2 i`

`f(z)= 1 + 1/2 i`

`:. |f(z) | = sqrt( 1 + 1/4) = sqrt(( 4+ 1)/4) = sqrt(5)/2 = (|z|)/2`

Correct Answer is `=>` (A) ` (|z|)/2`

Q 2418280100

For two complex numbers `z_1, z_2` the relation
`| z_1 + z_2| =| z_1| +| z_2|` hold, if
BCECE Stage 1 2016

(A)

`arg (z_1)=arg(z_2)`

(B)

`arg(z_1)+arg(z_2)=pi/2`

(C)

`z_1z_2=1`

(D)

`|z_1|=|z_2|`

Solution:

Since, `|z_1 + z_2| = |z_1| + |z_2|`

`|z_1|^2+|z_2|^2+2|z_1||z_2| cos (theta_1-theta_2)`

`=|z_1|^2+ +\z_2|^2+2z_1z_2`

`cos (theta_1-theta_2)=1=cos 0`

`theta_1-theta_2=0`

`=> theta_1=theta_2`

`arg (z_1) = arg (z_2 )`

Correct Answer is `=>` (A) `arg (z_1)=arg(z_2)`

Q 2336134972

What is the modulus of `(sqrt2 + i)/(sqrt2 - i)` where `i = sqrt(-1)` ?
NDA Paper 1 2012

(A)

`3`

(B)

`1//2`

(C)

`1`

(D)

None of these

Solution:

`(sqrt2 + i)/(sqrt2 - i) = (sqrt2 + i)/(sqrt2 - i) xx (sqrt2 + i)/(sqrt2 + i)`

` = (sqrt2 + i)^2/((sqrt2)^2 - (i)^2) = ( 2 + i^2 + 2sqrt(2)i)/(2 - i)`

` = ( 2 + 1 + 2sqrt(2)i)/(2 - (-1)) = (1 + 2sqrt(2)i)/3`

`:. (sqrt2 + i)/(sqrt2 - i) = ( 1 + 2sqrt(2))/3 => (sqrt2 + i)/(sqrt2 - i) = 1/3 + (2sqrt(2))/3 i`

` => |(sqrt2 + i)/(sqrt2 - i)| = | 1/3 + ((2sqrt(2))/3 i)|`

` = sqrt((1/3)^2 + ( (2sqrt(2))/3)^2) = sqrt( 1/9 + 8/9 ) = sqrt(9/9) = 1`

Correct Answer is `=>` (C) `1`

Q 2436445372

If `iz^3 + z^2- z + i = 0`, then `| z|` is equal to
UPSEE 2010

(A)

`0`

(B)

`1`

(C)

`2`

(D)

None of these

Solution:

`iz^3+z^2-z+i =0`

Dividing both side by `i` and using `1/i=-i`

We have

`z^3-iz^2 +iz+1=0`

`=> z^2(z-i)+i (z-i)=0`

`=> (z-i)(z^2+i)=0`

`:. z=i` or `z^2=-i`

`:. |z|=|i|=1`

and `|z^2| =|z|^2=|-i|=1`

`:. |z|=1`

Correct Answer is `=>` (B) `1`

Q 1762380235

Let, `z = x + iy` where `x, y` are real variables and `i = sqrt(-1)`.
If `| 2z - 1 | = | z - 2 | ` then the point `z` describes
NDA Paper 1 2014

(A)

a circle

(B)

an ellipse

(C)

a hyperbola

(D)

a parabola

Solution:

Given that, `z = x + iy,` and `x, y in R`

We have, `| 2z -1| = | z - 2|`

`=> | 2 (x + iy)- 1| = | x + iy - 2|`

`=> | (2x - 1) + 2 iy | = | (x - 2) + iy |`

`=> | (2x - 1) + 2iy |^2 = | (x - 2) + iy | ^2`

`=> (2x - 1)^2 + (2y)^2 = (x - 2) + y^2`

`=> 4x^2 + 1 - 4x + 4y^2 = x^ 2 + 4 - 4x + y^2`

`=> 3x^2 + 3y^2 - 3 = 0`

` => x^2 + y^2 = 1`

which represents a circle.

Hence, point `z` describes a circle.

Correct Answer is `=>` (A) a circle

Q 2346280173

What is arg `(bi)`, where `b > 0`?
NDA Paper 1 2008

(A)

`0`

(B)

`pi/2`

(C)

`pi`

(D)

`(3pi)/2`

Solution:

` ∵ theta = tan^(-1) (b/0) = tan^(-1) (oo) = tan^(-1) (tan \ pi/2)`

` theta = pi/2`

`:. arg (bi) = pi/2`

Correct Answer is `=>` (B) `pi/2`

Q 2306180078

Let `C` be the set of complex numbers and `z_1 ,z_2` are
in `C`.
I. `arg (z_1) = arg (z_2 ) => z_1 = z_2`
II. `| z_1| = | z_2 | => z_1 = z_2`
Which of the statements given above is/are correct?
NDA Paper 1 2008

(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

None of the given statements is correct.

Because

I. `z_1 = z_2 => arg (z_1) = arg (z_2)`

II. `z_1 = z_2 => | z_1 | = | z_2 |`

Correct Answer is `=>` (D) Neither I nor II

Q 2346667573

What is the modulus of `| ( 1 + 2i)/(1 - (1 - i)^2) |` ?
NDA Paper 1 2010

(A)

`1`

(B)

`sqrt(5)`

(C)

`sqrt(3)`

(D)

`5`

Solution:

`∵ ( 1 + 2i)/(1 - (1 - i)^2) = ( 1 + 2i)/(1 - (1 - 1 - 2i)) = ( 1 + 2i)/(1 + 2i) = 1`

`:. | ( 1 + 2i)/(1 - (1 - i)^2) | = 1`

Correct Answer is `=>` (A) `1`

Q 2306156078

What is modulus of `1/(1 + 3i) - 1/(1 - 3i)` ?
NDA Paper 1 2011

(A)

`3/5`

(B)

`9/(25)`

(C)

`3/(25)`

(D)

`5/3`

Solution:

`:. 1/(1 + 3i) - 1/( 1- 3i) = (1- 3i)/(1 + 9) - (1+ 3i)/(1 + 9)`

` = (1- 3i - 1- 3i)/(10)`

`= - (6i)/(10) = - 3/5 i`.

`=>` Modulus `| - 3/5 i | = sqrt((- 3/5)^2) = 3/5`

Correct Answer is `=>` (A) `3/5`

Q 2346045873

If `z = 1 + cos \ pi/5 + i sin \ pi/5` , then `| z |` is equal to
NDA Paper 1 2011

(A)

`2 cos \ pi/5`

(B)

`2 sin \ pi/5`

(C)

`2 cos \ pi/(10)`

(D)

`2 sin \ pi/(10)`

Solution:

`∵ z = ( 1 + cos \ pi/5) + i sin \ pi/5`

` = 2 cos^2 \ pi/(10) + i 2 sin \ pi/(10) cos \ pi/(10)`

`= 2 cos \ pi/(10) [cos \ pi/(10) + i sin \ pi/(10)] = 2 cos\ pi/(10) ·e^(i pi//10)`

`{ ∵ e^(i theta) = cos theta + i sin theta}`

{and `| e^(i theta) | = 1`

`| z | = | 2 cos pi/(10) . e^( i pi //10) | = 2 cos \ pi/(10)`

Correct Answer is `=>` (C) `2 cos \ pi/(10)`

Q 2356845774

If `z = 1 + i tan alpha`, where `pi < alpha < (3pi)/2 `, then `| z |` is
equal to'?
NDA Paper 1 2011

(A)

`sec alpha`

(B)

`- sec alpha`

(C)

`sec^2 alpha`

(D)

`- sec^2 alpha`

Solution:

Given, `z = 1 + i tan alpha`, where `pi < a < (3pi)/2`.

`=> | z | = sqrt(1 + tan^2 alpha)`

`=> | z | = sqrt(sec^2 alpha)`

`=> | z | = - sec alpha`

Correct Answer is `=>` (B) `- sec alpha`

Q 2386445377

What is the argument of `(1 - sin theta ) + i cos theta`?
(where, `i = sqrt(-1)`)
NDA Paper 1 2011

(A)

`pi/2 - theta/2`

(B)

`pi/2 + theta/2`

(C)

`pi/4 - theta/2`

(D)

`pi/4 + theta/2`

Solution:

Let `z = (1 - sin theta) + i cos theta`

`=> arg (z) = tan^(-1) ((Im(z))/(Re (z))) = tan^(-1) ( (cos theta) /(1 - sin theta ))`

`= tan^(-1) { ((cos^2 theta//2 - sin^2 theta//2))/( sin^2 theta//2 + cos^2 theta//2 - 2 sin theta//2 . cos theta//2)}`

`= tan^(-1) { ((cos theta//2 - sin theta//2) (cos theta//2 + sin theta//2))/(cos theta//2 - sin theta//2)^2 }`

` = tan^(-1) {( cos theta//2 + sin theta//2)/(cos theta//2 - sin theta//2)}`

`= tan^(-1) ((1 + tan theta//2)/(1 - tan theta//2))`

`= tan^(-1) tan (pi/2 + theta/2 )`

` = pi/2 + theta/2`

Correct Answer is `=>` (D) `pi/4 + theta/2`

Q 2336334272

What is the argument of the complex number
`( -1- i )`, where `i = sqrt(-1)`?
NDA Paper 1 2013

(A)

`(5 pi)/4`

(B)

`- (5 pi)/4`

(C)

`(3 pi)/4`

(D)

None of these

Solution:

Let `z = -1 - i = r (cos theta + i sin theta )`

On comparing real and imaginary partsj on both sides, we get

`r cos theta = -1` ... (i)

and `r sin theta = -1` ... (ii)

On dividing Eq. (ii) by Eq. (i), we get

`(rsin theta)/(r cos theta) = (-1)/(-1)`

`=> tan theta = 1 = tan pi/4`

` => theta = pi/4`

Since, argument of `z` lies in the IIIrd quadrant.

