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Must Do Problems of Complex Number for NDA

Must Do Problem For NDA

Must Do Problem For NDA
Q 2418245100

The points `0, 2 + 3i, i, -2- 2i` in the argand
plane are the vertices of a
UPSEE 2009
(A)

rectangle

(B)

rhombus

(C)

trapezium

(D)

parallelogram

Solution:

Let `A = 0, B = 2 + 3i, C = i, D = 2- 2i`

Now, `AB = sqrt(2^2 + 3^2) = sqrt(13)`

`BC = sqrt(2^2 + 2^2) = sqrt8`

`CD = sqrt(2^2 + 3^2) = sqrt(13)`

` DA = sqrt(2^2 + 2^2) = sqrt(8)`

Also, `AC = sqrt(1^2) = 1`

`BD = sqrt((4)^2+ (5)^2) = sqrt(16 + 25)`

`= sqrt(41)`

`=> AB != BC`

Hence, it is a parallelogram
Correct Answer is `=>` (D) parallelogram
Q 1987767687

Convert the complex numbers given in the polar form.

`sqrt(3) + i`
Class 11 Exercise 5.2 Q.No. 7
Solution:

`r= sqrt(3) + i =r (cos theta + i sin theta)`

`:. r cos theta = sqrt(3) , r sin theta= 1`

Squaring and adding `r^2 = 3 + 1 = 4, r = 2`

Also `tan theta =1/sqrt(3) , sin theta` and `cos theta` both are positive.

`:. theta` lies in the `I ` quadrant.

`:. theta = 30^(circ) = pi/6`

`:. ` polar form of `z` is `2 (cos (pi/6 )+ i sin (pi/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 2610356219

Express the complex number in polar form `5+2i`

Solution:

So, first find the absolute value of r

`r = |z| = sqrt(a^2 + b^2) = sqrt(5^2+2^2)`

`=5.39 `

Now find the argument `θ`

Since `a>0` , use the formula `θ= tan^(-1)(b/a)`

`θ= tan^(-1)(2/5) = 0.38` (θ is measured in radians.)

Therefore, the polar form of `5 + 2 i` is about `5.39(cos(0.38)+isin(0.38))`
Q 1365412365

Convert the given complex number in polar form `-1-i`

(A)

`sqrt{2} [cos (-frac{ pi}{2})+i sin(-frac{pi}{2})]`

(B)

`sqrt{2} [cos (frac{3 pi}{2})+i sin(frac{3 pi}{2})]`

(C)

`sqrt{2} [cos (-frac{3 pi}{4})+i sin(-frac{3 pi}{4})]`

(D)

`sqrt{2} [cos (frac{ pi}{2})+i sin(frac{pi}{2})]`

Solution:

Explanation :

Let `r cos theta =-1 ` and `r sin theta = -1`

on squaring and adding , we obtain

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

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

`r^{2} = 2` `[cos^{2} theta + sin^{2} theta=1]`

`r =sqrt{2} ` [conventionally ,` r > 0` ]

Therefore , `sqrt{2} cos theta =- 1 ` and `sqrt{2} sin theta = - 1`

`cos theta =frac{-1}{sqrt{2}}` and `sin theta = frac{-1}{sqrt{2}}`

`theta = - (pi-frac{pi}{4})=-frac{3 pi}{4}` [`theta` lies in the `3^{rd}` quadrant]

`-1-i = r cos theta + i r sin theta`

`=sqrt{2} [cos (-frac{3 pi}{4})+i sin(-frac{3 pi}{4})]` this is the required polar form .
Correct Answer is `=>` (C) `sqrt{2} [cos (-frac{3 pi}{4})+i sin(-frac{3 pi}{4})]`
Q 1715345260

If `| z | = 4` and `arg (z) = (5pi)/6` then `z =` ......
NCERT Exemplar
Solution:

Given that, `| z | = 4` and `arg (z) = (5pi)/6`

Let `z = x + iy = r (cos theta + i sin theta)`

`=> | z | = r = 4` and `arg (z) = theta`

`∵ tan theta = (5pi)/6`

`=> z = 4 ( cos ((pi)/6) + i sin (pi/6) ) = 4 [cos(pi - pi//6) + i sin (pi - pi//6)]`

` = 4[ -cos (pi/6) + i sin (pi/6) ) = 4 [ (-sqrt(3))/2 + i (1/2) ] = -2sqrt(3) + 2i`
Q 2513156049

Find the number of solutions of the equation `z^2 + | z |^2 - 0`.



