Please Wait... While Loading Full Video

previous year question of Complex Number for NDA

Previous Year Complex Number Question For NDA

Previous Year Complex Number Question For NDA
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 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 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 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 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 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 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`

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

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`

As, `z- 6 = 0 =>| (z-6)/(z+6) | = 0`
Correct Answer is `=>` (D) `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)

1

(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 2166478375

If `z = x + iy = (1/sqrt(2) - i/sqrt(2) )^(-25)` , where `i = sqrt(-1)`, then what

is the fundamental amplitude of ` (z - sqrt(2) )/ (z - isqrt(2) )`?

NDA Paper 1 2016
(A)

`pi`

(B)

`pi/2`

(C)

`pi/3`

(D)

`pi/4`

Solution:

We have, `z = x + iy =(1/sqrt(2) - i/sqrt(2) )^(-25)`

` = (e^(-i quad pi/4))^(-25) = e^(-i quad 25/4 quad pi)`

` = e ^(i (6pi +pi/4) = e ^(i(pi/4))`

` = cos pi/4 + i sin pi/4 = 1/sqrt(2) +( i 1)/sqrt(2)`

`:. (z - sqrt(2) )/ (z - isqrt(2) ) = ( 1/sqrt(2) + (i 1)/sqrt(2) - sqrt(2) )/ ( 1/sqrt(2) + (i 1)/sqrt(2) - isqrt(2) )`

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

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

` = -1 + 0i`

Now, `alpha= tan^(-1)|0/-1| = tan^(- 1)(0)= tan^(-1) | tan 0 |= 0`

`:. theta` (argument)= ` pi -0 = pi`
Correct Answer is `=>` (A) `pi`
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 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 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 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 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 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 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 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 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 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 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 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 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 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 2306134078

What is the value of `sqrt(-1)`, where `i = sqrt(-1)` ?
NDA Paper 1 2013
(A)

`( ± (1 - i)/sqrt(2))`

(B)

`( ± (1 + i)/sqrt(2))`

(C)

`( ± (1 - i)/2)`

(D)

`( ± (1 + i)/2)`

Solution:

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

Here, by comparing real and imaginary parts on both sides, we

get

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

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

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

`r^2 sin^2 theta + r^2 cos^2 theta = 0 + 1`

`=> r^2(sin^2 theta + cos^2 theta ) = 1`

`=> r^2 = 1`

`:. r = pm 1`

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

`tan theta = oo = tan90°`

` theta = 90°= pi/2`.

But principle argument of `z =- theta` (since, `z` lies in IVth quadrant)
`= (-pi)/2`

`:. z = -i = ± 1{ cos( (-pi)/2 ) + i sin ( (- pi)/2)}`

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

Now, `sqrt z = sqrt(-1) = ± (1)^(1//2) (cos pi/2 - i sin pi/2)^(1//2)`

(using De - Moivre theorem)

` = pm 1 ( 1/sqrt2 - i . 1/sqrt 2 ) = ± ((1 - i)/sqrt 2)`

`:. sqrt(-1) = pm ((1 - i)/sqrt 2)`

Alternate Method I

Let `x + iy = sqrt(-1)`

On squaring both sides, we get

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

`x^2 - y^2 + 2ixy = -i`

On equating real and imaginary parts on boths sides, we get

`x^2 - y^2 = 0` ... (i)

and `2xy = -1` ... (ii)

Using by identities,

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

`= 0 + 1 = 1`

`x^2 + y^2 = 1`

From Eqs. (i) and (iii),

`2x^2 = 1`

`=> x = + 1/sqrt2`

and ` y = pm 1/sqrt2`

From Eq. (ii), the product of `x` and `y` is negative.

`:. x = 1/sqrt2` and ` y = - 1/sqrt2`

` x = - 1/sqrt2` and ` y = 1/sqrt2`

`:.` Required value, `sqrt(-i) = pm ((1 - i)/sqrt 2)`

Alternate Method II ` sqrt(-i) = sqrt(e^(- i pi//2)) ( ∵ e^(-i 0) = cos theta - i sin theta )`

` = ± e^( - i pi/2 xx 1/2)`

`= ± e^( -i pi/4) = ± [ cos(pi/4) - i sin (pi/4 )]`

`= ± (1/sqrt2 - i/sqrt2 ) = ± ((1 - i)/sqrt2)`
Correct Answer is `=>` (A) `( ± (1 - i)/sqrt(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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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`
 
SiteLock