`:. a'g (z) = pi + theta = pi + pi/4 = (5pi)/ 4`

Correct Answer is `=>` (A) `(5 pi)/4`

Q 1743278143

What is the argument of the complex number
` ((1 + i ) (2 + i))/(3-i)` , where `i = sqrt(-1)`?
NDA Paper 1 2014

(A)

`0`

(B)

`pi/4`

(C)

`- pi/4`

(D)

`pi/2`

Solution:

Let `z = ( (1 + i) (2 + i))/(3-i)`

` = (2 + 3i + i^2)/(3-i) = ( 2 + 3i -1)/(3-i)`

` = (1+3i) /(3-i) xx ( 3 + i)/(3 + i) = ( 3+ 10i + 3i^2)/(9 - i^2)`

` = (3 + 10i -3)/ ( 9+1)`

` = (10i )/(10) = i`

` z = 0 + i .1`

So, arg `(z) = tan^(- 1) ( 1/0) = tan^(- 1)(oo) = pi/2`

Correct Answer is `=>` (D) `pi/2`

Q 1712434330

If `z` be a complex number such that `| z | = 4` and
arg `z = (5pi)/6`, then what is `z` equal to?

Where ` i = sqrt(-1)`
NDA Paper 1 2014

(A)

`2sqrt(3) + 2i`

(B)

`2sqrt(3) - 2i`

(C)

`- 2sqrt(3) + 2i`

(D)

`- sqrt(3) + i`

Solution:

Since, arg `z = (5pi)/6 `

So, `z` lies in second quadrant.

Let ` z = a + ib`

and `| z | = a^2 + b^2 = 4`

Also, `tan alpha = ( pi - (5pi)/6 ) = pi/6 => alpha = pi/6`

`:. tan alpha = 1/sqrt(3) quad [ ∵ tan alpha = (b/a)]`

` :. z =- sqrt(3) + i quad ( ∵ a < 0` and `b > 0)`

Correct Answer is `=>` (D) `- sqrt(3) + i`

Q 1713178049

If `| z + bar z |= | z- bar z |`, then the locus of `z` is
NDA Paper 1 2014

(A)

A pair of straight lines

(B)

A line

(C)

A set of four straight lines

(D)

A circle

Solution:

Given condition,

` | z+ bar z| = | z- bar z|`

Let `z = x + iy => bar z = x - iy`

`:. | x + iy + x - iy| = | x + iy - x + iy|`

` => | 2x| = | 2iy | quad (∵| a+ ib | = sqrt(a^2 + b^2))`

` => 2x = 2y`

`:. x = y`

which represent a line passing through the origin.

Hence, the locus of `z` is.a line.

Correct Answer is `=>` (B) A line

Q 2017301280

Find the principal argument of `(1 + i sqrt(3))^2`.

NCERT Exemplar

Solution:

Given that, ` z = (1 + isqrt(3))^2`

` => z = 1- 3 + 2i sqrt(3) => z = - 2 + i2 sqrt(3)`

` => tan alpha = |(2sqrt(3))/(-2)| = | - sqrt(3) | = sqrt(3)`

` => tan alpha = tan (pi/3) => alpha = pi/3`

` ∵ Re (z) < 0` and `Im (z) > 0`

`=> arg (z) = pi - pi/3 => = (2pi)/3`

Q 1775534466

Show that `|(z- 2)/(z-3)| = 2` represents a circle. Find its centre and radius.

NCERT Exemplar

Solution:

Let `z = x + iy`

Given, equation is `|(z- 2)/(z-3)| = 2 => |(z- 2)/(z-3)| = 2`

`=> | (x+ iy - 2)/(x+ iy - 3)| = 2`

` => | x - 2 + iy | =2 | x - 3 + iy |`

` => sqrt((x - 2)^2 + y^2) = 2 sqrt( (x- 3)^2 + y^2) quad [∵ | x + iy | = sqrt(x^2 + y^2)]`

On squaring both sides, we get

` x^2 - 4x + 4 + y^2 = 4 (x^2 - 6x + 9 + y^2)`

` => 3x^2 + 3y^2 - 20x + 32 = 0`

` => x^2 + y^2 - (20)/3 x + (32)/3 = 0` ........(1)

On comparing the above equation with `x^2 + y^2 + 2gx + 2fy + c = 0`, we get

` => 2g = (-20)/3 => g = (- 10)/3`

and `2f = 0 => f = 0` and `c = ( 32)/3`

`:.` Centre `= (-g ,- f) = (10//3, 0)`

Also radius `(r) = sqrt((10//3)^2 + 0 - 32//3) quad [∵ r = sqrt(g^2 + f^2 -c) ]`

` = 1/3 sqrt((100- 96)) = 2//3`

Q 1745534463

If `arg (z -1) = arg (z +3i)`, then find `x -1 : y`, where `z = x + iy`.
NCERT Exemplar

Solution:

Given that, `arg( z - 1 ) = arg ( z + 3i )`

and let `z = x + iy`

Now, `arg (z- 1) = arg (z + 3i)`

` => arg (x + iy - 1) = arg (x + iy + 3i)`

` => arg (x- 1 + iy) = arg [x + i (y + 3)]`

`=> tan^(-1) (y/(x-1)) = tan^(-1) ((y + 3)/x)`

` => y/(x-1) = (y+3)/x => xy = (x - 1) (y + 3)`

`=> xy = xy - y + 3x - 3 => 3x - 3 = y `

` => (3(x - 1))/y = 1 => (x -1)/y = 1/3`

`:. (x -1) : y = 1 : 3`

Finding maximum and minimum value

Solve with the help of following inequalities :

(vi) `|z_1 + z_2| <= | z_1| + | z_2|`
(vii) `|z_1 - z_2 | >= |z_1| -|z_2|`
(viii) `|z_1| - |z_2 | <= | z_1 +z_2| <= |z_1| + |z_2|`

Q 2763880745

If `| z + 4 | <= 3`, then the maximum value of `| z + 1 |` is
NDA Paper 1 2017

(A)

`0`

(B)

`4`

(C)

`6`

(D)

`10`

Solution:

`|z +1|`

`=|z+4-3|`

`le |z+4|+ |-3|`

` le 3+3`

`le 6`

Correct Answer is `=>` (C) `6`

Q 1785423367

For a positive integer `n`, find the value of `(1 - i)^n ( 1 - 1/i)^n`.
NCERT Exemplar

Q 2459891714

If `|z- 4/z| =2` ,then the maximum value of |z| is equal to
BCECE Stage 1 2014

(A)

`sqrt3 +1`

(B)

`sqrt5 +1`

(C)

(D)

`2 + sqrt2`

Solution:

Now, `|z| = | (z -4/z ) + 4/z |`

`=> |z| <= | z - 4/z | + |4/z|`

`=> |z| <= 2 + 4/|z|`

`=> |z|^2 - 2 |z| - 4 <= 0`

`=> (|z| - (sqrt5 +1 ) ) ( |z| -(1- sqrt5)) <= 0`

`=> 1-sqrt5 <= |z| <= sqrt5 +1`

Correct Answer is `=>` (B) `sqrt5 +1`

Q 2511291129

The maximum value of `| z|`, when the complex number `z` satisfies the condition `| z + 2/z | = 2` is
WBJEE 2012

(A)

`sqrt3`

(B)

`sqrt3 + sqrt2`

(C)

`sqrt3 + 1`

(D)

`sqrt3 -1`

Solution:

We have the identity,

` | z| = | z + 2/z - 2/z | <= | z + 2/z | + | 2/z|`

` => |z| <= 2 + 2/(|z|)`

` => |z|^2 <= 2 |z| + 2`

`=> |z|^2 - 2 |z| +1 <= 3`

`=> ( |z| -1 )^2 <= 3`

`=> - sqrt3 <= |z| - 1 <= sqrt3`

`=> - sqrt3 + 1 <= |z| <= sqrt3 + 1`

Hence, the maximum value `= sqrt3 + 1`

Correct Answer is `=>` (C) `sqrt3 + 1`

Question Based on algebra of complex number

Q 2316145079

If `A + iB = (4 + 2i)/(1 - 2i)` ,where `i = sqrt(-1)`, then what is the
value of `A`?
NDA Paper 1 2012

(A)

`-8`

(B)

`0`

(C)

`4`

(D)

`8`

Solution:

`A + iB = (4 + 2i)/(1 - 2i) xx (1+ 2i)/( 1 + 2i)`

` = (4 + 2i + 8i + 4i^2)/(1 - 4i^2)`

`= (4 + 10i - 4 )/(1 + 4) = (10 i)/5 = 2i`

` => A + iB = 0 + i · 2`

`:. A = 0` and `B = 2`

Correct Answer is `=>` (B) `0`

Q 2366245175

If `z = - bar(z)`, then which one of the following is
correct'?
NDA Paper 1 2012

(A)

The real part of z is zero

(B)

The imaginary part of z is zero

(C)

The real part of z is equal to imaginary part of z

(D)

The sum of real and imaginary parts of z is z

Solution:

Given that, `z = -z`

Let `z = x + iy`

`(x + iy) = - (bar x + i y)`

`(x + iy) = - (x - iy)`

`=> (x + iy) = (-x + iy)`

`=> 2x = 0`

`=> x = 0`

`:. z = x + iy = 0 + iy = iy`

Hence, the real part of `z` is zero.