Solution:

`∵ z^2 + | z |^2 = 0`

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

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

`:. z = 0` ......(i)

and `z + bar z = 0`

`=> 2Re (z) =0`

`:. Re (z) =0`

If `z = x + iy` [∵ x = Re (z)]

`= 0 + iy, y in R`

and `i = sqrt(-1)` ... (ii)

On combining Eqs. (i.) and (ii), then we can say that the given equation has

infinite solutions.
Q 2417156989

If `\alpha ,\beta` are the roots of `x ^ 2 -x+1=0` then the quadratic equation whose roots are `\alpha ^ 2015 , \beta ^ 2015` is
EAMCET 2016
(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:

`\alpha =-\omega ,\beta =- \omega ^{ 2 }`

`\therefore \alpha ^{ 2015 }=- \omega ^{ 2 }, \beta ^{ 2015 }=-\omega`
Correct Answer is `=>` (A) `x ^ 2 -x+1=0`
Q 1315191060

The integral solution of the equation `(1-i)^x=2^x` is?

(A)

`3`

(B)

`2`

(C)

`1`

(D)

`0`

Solution:

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

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

`(1-i)^8=(-4)^2=16=2^4`

Given `(1-i)^x=2^x`

`Rightarrow [(1-i)^x]^8=(2^x)^8`

`Rightarrow [(1-i)^8]^x=2^{8x}`

`Rightarrow (2^4)^x=2^{8x}`

`Rightarrow 2^{4x}=2^{8x}`

`Rightarrow 4x=8x` or `x=0`
Correct Answer is `=>` (D) `0`
Q 1416301270

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 `=>` (B) `2`
Q 2312480339

`sqrt(6 + 8i) + sqrt(6 − 8i)` is equal to :
BITSAT Mock
(A)

`3 sqrt(2) i`

(B)

`2 sqrt(2) i`

(C)

`4 sqrt(2) i`

(D)

None of these

Solution:

Let `sqrt(6 + 8i) = a + ib`

`6 + 8i = (a + ib)^2` [Squaring both sides]

`6 + 8i = a^2 + i^2b^2 + 2iab`

`6 + 8i = (a^2 − b^2) + 2iab`

Comparing the co-efficients of both sides

`a^2 − b^2 = 6` ...(1)

`2ab =8`

`a = 4/b`

Putting value of a in(1)

`(16)/b^2 − b^2 = 6`

`16 − b^4 = 6b^2`

`b^4 + 6b^2 − 16 = 0`

`b^4 + 8b^2 − 2b^2 − 16 = 0`

`b^2 (b^2 + 8) − 2 (b^2 + 8) = 8`

`b^2 − 2 = 0, b^2 + 8 = 0`

`b = √2`

we get `a = 4/sqrt(2) = 2sqrt(2)`

So, `sqrt(6 + 8i) = 2sqrt(2) + sqrt(2) i`

`sqrt(6−8i) = 2sqrt(2) − sqrt(2) i`

`∴ sqrt(6 + 8i) − sqrt(6 − 8i)`

`= (2 sqrt(2) + sqrt(2) i) − (2 sqrt(2) − sqrt(2) i)`

`= 2sqrt(2) i`.
Correct Answer is `=>` (B) `2 sqrt(2) i`
Q 2681201127

Number of solutions of the equation `z^3 = bar z i|z|` are `5`

(A) True
(B) False
Solution:

`z^3= bar z i |z| => | z^3| = | bar z| | i | |z| `

`=> |z| =1` or `|z|=0`

Thus `z = 0` is a solution

if `|z|=1, ` Let `z=e^(i theta)` then `e^ (3 i theta)=ie^(-i theta) `

`=> e^(4 i theta) =i`

`=> 4 theta =pi/2 ,( 5pi)/2 , (9 pi)/2 , (13 pi)/2`

`:. theta =pi/8 , (5 pi)/8 , (9 pi)/8 , (13 pi)/8` are solutions.