Correct Answer is `=>` (A) The real part of z is zero

Q 2326445371

If `z = (1 + 2i)/(2 - i) - (2 - i)/(1 + 2i)` then what is the value of
`z^2 + z bar z`? (where, `i = sqrt(-1))`
NDA Paper 1 2011

(A)

`0`

(B)

`-1`

(C)

`1`

(D)

`8`

Solution:

Given, `z = (1 + 2i)/(2 - i) - (2 - i)/(1 + 2i)`

` => z = ((1 + 2i)^2 - (2 - i)^2)/( (2 - i) (1 + 2i))`

` => z = ( (1 + 2i + 2 - i)(1 + 2i - 2 + i))/((2 - i + 4i - 2i^2))`

` => z = ((3 + i) (-1 + 3i))/(4 + 3i) xx ((4 - 3i))/((4- 3i))`

` => z = ((- 3 + 9i - i - 3) (4 - 3i))/(16- 9i^2)`

` => z = ( (-6 + 8i) (4 - 3i))/(16 + 9)`

`= ( (-24 + 32i + 18i - 24i^2))/(25)`

` => z = ( -24 + 50i + 24)/(25) = (50i)/(25) = 2i`

`=> z = 0 + 2 i` and `z = 0 - 2 i`

Now, `z bar z = (0 + 2i)(0 - 2i) = - 4i^2 = 4` ...(i)

and `z^2 = | z |^2 = z bar z = 4` ..........(ii)

From Eqs. (i) and (ii),

`z^2 + (z bar z) = 4 + 4 = 8`

Correct Answer is `=>` (D) `8`

Q 2146012873

Let `z_(1), z_(2)` and `z_(3)` be non-zero complex numbers satisfying `z^(2) = i bar (z),`
where `i = sqrt (-1)`

Consider the following statements
1. `z_(1)z _(2)z_(3)` is purely imaginary.
2. `z_(1)z _(2) + z_(2)z_(3) + z _(3)z_(1)` is purely real.

Which of the above statements is/are correct?
NDA Paper 1 2016

(A)

Only 1

(B)

Only 2

(C)

Both 1 and 2

(D)

Neither 1 nor 2

Solution:

We have, `z^(2) = i bar(z)`... (i)

`| z |^(2) =| i bar(z) |`

` => | z |^(2) =| z |`

` => | z |^(2) -| z |= 0`

`=> | z | (| z |- 1) = 0`

As,` z` is non-zero complex number.

`:. | z | = 1 => | z |^(2) = 1`

`=> z bar(z) =1 => bar(z) = 1/z`

From Eq. (i), we have

`z^(2) = i bar(z) => z^(2) = i 1/z`

`=> z^( 3) = i => z^(3) - i = 0 ... (ii)`

If `z_(l), z_(2)` and `z_(3)` satisfying the Eq. (ii), then

`z_(1) + z_(2) + z_(3) = 0`

`z_(1)z_(2) + z_(2) z_(3) + z_(1)z_(3) = 0`

and `z_(1) z_(2)z_(3) = i`

We have, `z_(1)z_(2)z_(3) = i`

`=> z_(1) z_(2)z_(3)` is purely imaginary.

:. Statement 1 is correct and .`z_(1) z_(2) + z_(2) z_(3) + z_(3) z_(1) = 0`

`=> z_(1) z_(2) + z_(2) z_(3) + z_(3) z_(1)` is purely real.

:. Statement 2 is correct.

Correct Answer is `=>` (C) Both 1 and 2

Q 2601756628

`(x - iy) ( 3 + 5i)` is the conjugate of `( -6 - 24i)`, then `x` and `y` are

(A)

`x= 3, y = - 3`

(B)

`x = - 3, y = 3`

(C)

`x = - 3, y = -3`

(D)

`x = 3, y = 3`

Solution:

`(a)(x- iy) (3 + 5i) = 3x + 5x i- 3yi- 5yi^2`

`=3 x +(5x-3y)i+5y` `\ \ \ \ ( :. i^2=-1)`

`= (3x + 5y) + (5x - 3y)i`...........(i)

Given, `(x- iy) (3 + 5i) =- 6 + 24i`

[using Eq.(i) and `z =(a+ ib) => bar z =(a- ib)]`

On comparing the real and imaginary parts of both sides,
we get

`3x + 5y = - 6` and `5x - 3y = 24`

Solving the above equations by substitution or elimination method, we get

`x= 3,y= -3`

Correct Answer is `=>` (A) `x= 3, y = - 3`

Q 2336567472

If `x^2 + y^2 = 1`, then what is `(1 + x + iy)/(1 + x - iy)` equal to?
NDA Paper 1 2010

(A)

`x - iy`

(B)

`x + iy`

(C)

`2x`

(D)

`-2iy`

Solution:

` (1 + x + iy)/(1 + x - iy) = ( (1 + x + iy)(1 + x + iy))/((1 + x - iy)(1 + x + iy))`

` = ((1 + x)^2 + iy(1 + x) + iy(1 + x)- y^2)/(1 + x^2 + 2x + y^2)`

` = (1 + x^2 + 2x - y^2 + 2iy(1 + x))/(2(1 + x))`

` = (1 - y^2 + 2x + x^2 + 2iy (1 + x))/(2(1 + x))`

` = (2x^2 + 2x + 2iy(1 + x))/(2(1 + x))`

` = (2(x + 1)x + 2(x + 1)yi)/(2(x + 1)) = x + iy`

Correct Answer is `=>` (B) `x + iy`

Q 2326478371

What is `(sqrt3 + 1)/(1 + sqrt 3 i)` equal to?
NDA Paper 1 2009

(A)

`1 + i`

(B)

`1 - i`

(C)

` (sqrt(3)(1 - i))/ 2`

(D)

` (sqrt(3) - i)/2`

Solution:

`(sqrt3 + 1)/(1 + sqrt 3 i) = (( sqrt3 + 1) ( 1 - sqrt3 i))/( (1 + sqrt3 i) (1 - sqrt3 i))`

`= ( sqrt3- 3i + i + sqrt3)/(1 + 3) = ( 2sqrt(3) - 2i)/4 = ( sqrt(3) - i)/2`

Correct Answer is `=>` (D) ` (sqrt(3) - i)/2`

Q 2376478376

If `2x = 3 + 5i`, then what is the value of
`2x^3 + 2x^2 - 7 x + 72`?

NDA Paper 1 2009

(A)

`4`

(B)

`-4`

(C)

`8`

(D)

`-8`

Solution:

`∵ x = ( 3 + 5i)/2`

`:. x^3 = (27 + 125 i^3 + 225 i^2 + 135 i)/8`

` = (27 - 125i - 225 + 135i)/8`

` = (-198 + 10i)/8 = ( -99 + 5i)/4`

and `x^2 = (9 + 25i^ 2 + 30 i)/4 = ( 9- 25 + 30i)/4 = ( -8 + 15i)/2`

Now, `2x^3 + 2x^2 - 7x + 72`

`= ( (-99 + 5i)/2) + (-8 + 15i) - (7(3 + 5i))/2 + 72`

`= -(99)/2 + (5i)/2 - 8 + 15i - (21)/2 - (35)/2 i + 72`

`= (- (99)/2 - 8 - (21)/2 + 72) + (5/2 + 15 - (35)/2 )i`

` = (-99 - 16 - 21+ 144)/2 = 8/2 = 4`

Correct Answer is `=>` (A) `4`

Finding number of solution of a complex equation

Q 2783880747

The number of roots of the equation `z^2 = 2 bar z` is
NDA Paper 1 2017

(A)

`2`

(B)

`3`

(C)

`4`

(D)

zero

Solution:

`z^2 = 2 bar z`

`z= x+iy`

`(x+iy)^2 = 2(x-iy)`

on comparing

`x^2 -y^2 = 2x`...................(i)

`2xy= -2y`

`x =-1 , y=0`....................(ii)

so, `x=-1 , y= + sqrt 3 ` & `- sqrt 3, `

on ` y = 0, x= 0 & x = 2`

Total No. of Sol `=4`

Correct Answer is `=>` (C) `4`

Q 2781045827

What is the number of distinct solutions of the equation `z^2 + |z | = 0` (where z is a complex number ) ?

NDA Paper 1 2016

(A)

one

(B)

two

(C)

three

(D)

Five

Solution:

`z^2 +|z| =0`

Let `z= x+iy`

`=> (x+iy)^2 + sqrt (x^2+y^2)=0`

`=> x^2 -y^2 + 2i xy + sqrt(x^2+y^2)=0`

on comparing both side real & imaginary part should be zero saprately.

`=> xy=0`

`x=0` or `y=0`

`=> x^2 -y^2 pm sqrt(x^2 +y^2) =0`

Case - 1 If `x=0`

`-y^2 + |y| = 0`

`y=0, y=1, y = -1`

Case-2 If `y=0`

`x^2+ |x|=0`

No real solution exist for `x`

so no of solution `x=0, y=0`

its two `x=0, y=1`

Correct Answer is `=>` (B) two

Q 2116712679

Let `z_(1), z_(2)` and `z_(3)` be non-zero complex numbers satisfying `z^(2) = i bar (z),`
where `i = sqrt (-1)`

What is `z_(1) + z_( 2) + z_(3)` equal to?
NDA Paper 1 2016

(A)

`i`

(B)

`-i`

(C)

`0`

(D)

`1`

Correct Answer is `=>` (C) `0`

Q 1530591412

The number of solutions for the equations

`|z-1 | = | z- 2|= | z-i |` is
BITSAT 2005

(A)

one solution

(B)

3 solutions

(C)

2 solutions

(D)

no solution

Solution:

Let `z = x + iy`

:. `|z-1 | = | z- 2|= | z-i |`

`=> | (x - 1) + i y |= | (x - 2) + i y |`

`= | (x + i (y- 1) |`

`=> x ^2- 2x + 1 + y^ 2 = x^2 + 4 - 4x + y ^2`

`= x ^2 + y ^2 + 1 - 2y`

Taking 1st and 2nd term

`=> - 2x + 1 = 4 - 4x`

`=> 2x = 3` .................(1)

Taking 2nd and 3rd term

`=> 4- 4x = 1 - 2y`

`=> 4x-2y=3` ..................(2)

Taking 1st and 3rd term

`=> -2x + 1 = 1 - 2y`

`=> 2x - 2y = 0`

`=> x = y`.........................(3)

From (1) `x -3/2`

On putting value of `x` in Eq. (iii), we get

`y =3/2`

On putting the value of `x` and `y` in Eq. (ii), we

get `4 (3/2) -2 (3/2) =3`

`=> 3=3`

`:.` One solution exist.

Correct Answer is `=>` (A) one solution

Q 1416323270

The number of solutions of the equation `z^ 2 +|z |^2 = 0`, where `z in C` is

(A)

`1`

(B)

`2`

(C)

`3`

(D)

`oo`

Solution:

`z^2 +| z |^2 = 0`

`=> z^2 + z bar z = 0`

`=> z (z + bar z) = 0`

or `z·2Re(z)=0`

`z = 0` and `Re (z) = 0`

If `z=a+ib`

`:.` Solutions are `z = 0, ib (b in R)`

Correct Answer is `=>` (D) `oo`

Q 2865423365

The solution of the equation `|z| -z = 1+2i ` is

(A)

`3/2-2i`

(B)

`3/2+2i`

(C)

`2-3/2i`

(D)

None of these

Solution:

Let `z = x+ iy , |z| -z = 1+2i`

`sqrt(x^2+y^2) - (x+iy) = 1+2i`

`=> ( sqrt(x^2+y^2) - x) -iy = 1+2i`

`=> sqrt(x^2+y^2) -x = 1` and `y = -2`

Now `sqrt(x^2+y^2) -x = 1`

`=> sqrt(x^2+4) = 1+x \ \ [ because y = -2]`

`=> x^2+4 = 1+x^2+2x`

`=> x = 3/2`

`therefore ` The solution is `3/2-2i`

Correct Answer is `=>` (A) `3/2-2i`

Q 2875323266

The number of solutions of the equation `z^2 = barz` is

(A)

`2`

(B)

`3`

(C)

`4`

(D)

None of these

Solution:

`z^2 = barz => ( x+ i y)^2 = x - iy`

`=> x^2 - y^2 - x + i ( 2xy+y) = 0`

`=> x^2-y^2 - x = 0` and `2xy + y = 0`

Now `2xy + y = 0` gives `y = 0 ` or `x = -1/2`

when `y = 0 , x^2 - y^2 - x = 0`

Gives `x^2 - x = 0 ` or `x = 0 , 1`

When`x = -1/2 , x^2 -y^2 -x = 0` gives `1/4 - y^2+1/2 = 0`

`=> y = pm ( sqrt3/2)`
Hence, there are four solutions

Correct Answer is `=>` (C) `4`

Q 2416645579

The nmnber of the solutions of the equation
`z^2 +bar z = 0` is
UPSEE 2010

(A)

`1`

(B)

`2`

(C)

`3`

(D)

`4`

Solution:

Let `z = x + iy`, .. then

`z^2 + bar z = 0`

`=> (x^2 -y ^2 + 2ixy) + (x - iy) = 0`

`=> x^ 2- y ^2 + x = 0` and `2.xy -y = 0`

Now, `2.xy- y = 0=> y (2x -1) = 0`

`=> y = 0` or `x = 1// 2`

If `y = 0`, then `x ^2- y^ 2 + x = 0 => x ^2 + x = 0`

`=> x = 0` or `x =-1`

If `x=1//2` ,then `x^2 -y ^2 +x =0`

`=> y = pm sqrt 3/2`

Thus, the given equations has four solutions.

Correct Answer is `=>` (D) `4`

Question based on cube root of unity

`omega = (-1 + sqrt3 i )/2 ; omega^2 = (-1 -sqrt3i )/2` and `omega^3 =1`

`1 + omega + omega^2 = 0`

Q 2713680549

The value of ` ((-1 + i sqrt3)/2)^n + ((-1 - i sqrt3)/2)^n` where `n` is not a multiple of `3` and `i = sqrt(-1)`, is
NDA Paper 1 2017

(A)

`1`

(B)

`-1`

(C)

`i`

(D)

`-i`

Solution:

`((-1)/2 + (sqrt 3)/2 i) ^n +((-1)/2 -(sqrt 3)/2 i)^n`

`= omega^n +(omega^2)^n`

Using NDA Special Trick let us assume n = 1 ( we can do that as 1 is not a multiple of 3 )

`= omega +omega^2`

`= 1`

As n is not a multiple of 3 it must be of the form 3m+1 or 3m +2

For both 3m+1 and 3m+2 the value of given expression would be -1

Correct Answer is `=>` (B) `-1`

Q 2773780646

If `1, omega , omega^2` are the cube roots of unity, then `(1 + omega )(1 + omega^2 ) (1 + omega^3) (1 + omega + omega^2)` is equal to
NDA Paper 1 2017

(A)

`-2`

(B)

`- 1`

(C)

`0`

(D)

`2`

Solution:

`(1+ omega)(1+ omega^2) (1+omega^3) (1+ omega+ omega^2)`

`=(1+ omega)(1+ omega^2) (1+ omega^3) (0)`

`=0`

Correct Answer is `=>` (C) `0`

Q 2771445326

What is `omega^100 +omega^200 + omega^300 ` equal to, where `omega` is the cube root of unity ?
NDA Paper 1 2016

(A)

`1`

(B)

`3omega`

(C)

`3 omega`

(D)

`0`

Solution:

`omega^100 +omega^200 + omega^300 `

` => omega^99 * omega + (omega^99 * omega)^2 +(omega^99 * omega)^3`

`=> omega + omega^2 + omega^3 = omega (1+ omega + omega^2)`

`=> 0`

Correct Answer is `=>` (D) `0`

Q 2741745623

If `z = ( sqrt3 /2 + i /2 ) ^107 + ( sqrt3/2 - i/2 )^107` , then what is the imaginary part of z equal to ?
NDA Paper 1 2016

(A)

`0`

(B)

`1/2`

(C)

`sqrt3/2`

(D)

`1`

Solution:

`z= (sqrt 3 /2 = i/2)^(107) + ( sqrt 3 /2 -i/2)^(107)`

`=[ -i *-1/2 + sqrt 3 /2 i)]^(107) +[ i (-1/2 - sqrt 3/2 i)]^(107)`

`= (- i omega)^(107) +(i omega^2)^(107)`

`= i^(107) omega^(107) (omega^(107) -1)`

`=-i * omega^2 [ omega^2-1]`

`= i [ omega^2 - omega]= i [ -1 - 2 omega]= i [- sqrt 3 i]= sqrt 3`

`Im (2) =0`

Correct Answer is `=>` (A) `0`

Q 2711667520

What is `sqrt ( (1 + omega^2 ) / ( 1 + omega ) )` equal to , where `omega` is the cube root of unity ?
NDA Paper 1 2016