`:.` In all there are 5 solutions
Correct Answer is `=>` (A)
Q 2406445378

If `x ^2- 2x cos theta + 1 = 0`, then `x^(2n) -2x^n cos n theta+1` is equal to
UPSEE 2010
(A)

`cos 2 n theta`

(B)

`sin 2 n theta`

(C)

`0`

(D)

`R-{0}`

Solution:

From given relation, we have

`x = cos theta ± i sin theta`

take `x = cos theta + i sin theta`

`x^n = cos ntheta + i sin n theta`

and `1/ x^n =cos n theta- i sin n theta`

`:. x^n + (1// x^n) = 2cos n theta`

`:. 2x^(2n) cos ntheta + 1 = 0`
Correct Answer is `=>` (C) `0`
Q 2434480352

The real part of `(1 - i )^(-i)` is
UPSEE 2012
(A)

`e^(-pi/4) cos (1/2 log 2)`

(B)

`-e^(-pi/4) sin (1/2 log 2)`

(C)

`e^(pi/4) cos (1/2 log 2)`

(D)

`e^(-pi/4) sin (1/2 log 2)`

Solution:

Let `z= (1-i)^(-i)`

On taking log on both sides, we get

`=> log z = -i log (1-i)`

`= - i log sqrt(2) (cos pi/4 - i sin pi/4)`

`= -i log (sqrt (2) * e^((-i pi)/4))`

`= -i [1/2 log 2 + log e^ ((-i pi)/4)]`

`= -i [1/2 log 2 - (i pi)/4] = -i/2 log 2 - pi/4`

`=> z = e^(-pi/4) * e^(-i/2 log 2)`

On taking real part only,

`=> Re (z) = e^(-pi/4) * cos (1/2 log 2)`
Correct Answer is `=>` (A) `e^(-pi/4) cos (1/2 log 2)`
Q 2541556423

If `z_1 = 2 + 3i` and `z_2 = 3 + 4i` be two points on the complex plane. Then, the set of complex number `z` satisfying `| z- z_1 |^2 +| z- z_2|^2 = | z_1 - z_2 |^2` represents
WBJEE 2013
(A)

a straight line

(B)

a point

(C)

a circle

(D)

a pair of straight line

Solution:

Given `z_1 = 2+3i` and `z_2 = 3+4i`

Now we have


`|z-z_1|^2+|z-z_2|^2 = |z_1-z_2|^2` (let `z = x+iy`)


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


`=> | (x-2)+i(y-3)|^2+|(x-3)+i(y-4)|^2 = |-1-i^2|`


`=> (x-2)^2+(y-3)^2+(x-3)^2+(y-4)^2 = 1+1`


`=> x^2+4-4x+y^2+9-6y`


`=> 2x^2+2y^2-10x-14y+36 = 0`


`=> x^2+y^2-5x-7y+18 = 0`

which represent a drcle with centre `(5/2 , 7/2)` and radius `sqrt(25/4+49/4-18) = 1/sqrt2`
Correct Answer is `=>` (C) a circle
Q 2365867765

If `z` is complex number, then the

equation `| (2z-i)/(z+1) | =` a represents a circle when
BITSAT Mock
(A)

` a = 1` alone

(B)

`a = 2` alone

(C)

for all values of a except `2`

(D)

for no value of `a`

Solution:

`| z- i/2 | = a/z | z+1 |`

When `a = 2`, the equation is

`| z- i/2 | = | z+1 |`, which represents

a line which is the perpendicular of

the segment joining `i/2` and `-1`

When `a ne 2`, when we substitute

`z = x + iy`, the coefficient of `x^2`

`=` coefficient of `y^2= a^2/4` and `xy`

terms is absent.

`:.` the given equation represents a
circle when `a ne 2`.
Correct Answer is `=>` (C) for all values of a except `2`
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 2542434333

If `z= | (1, 1+2i , -5 i), (1-2i , -3, 5+3i), (5i , 5-3i , 7) |`

then `(i= sqrt (-1) )`
WBJEE 2011
(A)

`z` is purely real

(B)

`z` is purely imaginary

(C)

`z = bar (z) =0`

(D)

`(z- bar (z))i` is purely imaginary

Solution:

Given, `z= | (1, 1+2i , -5 i), (1-2i , -3, 5+3i), (5i , 5-3i , 7) |`


`=1 (-21- (25 + 9)) -(1 + 2i)`

`(7 -14i - (25i -15)) -- Si [(-1 -13i) + 15 i]`

`= -55- (1 + 2i) (22- 39i)- 5i ( -1 + 2i)`

`= -55 -100- 5i + 5i + 10`

`=-145`

Hence, `z` is purely real.
Correct Answer is `=>` (A) `z` is purely real
Q 2572491336

If `z_1 , z_2 , z_3 ` are imaginary numbers such that `| z_1 | = | z_2| = |z_3|` ` = | 1/z_1 + 1/z_2 + 1/z_3 | = 1` is
WBJEE 2016
(A)

equal to 1

(B)

less than 1

(C)

greater than 1

(D)

equal to 3

Solution:

We have,

`| z_1 | = | z_2 | = | z_3| = 1`

`=> | z_1 |^2 = | z_2 |^2 = | z_3 |^2 = 1`

`=> z_1 bar z_1 = z_2 bar z_2 = z_3 bar z_3 = 1`

` => 1/z_1 = bar z_1 , 1/z_2 = bar z_2 , 1/z_3 = bar z_3`

Now, ` | 1/z_1 + 1/z_2 + 1/z_3 | = 1`

` => | bar z_1 + bar z_2 + bar z_3 | = 1`

` => | bar (z_1 + z_2 + z_3 ) | = 1`

`:. | z_1 + z_2 + z_3 | = 1`
Correct Answer is `=>` (A) equal to 1
Q 2583023847

If `z` is a complex number, then which of the following statement is true?
UPSEE 2016
(A)

`(z bar(z))` is purely imaginary

(B)

`(z. bar(z))` is non-negative real

(C)

`(z- bar(z))` is purely real

(D)

`(z + bar (z) )` is purely imaginary

Solution:

Let `z = x + iy`, then `bar(z) = x - iy`


Now,


(i) `z bar(z) = x^2 + y^2` [purely real]


(ii) `z bar (z) = x^2 + y^2` [non-negative real]

(iii) `z- bar ( z) = 2i lm (z)` [purely imaginary]

(iv) `z + bar (z) = 2 Re (z)` [purely real)
Correct Answer is `=>` (B) `(z. bar(z))` is non-negative real
Q 1605512468

Let `omega = (-1)/2 + i (sqrt3)/2.` Then, the value of the determinant

` | (1, 1, 1) , (1, -1-omega^2, omega^2) , (1, omega^2, omega^4) |` is
BITSAT 2016
(A)

`3 omega`

(B)

`3 omega (omega - 1)`

(C)

`3 omega^2`

(D)

`3 omega (1 - omega)`

Solution:

`Delta = |(1, 1, 1) , (1, -1-omega^2, omega^2) , (1, omega^2, omega^4)|`

` => Delta = |(1, 0, 0) , (1, -2-omega^2, omega^2 -1) , (1, omega^2 - 1, omega^4 - 1)|`

[Applying `C_2 -> C_2 - C_1, C_3 -> C_3 - C_1`] `[ :. omega^3 = 1]`

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

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

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

` => Delta = - (2omega + omega) - (-3omega^2) = -3omega + 3 omega^2`

` = 3omega (omega -1)`
Correct Answer is `=>` (B) `3 omega (omega - 1)`
Q 2475891766

Common roots of the equations
`z^3 + 2z^2 + 2z + 1 = 0` and `z^(1985) + z^(100) +1 = 0`
are
UPSEE 2014
(A)

` omega , omega^2`

(B)

` omega , omega^3`

(C)

` omega^2 , omega^3`

(D)

None of these

Solution:

The given equation `z^3 + 2z^2 + 2z + 1 = 0` can be

rewritten as `(z + 1) (z^2 + z + 1) = 0`. Its roots are

`- 1, omega` and `omega ^2`.

Let `f(z) = z^(1985) + z^(100) + 1`

Putting `z - 1, omega ` and `omega ^2` respectively, we get

`f(-1) = (-1)^(1985) + (-1)^(100) + 1 != 0`

Therefore, `- 1` is not a root of the equation

`f(z) = 0`.

Again, `f(omega ) = omega ^(1985) + omega ^(100) + 1`

`= (omega ^3)^(661) omega ^2 + (omega ^3)^(33) omega + 1`

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

Therefore, `omega ` is a root of the equation `f(z) = 0`.

Similarly, `f(omega ^2) = 0`

Hence, `omega ` and `omega ^2` are the common roots.
Correct Answer is `=>` (A) ` omega , omega^2`
Q 2584645557

Let `z_1, z_2` and `z_3` be the affixes of the veltices of a triangle having the circumcentre at the origin. If `z`
is the affix of its orthocentre, then `z` is equal to
UPSEE 2008
(A)

`(z_1 + z_2 +z_3)/3`

(B)

`(z_1 +z_2 +z_3)/2`

(C)

`z_1 + z_2 +z_3`

(D)

None of these

Solution:

Centroid divides the line, joining
orthocentre and circumcentre in the ratio `2 : 1`.

Let `O, G` and `C` be the orthocentre, centroid and
circumcentre respectively, then

`:. (z_1 +z_2 +z_3)/3 = (2 xx 0 + 1(z) )/3`

`=> z= z_1 +z_2 +z_3`
Correct Answer is `=>` (C) `z_1 + z_2 +z_3`
 
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