(A)

(B)

`omega`

(C)

`omega^2`

(D)

`i omega ` , where ` i = sqrt (-1)`

Solution:

`sqrt( ( 1+w^2)/(1+w) ) = sqrt( ( w+w^3)/(w^2 +w) ) ` `{ tt ( (1+w+w^2 = 0 ), (w^3 =1) )`

`= sqrt ( (1+w)/(-1) )`

`= sqrt ( (-w^2)/(-1) )`

`=w`

Correct Answer is `=>` (B) `omega`

Q 2356656574

If `omega` is a complex cube root of unity, then what is
`omega^(10) + omega^(-10)` equal to?
NDA Paper 1 2010

(A)

`2`

(B)

`-1`

(C)

`-2`

(D)

`1`

Solution:

`omega^(10) + omega^(-10) = omega^(10) + 1/omega^(10)`

` = omega + 1/omega = omega + omega^2/omega^3`

` = omega + omega^2 = -1 ( ∵ omega^3 = 1 , omega^2 + omega + 1 = 0)`

Correct Answer is `=>` (B) `-1`

Q 2107823788

Suppose, `omega_(1)` and `omega_(2)` are two distinct cube roots of unity
different from `1.` Then, what is `( omega_(1) - omega_(2))^(2)`
equal to ?
NDA Paper 1 2016

(A)

`3`

(B)

`1`

(C)

`-1`

(D)

`-3`

Solution:

Given , `( omega_(1) - omega_(2))^(2) = omega_(1)^(2) + omega_(2)^(2) - 2omega_(1)omega_(2)`

Let ` omega_(1) = omega^(2)` and `omega_(2) = omega`

`:. ( omega_(1) - omega_(2))^(2) = (omega^(2))^(2) + omega^(2) - 2omega^(2) .omega`

` = omega^(4) + omega^(2) - 2omega^(2)`

` = omega^(3) . omega + omega^(2) - 2omega^(3)`

` = omega + omega^(2) - 2 = -1 -2 `

` => ( omega_(1) - omega_(2))^(2) = -3 `

Correct Answer is `=>` (D) `-3`

Q 2424712651

If `x=(-3 + i sqrt 3)/2` is a complex number, then the
value of `(x^2+3x)^2(x^2+3x+1)` is
UPSEE 2011

(A)

`-9/8`

(B)

`6`

(C)

`-18`

(D)

`36`

Solution:

Given, `x =( -3+i sqrt 3)/2`

`=> x=(-1 +i sqrt 3)/2-1=omega-1`

`:. (x^2 + 3x)^2(x^2 + 3x + 1)`

`=[(omega -1)^2 + 3(omega -1)]^2[(omega -1)^2
+ 3 (omega - 1) +i1]`

`= (omega ^2 + 1 - 2omega + 3omega- 3)^2
(omega ^2 + 1 - 2omega + 3omega - 3 + 1)`

`= (omega^2 +omega - 2)^2(omega^2 + omega - 1)`

`= (-1- 2)^2(-1-1)` `( 1 +omega+ omega^2 = 0)`

`= -18`

Correct Answer is `=>` (C) `-18`

Q 1618423309

If `1, omega ` and `omega^2` are the cube roots of unity, then the
value of `(1 + omega )(1 + omega)^2(1 + omega)^4 (1 + omega)^8`
is
NDA Paper 1 2015

(A)

`-1`

(B)

`0`

(C)

`1`

(D)

`2`

Solution:

`(1 + omega )(1 + omega^2 )(1 + omega^4 )(1 + omega^8)`

` = (1 + omega + omega^2 + omega^3 ) (1 + omega^3 . omega ) [ 1 + (omega^3)^2 . omega^2]`

` = 1. (1 + omega)(1 + omega^2)`

`= ( 1 + omega + omega^2 + omega^3 ) = 1`

Correct Answer is `=>` (C) `1`

Q 1688723607

`(x^3 - 1)` can be factorised as

where, `omega` is one of the cube roots of unity.
NDA Paper 1 2015

(A)

`(x -1) (x - omega ) (x + omega^2 )`

(B)

`(x -1) (x - omega ) (x - omega^2 )`

(C)

`(x -1) (x + omega ) (x + omega^2 )`

(D)

`(x -1) (x + omega ) (x - omega^2 )`

Solution:

`(x^3- 1) = (x -1)(x^2 + 1 + x)`

`= (x- 1)(x^2 + x - omega - omega^2 )`

`= (x- 1) (x^2 - omega^2 + x - omega)`

`= (x- 1) [(x +omega) (x - omega) (x- omega)]`

`= (x -1) (x- omega) (x + omega + 1)`

` = (x -1) (x- omega) (x - omega^2)`

Correct Answer is `=>` (B) `(x -1) (x - omega ) (x - omega^2 )`

Q 2326556471

What is `( (sqrt3 + i)/(sqrt3 - i) )^6` equal to'?
NDA Paper 1 2010

(A)

`-1`

(B)

`0`

(C)

`1`

(D)

`2`

Solution:

`( (sqrt3 + i)/(sqrt3 - i) ) = ( (sqrt3 + i) (sqrt3 + i))/(3 - i^2)`

` = (3 + i^2 + 2sqrt(3)i)/4`

` = ( 2 + 2 sqrt(3))/4 = ( 1 + sqrt(3)i)/2 = - omega^2`

Now, `( (sqrt3 + i)/(sqrt3 - i) )^6 = ( - omega^2)^6 = omega^(12) (omega^3)^4 = (1)^4 = 1 ( ∵ omega^3 = 1)`

Correct Answer is `=>` (C) `1`

Q 2316480370

If `1, omega, omega^2` are the three cube roots of unity, then
what is `(a omega^6 + b omega^4 + c omega^2)/(b + c omega^(10) + a omega^(8) )`
equal to?
NDA Paper 1 2007

(A)

`a/b`

(B)

`b`

(C)

`omega`

(D)

`omega^2`

Solution:

Since `1, omega` and `omega^2` are the three cube roots of unity.

`:. 1 + omega + omega^2 = 0` and `omega^3 = 1`

` (a omega^6 + b omega^4 + c omega^2)/(b + c omega^(10) +a omega^8) = (a+ b omega + c omega^2)/(b + c omega + a omega^2)`

`= (omega (a + b omega + c omega^2))/(omega(b + c omega + a omega^2))`

`= ( omega (a + b omega + c omega^2))/(a omega^3 + b omega + c omega^2)`

` = ( omega (a + b omega + c omega^2))/(a+ b omega + c omega^2) = omega`

Correct Answer is `=>` (C) `omega`

Q 2316280179

If `omega` is a complex non-real cube root of unity, then
`omega` satisfies which one of the following equations?
NDA Paper 1 2008

(A)

`x^2 - x + 1 = 0`

(B)

`x^2 + x + 1 = 0`

(C)

`x^2 + x - 1 = 0`

(D)

`x^2 - x - 1 = 0`

Solution:

Since, `omega` is cube root of unity, then

`omega^2 + omega + 1 = 0, omega^3 = 1`

From above it is clear that, `omega` satisfies the equation

`x^2 + x + 1 = 0 .`

Correct Answer is `=>` (B) `x^2 + x + 1 = 0`

Q 2316867779

If `alpha` is a complex number such that `alpha^2 + alpha + 1 = 0`,
then what is `alpha^(31)` equal to?
NDA Paper 1 2009

(A)

`alpha`

(B)

`alpha^2`

(C)

`0`

(D)

`1`

Solution:

Since, `alpha` is a complex root such that

`alpha^2 + a + 1 = 0`

`:. alpha = omega` or `omega^2` (since, `alpha` is a cube root of unity)

`:. alpha^(31) = (omega)^(31) = (omega^3)^(10) . omega = omega = alpha`

Correct Answer is `=>` (A) `alpha`

Q 2326178971

Which one of the following is correct? If `z` and `omega`
are complex numbers and `bar omega` denotes the conjugate
of `omega` then `| z + omega | = | z - omega |` holds only, if
NDA Paper 1 2008

(A)

`z = 0` or `omega = 0`

(B)

`z = 0` and `omega = 0`

(C)

`z. bar omega` is purely real

(D)

`z · bar omega` is purely imaginary

Solution:

We know that, `| z + omega | = | z - omega |` holds only if one of `z`

and `w` is equal to zero.

Correct Answer is `=>` (A) `z = 0` or `omega = 0`

Q 2356878774

If `omega` is complex cube root, then what is the value of
`1 - 1/(1 + omega) - 1/(1 + omega^2) ` ?

NDA Paper 1 2008

(A)

`1`

(B)

`0`

(C)

`omega`

(D)

`omega^2`

Solution:

`1 - 1/ (1+omega) - 1/( 1+ omega^ 2) = 1 - 1/ (-omega^2 ) - 1/(-omega)`

` = (omega^ 2 +1 + omega)/omega^2 = 0/omega^2 = 0 ( ∵ 1 + omega + omega^2 = 0)`

Correct Answer is `=>` (B) `0`

Q 2316578479

NDA Paper 1 2009

Assertion : ` ( ( -1 + sqrt(-3))/2)^(29) + ( ( -1 - sqrt(-3))/2)^(29) = -1`

Reason : `omega^2 = -1`

(A)

Both A and R individually true and R is the correct explanation of A

(B)

Both A and R are individually true but R is not the correct explanation of A

(C)

A is true but R is false

(D)

A is false but R is true

Solution:

` [ ( -1 + sqrt(-3))/2]^(29) + [ ( -1 - sqrt(-3))/2]^(29) `

` = [ ( -1 + i sqrt(+3))/2]^(29) + [ ( -1 - i sqrt(-3))/2]^(29) `

` = (omega)^(29) + ( omega^2)^(29)`

` = omega^2 + omega = -1 ( ∵ omega^3 = 1` and `omega^2 + omega + 1 = 0`)

`R ∵ omega^ 2 != -1`

Hence, A is true but R is false.

Correct Answer is `=>` (C)

Q 2356378274

If `z` is a complex number such that `z + z ^(-1) = 1`,
then what is the value of `z^(99) + z^(-99)` ?
NDA Paper 1 2009

(A)

`1`

(B)

`-1`

(C)

`2`

(D)

`-2`

Solution:

Given, `z + z^(-1) = 1`

`=> z^2 - z + 1 = 0 => z = - omega` and `- omega^2`

When `z = - omega`, then

`z^(99) + z^(-99) = (- w)^(99) + (- omega)^(-99) = -1 - 1 = - 2 ( ∵ omega^3 = 1)`

When `z = - omega^2`, then

`z^(99) + z^(-99) = (- omega ^2)^(99) + (- w^2)^(-99) = -1 - 1 = - 2`

Correct Answer is `=>` (D) `-2`

Q 2376178076

If `omega` is the cube root of unity, then what is the
conjugate of `2 omega^2 + 3i`?
NDA Paper 1 2009

(A)

`2 omega - 3i`

(B)

`3 omega - 2i`

(C)

`2 omega + 3i`

(D)

`3 omega - 2i`

Solution:

let `z = 2 omega^2 + 3i`

` = 2 (-1 - sqrt(3)i) /2 + 3i`

`= - 1 + (3 - sqrt(3))i`

`:. bar z = -1 - (3- sqrt(3)) i`

`= (-1 + sqrt(3) i)- 3i = 2 omega - 3i`

` ( ∵ omega = ( -1 + i sqrt3)/2 )`

Correct Answer is `=>` (A) `2 omega - 3i`

Q 2366534475

Consider the following statements
I. `(omega ^(10) +1)^7 + omega = 0`
II. `(omega^( 105) + 1)^(10) = p^(10)` for some prime number `p`.
where, `omega = 1` is a cubic root of unity.
Which of the above statements is/are correct?
NDA Paper 1 2012

(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

`L.H.S = (omega^(10) + 1 )^7 + omega = 0`

`= [( omega^3)^3 omega + 1)^7 + omega = 0 (∵omega^3 = 1)`

`= (1 + omega)^7 + omega = 0`

`= (- omega^2)^7 + omega = 0`

`[ ∵ 1 + omega + omega^2 = 0 => 1 + omega = - omega^2]`

`= - omega^(14) + omega = 0`

`= - (omega^3)^4 omega^2 + omega = 0`

`= - omega^2 + omega = 0 = omega (-omega + 1) = 0`

`∵ omega != 0`

`:. -omega + 1 = 0 => omega = 1`

Hence, Statement I is false.

Statement II

`(omega^(105) + 1)^(10) = p^(10)`

`=> [(m^3)^(35) + 1]^(10) = p^(10)`

`=> (1 + 1)^(10) = p^(10)`

`2^(10) = p^(10)` which is true for prime number `2`.

So, Statement I is false and Statement II is true .

Correct Answer is `=>` (B) Only II

Q 2376367276

If `omega` is a complex cube root of unity and
`x = omega^2 - omega - 2`, then what is the value of
`x^2 + 4x + 7`?
NDA Paper 1 2010

(A)

`-2`

(B)

`-1`

(C)

`0`

(D)

`1`

Solution:

`∵ x = omega^2 - omega - 2`

`=> x + 2 = omega^2 - omega`

On squaring both sides, we get

`(x + 2)^2 = (omega^2 - w)^2`

`=> x^2 + 4x + 4 = omega^4 + omega^2 - 2 omega^3`

`=> x^2 + 4x + 4 + 3 = omega + omega^2 - 2 + 3`

`=> x^2 + 4x + 7 = 1 + omega + omega^2`

`=> x^2 + 4x + 7 = 0`

Correct Answer is `=>` (C) `0`

Q 2386267177

What is the value of

` ( ( i + sqrt3)/(-1 + sqrt3))^(200) + ( ( i - sqrt3)/( 1 + sqrt3))^(200) + 1` ?
NDA Paper 1 2010

(A)

`-1`

(B)

`0`

(C)

`1`

(D)

`2`

Solution:

Now ` ( ( i + sqrt3)/(-1 + sqrt3)) = (i + sqrt(3))^2/((sqrt3 - i)(sqrt3 + i))`

` = ( i^2 +3 + 2 sqrt(3) i)/(3 + 1)`

` = (- 1 + 3 + 2 sqrt(3)i)/4 = (1 + sqrt(3)i)/ 2 = - omega^2`

and ` ( ( i - sqrt3)/(1 + sqrt3)) = (i - sqrt(3))^2/(i^2 - (sqrt(3))^2)`

` = ( i^2 + 3 - 2 i sqrt(3))/(-4) = ( 2 - 2 i sqrt(3))/(-4)`

` = ( -1 + i sqrt(3))/2 = omega`

` :. ( ( i + sqrt3)/(-1 + sqrt3))^(200) + ( ( i - sqrt3)/( 1 + sqrt3))^(200) + 1`

`= (- omega^2)^(200) + omega^(200) + 1 = omega^( 400) + omega^(200) + 1`

`(∵ omega^3 = 1` and `omega^2 + omega + 1 = 0`)

`= omega^(3 xx 133 + 1) + omega^( 3 xx 66 + 2) + 1`

`= omega + omega^2 + 1 = 0`

Correct Answer is `=>` (B) `0`

Q 2376756676

What is the value of `( -1 + i sqrt(3))^( 48)`?
NDA Paper 1 2010

(A)

`1`

(B)

`2`

(C)

`2^(24)`

(D)

`2^(48)`

Solution:

`∵ omega = - ( 1 + i sqrt(3))/2`

`:. (-1 + i sqrt(3) )^(48) = (2 w)^(48) = 2^(48) omega^(48) = 2^( 48) (omega^3)^(16) = 2^(48)`

`(∵ omega^3 = 1)`

Correct Answer is `=>` (D) `2^(48)`

Q 2336356272

If `omega` is the imaginary cube root of unity, then
`(2 - omega + 2 omega^2)^( 27)` is equal to
NDA Paper 1 2011

(A)

`3^(27) omega`

(B)

`- 3^(27) omega^2`

(C)

`3^(27)`

(D)

`- 3^(27)`

Solution:

`(2 - omega + 2 omega^2)^(27) = [2 (1 + omega^2) - omega]^(27)`

(`∵ 1 + omega + omega^2 = 0` and `omega^3 = 1` )

`= (-2 omega - omega)^(27) = (-3 omega )^(27)`

`= - 3^(27). omega^(27)`

`= - 3^(27)(omega^3)^9 = - 3^(27). 1 = - 3^(27)`

Correct Answer is `=>` (D) `- 3^(27)`

Q 2460278115

What is the value of ` |(1,omega,2omega^2),(2,2omega^2,4omega^3),(3,3omega^3,6omega^4)|`, where `omega` is
the cube root of unity?
NDA Paper 1 2011

(A)

`0`

(B)

`1`

(C)

`2`

(D)

`3`

Solution:

` |(1,omega,2omega^2),(2,2omega^2,4omega^3),(3,3omega^3,6omega^4)| = |(1,omega,2omega^2),(2,2omega^2,4),(3,3,6omega)|`

`= 1 (12 omega^3 - 12)- omega(12 omega - 12) + 2 omega^2 (6 - 6 omega^2) (∵ omega^3 = 1)`

`= 0 - 12 omega^2 + 12 omega + 12 omega^2 - 12 omega = 0`

Correct Answer is `=>` (A) `0`

Q 2386545477

If `alpha` and `beta` are the complex cube roots of unity, then
what is the value of `(1 + alpha) (1 + beta ) (1 + alpha^2 )(1 + beta^2 )`?
NDA Paper 1 2011

(A)

`-1`

(B)

`0`

(C)

`1`

(D)

`4`

Solution:

Given, `alpha` and `beta` are the complex cube roots of unity,

then

`1 + alpha + beta = 0`

`=> alpha + beta = -1` ......(i)

and `alpha . beta = 1` .......(ii)

Now,

`(1 + alpha ) (1 + beta ) (1 + alpha^2) (1 + beta^2)`

`= {1 + (alpha + beta) + alpha beta } {1 + (alpha beta )^2 + (alpha^2 + beta^2)}`

`= {1 + (alpha + beta ) + alpha beta} {1 + ( alpha beta)^2 + (alpha + beta)^2 - 2 alpha beta }`

From Eqs. (i) and (ii),

`= {1 - 1 + 1}. {1 + 1 + (-1)^2 - 2}`

`=1·(2 + 1 - 2)`

`= 1·1 = 1`

Correct Answer is `=>` (C) `1`

Q 2318023809

If `p, q` and `r` are positive integers, `omega` is the cube root
of unity and `f (x) = x^(3P) + x^(3q + 1) + x ^(3r + 2)` , then
what is `f(omega)` equal to?
NDA Paper 1 2011

(A)

`omega`

(B)

`-omega^2`

(C)

`-omega`

(D)

`0`

Solution:

Given, `p, q` and `r in z^+`

and `omega` is the cube root of unity.

Then, `f(x) = x^(3p) + x^(3q + 1) + x^(3r+ 2)`

`=> f(omega) = omega^(3p) + omega^(3q + 1) + omega^(3r + 2)`

`= (omega^3)^p + (omega^3)^q * omega + (omega^3)^r* omega^2`

`= (1)^p + (1)^q * omega + (1)^r *omega^2`

`=1+omega+omega^2`

`=0`

Correct Answer is `=>` (D) `0`

Q 2513401349

If `omega ne 1 ` is a cube root of unity, then the sum of

the series `S = 1+ 2omega + 3 omega^2 +....+ 3 n omega^(3n -1)` is
WBJEE 2011

(A)

`(3n)/(omega -1)`

(B)

`3n (omega -1)`

(C)

`(omega -1)/(3n)`

(D)

`0`

Solution:

Given, `S = 1 + 2 omega + 3 omega^2 + ... + 3 n omega^(3n- 1)`

`:. S omega = omega +2 omega^2 + ......+ (3n-1) omega^(3n) + 3n omega^(3n)`

`=> S (1- omega) =1 + omega + omega^2 +....+ omega^(3n-1) + 3 n omega^(3n)`

`=> S(1-omega) = 0 - 3n`

`=> S= - (3n)/(1- omega) =(3n )/(omega -1)`

Correct Answer is `=>` (A) `(3n)/(omega -1)`

Q 2520856711

The value of `((1+ sqrt (3)i)/(1- sqrt (3)i))^(64) + ((1- sqrt(3)i)/(1+sqrt(3)i))^(64)` is
WBJEE 2015

(A)

`0`

(B)

`-1`

(C)

`1`

(D)

`i`

Solution:

We know that, `omega = (-1+ sqrt (3)i )/2 1-sqrt(3) i = -2 omega`

and `omega^2 = (-1 - sqrt (3) i)/2 1+ sqrt() i = -2 omega^2`

Now, `((1+sqrt(3) i)/(1- sqrt(3) i))^(64) + ((1- sqrt(3) i)/(1+ sqrt (3) i))^(64)`

`= ((-2 omega^2)/(-2 omega))^(64) + ((-2 omega)/(-2 omega))^(64)`

`= omega^(64) + 1/omega^(64) = omega + omega^2` `[ :. omega^3 = 1 ]`

`=-1` `[ :. 1+ omega + omega^2 = 0 ]`

Correct Answer is `=>` (B) `-1`

Q 1622334231

If ω is the cubic root of unity, then value
of the `(1+omega-omega^2)^2+(1-omega+omega^2)^2+1` is
UPSEE 2016

(A)

`1`

(B)

`-3`

(C)

`-1`

(D)

`7`

Solution:

`1+omega+omega^2 =0`

`(1+omega-omega^2)^2+(1-omega+omega^2)^2+1 =(-omega^2-omega^2)^2+(-omega-omega)^2+1`

`=4omega^4+4omega^2+1 =4(omega+omega^2)+1 =4(-1)+1 =-4+1 =-3`

Correct Answer is `=>` (B) `-3`

de-moivre's theorem

If n is an integer, positive, negative or a rational number, then

`(cos theta + i sin theta)^n =cos ntheta + i sin ntheta`

Q 2372512436

What is `[(sin \ \ pi/6 + i (1 - cos \ \ pi/6))/(sin\ \ pi/6 - i (1 - cos pi/6)) ]^3 `, where `i = sqrt(-1) `, equal to ?
NDA Paper 1 2015

(A)

`1`

(B)

`-1`

(C)

`i `

(D)

`-i`

Solution:

Let `Z = [(sin \ \ pi/6 + i (1 - cos \ \ pi/6))/(sin\ \ pi/6 - i (1 - cos pi/6)) ]^3 `

`= [(2 sin \ \ pi/12 cos \ \pi/12 + i 2 sin^2 \ \ pi/2)/(2 sin \ \ pi/12 cos \ \pi/12 - i 2 sin^2 \ \pi/12)]^3`

`=[ (cos \ \pi/2 + i sin \ \pi/12)/(cos\ \pi/2 - i sin \ \ pi/12)]^3`

`= [((cos \ \pi/2 + i sin \ \ pi/12 ) ( cos \ \ pi/12 + i sin \ \ pi/12))/((cos \ \ pi/12 - i sin \ \pi/12)( cos\ \ pi/12 + i sin \ \pi/12))]^3`

`=[((cos \ \pi/12 + i sin \ \pi/12)^2)/(cos^2 \ \pi/12 + sin^2 \ \ pi/12)]^3`

`= (cos \ \ pi/12 + i sin^2 \ \pi/12)^6`

`= cos 6 xx pi/12 + i sin 6 xx pi/12`

`= cos \ \pi /2 + i sin \ \pi/2 = i`

Correct Answer is `=>` (C) `i `

Q 1658023804

What is the real part of `(sin x + i cos x)^ 3`, where
`i = sqrt(-1)?`
NDA Paper 1 2015

(A)

`-cos 3x`

(B)

`-sin 3x`

(C)

`sin 3x`

(D)

`cos 3x`

Solution:

` ( sinx + i cos x)^3`

` [ cos ( pi/2 - x) + i ( sin \ pi/2 - x) ]^3`

` [ {e ^(i(pi/2 - x)) }^3= e ^(3i(pi/2 - x))]`

` = cos 3 (pi/2 - x) + i sin 3 (pi/2 - x)`

` = cos ((3pi)/2 -3x) + i sin ((3pi)/2 -3x)`

` = (-sin 3x - i cos 3x)`

Hence, the real part is `- sin 3x`.

Correct Answer is `=>` (B) `-sin 3x`

Q 1712334239

What is `( (sqrt(3) + i)/(sqrt(3) - i) )^6` equal to, where `i = sqrt(-1) ?`
NDA Paper 1 2014

(A)

`1`

(B)

`1//6`

(C)

`6`

(D)

`2`

Solution:

`( (sqrt(3) + i)/(sqrt(3) - i) ) = (sqrt(3) + i)/(sqrt(3) - i) xx (sqrt(3) + i)/(sqrt(3) + i)`

` = ( 3 + i^2 + 2sqrt(3) i)/ ( 3 - i^2) = (3 - 1 + 2 sqrt(3) i)/( 3 + 1)`

` = ( 2(1 + sqrt(3) i))/4 = 1/2 + i sqrt(3)/2`

` = ( cos (pi/3) + i sin (pi/3) ) = e ^(i pi/3)`

`:. ( (sqrt(3) + i)/(sqrt(3) - i) )^6 = (e ^(i pi/3))^6 = e ^(i2pi) = cos 2 pi + i sin 2 pi`

` = 1+ 0 . i = 1`

Correct Answer is `=>` (A) `1`

Q 1305801768

If `z_k=cos(frac{k pi}{10})+i sin(frac{k pi}{10})` , then `z_1z_2z_3z_4` equals

(A)

`-1`

(B)

`0`

(C)

`1`

(D)

`2`

Solution:

We have `z_k= omega^k`

where `omega=cos frac{ pi}{10}+i sin frac{pi}{10}`

Thus `z_1z_2z_3z_4= omega^1 . omega^2 . omega^3 . omega^4=omega^{10}`

`Rightarrow cos(frac{10 pi}{10})+i sin(frac{10 pi}{10})`

(By De-Moivre's theorem)

`Rightarrow cos pi`

`Rightarrow -1`

Hence `1` is the correct answer.

Correct Answer is `=>` (A) `-1`

Q 2444812753

`[(1+ cos (theta /2) -i sin (theta/2))/(1+ cos (theta/2) + i sin (theta/2))]^(4n)` is equal to
UPSEE 2011

(A)

`cos n pi- i sin n theta`

(B)

`cos n theta + i sin n theta`

(C)

`cos 2n theta - i sin 2n theta`

(D)

`cos 2ntheta + i sin 2n theta`

Solution:

`[(1+ cos (theta /2) -i sin (theta/2))/(1+ cos (theta/2) + i sin (theta/2))]^(4n)`

`[(2 cos^2 (theta/4)-2i sin (theta/4) cos (theta/4))/(2 cos ^2 (theta/4)+2i sin(theta/4) cos (theta/4))]^(4n)`

`=[(cos (theta/4)-i sin(theta/4))/(cos(theta/4)+ i sin (theta/4))]^(4n)`

`= [(e^(-itheta/4))/(e^(i theta/4))]^(4n)`

`=[e^(-i theta/2)]^(4n)`

`=e^(-2nitheta)`

`=cos 2 n theta-i sin 2 n theta`

Correct Answer is `=>` (C) `cos 2n theta - i sin 2n theta`

Q 2510523419

If `x_r = cos ( pi/3^r) - i sin( pi/3^r) `, (where `i = sqrt(-1) )`
then the value of `x_1 · x_2 .................oo `,is
BCECE Stage 1 2013

(A)

`1`

(B)

`-1`

(C)

`-i`

(D)

`i`

Solution:

since , `x_r = cos ( pi/3^r) - i sin ( pi/3^r)`

` x_1 . x_2 .x_2 .. oo = cos ( pi/3^1 + pi/3^2 + pi/3^3 + ... oo )`

` - i sin ( pi /3^1 + pi/3^2 + pi/3^3 + ... oo )`

` = cos ( (pi/3)/(1 - 1/3)) - i sin ( (pi/3)/(1 - 1/3))`

` = cos (pi/2) - i sin ( pi/2) = -i`

Correct Answer is `=>` (C) `-i`

Square Roots of a Complex Number

Q 2356023874

What is one of the square roots of `3 + 4 i`, where
`i = sqrt(-1)`?
NDA Paper 1 2013

(A)

`2 + i`

(B)

`2 - i`

(C)

`- 2 + i`

(D)

`- 3 - i`

Solution:

Let `x + iy = sqrt(3 + 4i)`

On squaring both sides, we get

`(x + iy)^2 = 3 + 4i`

`=> x^2 - y^2 + 2x yi = 3 + 4 i`

Equating real and imaginary parts on both sides, we get

`x^2 - y^2 = 3` ... (i)

and `2xy = 4` ... (ii)

Now, we use the following identity,

`(x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2`

`=> (x^2 + y^2)^2 = (3)^2 + (4)^2 = 9 + 16 = 25`

`=> x^2 + y^2 = 5` ... (iii)

From Eqs. (i) and (iii),

`x^2 = 4` and `y2 = 1`

`=> x = ± 2` and `y = ± 1`

Since, the product of `x y` is positive.

`:. x = 2` and `y = 1`

or `x = - 2` and `y = - 1`

Thus, the square root of the complex number `3 + 4i` is `± 2 ± i`.

Correct Answer is `=>` (D) `- 3 - i`

Q 2336745672

What are the square roots of `-2 i` ?(where, `i = sqrt(-1))`
NDA Paper 1 2011

(A)

`± (1 + i)`

(B)

`± (1 - i)`

(C)

`± i`

(D)

`± 1`

Solution:

Square root of `(-2i)` i.e., `(-2i)^(1//2)` ... (i)

Let `z = r (cos theta + i sin theta) = 0 - 2i`

On comparing

`r cos theta = 0` .. (ii)

`r sin theta = - 2` ... (iii)

On squaring Eqs. (ii) and (iii) and then adding, we get

`r^2 = 4`

`=> r = ±2`

On dividing Eq. (iii) by Eq. (ii), we get

`tan theta = oo = tan \ pi/2`

`:. theta = pi/2`

For principal value of `theta = - pi/2`

From Eq. (i),

`(-2i)^(1//2) = {± 2 (cos pi//2 - i sin pi//2)}^(1//2)`

`= ± 2^(1//2) {cos (pi//2) - i sin (pi//2)}^(1//2)`

`= ± sqrt(2) {cos pi//4 - i sin pi//4}`

(by De-Moivre theorem)

`= ± sqrt(2) {1/sqrt(2) -i 1/sqrt(2)}`

`= ± sqrt(2) (1 - i)/sqrt(2) = ± (1 - i)`

Correct Answer is `=>` (B) `± (1 - i)`

Q 2356180074

What is the square root of `1/2 - i sqrt(3)/2`?
NDA Paper 1 2008

(A)

`pm ( sqrt(3)/2 + i/2)`

(B)

`pm ( sqrt(3)/2 - i/2)`

(C)

`pm ( 1/2 + i sqrt(3)/2 )`

(D)

`pm ( 1/2 - i sqrt(3)/2 )`

Solution:

Let ` sqrt( 1/2 - (i sqrt(3))/2) = x + iy => 1/2 - (i sqrt(3))/2 = (x + iy)^2`

` => 1/2 - (i sqrt(3))/2 = x^2 - y^2 - 2 ixy`

On comparing real and imaginary parts, we get

`x^2 - y^2 = 1/2` ............(i)

and `2xy = sqrt(3)/2` ......(ii)

and `(x^2+y^2)^2 = (x^2 - y^2)^2 + 4x^2y^2`

`= 1/4 + 3/4 = 1`

`=> x^2 + y^2 = 1` .....(iii)

On solving Eqs. (i) and (iii), we get

`x^2 = 3/4` and `y^2 = 1/4`

` => x = pm sqrt(3)/2 `and ` y = pm 1/2`

Since, the product of `x` and `y` is positive.

`:. x = sqrt(3)/2` and `y = - 1/2`

`x = - sqrt(3)/2` and `y = - 1/2` is

` sqrt( 1/2 - (i sqrt (3))/2 ) = pm ( sqrt(3)/2 - i/2)`

Correct Answer is `=>` (B) `pm ( sqrt(3)/2 - i/2)`

Use of Complex Numbers in Coordinate Geometry

Q 2211701620

If the point `z_1 = 1 + i`, where `i = sqrt(-1)`.,is the reflection of a
point `z_ 2 = x + iy` in the line `ibarz - iz = 5`, then the point `z_ 2`
is
NDA Paper 1 2015

(A)

`1 + 4 i`

(B)

`4 + i`

(C)

`1- i`

(D)

`-1- i`

Solution:

Given equation of line is `i barz - iz = 5`.

Let `z = x + iy` and `z = x - iy`

`:. i(x- iy)- i(x + iy) = 5 => 2y = 5`

`=> 2y- 5 = 0` .........(1)

Since, it is given that reflection of point `x + iy`, i.e. `(x, x)`

about the line `(i)` is `(1 + i)`, i.e. `(1, 1)`.

`:. (1 - x)/0 = (1-y)/2 = (- 2(2y - 5))/4`

` => (1 - x)/0 = (- (2y - 5))/2` and `( 1 - y)/2 = - (2y - 5)/2`

`=> x = 1` and `1 - y = - 2y + 5`

`=> x = 1` and `y = 4`

`:.` Required point is `1 + 4i`.

Correct Answer is `=>` (A) `1 + 4 i`

Q 2281012827

`z bar(z) + (3 - i) z + (3 + i) bar (z) + 1 = 0` represents a circle with
NDA Paper 1 2015

(A)

centre `(- 3,- 1)` and radius `3`

(B)

centre `(- 3, 1)` and radius `3`

(C)

centre `(- 3,- 1)` and radius `4`

(D)

centre `(- 3, 1)` and radius `4`

Solution:

Given, `z bar(z) + (3 - i) z + (3 + i) bar (z) + 1 = 0`

Put `z = x + iy` and `bar(z) = x - iy`, we get

`(x + iy) (x -iy) + (3- i) (x + iy) + (3 + i) (x -iy) + 1 = 0`

`=> x^2 + y^ 2 + 3x + 3iy - ix + y + 3x- 3iy + ix + y + 1 = 0`

`=> x^2 + y ^2 + 6x + 2y + 1 = 0`

:. Centre `= (- g,- f) = (- 3, -1)`

and radius `= sqrt(g^2 +f^2 -c) = sqrt( 9 + 1 -1) - sqrt(9) = 3`

Correct Answer is `=>` (A) centre `(- 3,- 1)` and radius `3`

Q 2356078874

A straight line is passing through the points
represented by the complex numbers `a + ib` and
`1/(-a+ ib) `,where `(a, b) != (0,0)`.
Which one of the following is correct?
NDA Paper 1 2008

(A)

It passes through the origin

(B)

It is parallel to the X-axis

(C)

It is parallel to the Y-axis

(D)

It passes through (0, b)

Solution:

`1/(-a+ ib) = ( - a - ib)/(a^2 + b^2) = ( (-a)/(a^2 + b^2) - i b/(a^2 + b^2) )`

( `∵ A + iB` form)

Equation of line which passes through the point `(a, b)` and the

point ` ( (-a)/(a^2 + b^2) ,(- b)/(a^2 + b^2) )`

` (y - b) = ((-b)/(a^2 + b^2 )-b)/((-a)/(a^2 + b^2)-a) (x - a)= b/a (x - a)`

` => ay = bx`

Hence, a straight line is passing through the points represented

by the complex numbers ` a + ib` and ` 1/(-a + ib)` which passes

through the origin.

Correct Answer is `=>` (A) It passes through the origin