Practice Question On Complex Number for NDA , Airforce and IITJEE page no. 3

COMPLEX NUMBER

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Question no. 31

If z ≠1 and z²/(z-1) is real , then the point represented by complex number z lies

looks_one Either on the real axis or on circle not passing through the origin

looks_two On circle with center at the origin

looks_3 Either on real circle or on a circle passing through the origin

looks_4 On the imaginary axis

Option looks_one Either on the real axis or on circle not passing through the origin

Solution :

Question no. 32

If ω (≠1) is cube root of unity and (1+ω)⁷= A+ Bω. then (A,B) equals

looks_one (1,0)

looks_two (0,1)

looks_3 (-1,1)

looks_4 (1,1)

Option looks_4 (1,1)

Solution :
(1+ω)⁷= A+ Bω
(-ω²)⁷ = A + Bω
-ω¹⁴ = A + Bω
-(ω³)⁴ .ω² = A + Bω
- ω² = A+ Bω
1 + ω = A + Bω
A = 1 , B = 1

Question no. 33

Let α , β be real and z be complex number, if z² +αz+β=0 has two distinct roots on the line Re(z)=1, then it is neccessary that

looks_one |β|=1

looks_two β∈(1,∞)

looks_3 β∈(0,1)

looks_4 β∈(-1,0)

Option looks_two β∈(1,∞)

Solution :
Re(z)=1
roots of equation = 1 + ia , 1 -ia
Product of roots = β = (1+ia)(1 -ia) = 1 + a² > 0

Question no. 34

The number of complex number z such that |z-1|=|z+1|=|z-i| equals

looks_one 1

looks_two 2

looks_3 ∞

looks_4 0

Option looks_one 1

Solution :
|z-1|=|z+1|
|x + iy - 1| = |x + iy + 1|
( x - 1 )² + y² = ( x + 1 )² + y²
x² + 1 -2x + y² = x² + 1 + 2x + y²
x = 0
|z-1| =|z-i|
(x-1)² + y² = x² + (y -1)²
x² + 1 -2x + y² = x² + y² + 1 - 2y
x = y
y = 0
z = 0 + i0

Question no. 35

If | Z - 4/Z|=2, then the maximum value of |Z| is equal to

looks_one √3+1

looks_two √5+1

looks_3 2

looks_4 √2 + 2

Option looks_two √5+1

Solution :
| Z - 4/Z|=2
|z| ≤ | Z - 4/Z| + |4/Z|
|z| ≤ 2 + |4/Z|
|z|² - 2|z| - 4 ≤ 0
For the maximum value of z
|z|² - 2|z| - 4 = 0
|z| = 1 + √5 , 1- √5

Question no. 36

The conjugate of complex number is 1/(i-1), then that complex number is

looks_one 1/(i+1)

looks_two -1/(i+1)

looks_3 1/(i-1)

looks_4 -1/(i-1)

Option looks_two -1/(i+1)

Solution :
change the sign of imaginary term :
= 1/(-i -1)
=-1 /(i+1)

Question no. 37

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

looks_one 4

looks_two 10

looks_3 6

looks_4 0

Option looks_3 6

Solution:
|z₁ + z₂| ≤ |z₁| + |z₂|
|z + 4 -3| ≤ |z+4| + |-3|
|z + 1| ≤ 3 + 3
|z + 1| ≤ 6

Question no. 38

If z and w are two non zero complex numbers such that |z|=|w| and arg(z) + arg(w)= π then z equals

looks_one w¯

looks_two -w¯

looks_3 w

looks_4 -w

Option looks_two -w¯

Solution :
|z|=|w| = r
arg(z) + arg(w)= π
z = re^iθ
= re^i(π-φ)
= re^iπe^-iφ
= -re^-iφ
= -w

Question no. 39

If |z-4| < |z-2|, its solution is given by

looks_one Re(z)>0

looks_two Re(z)<0

looks_3 Re(z)>3

looks_4 Re(z)>2

Option looks_3 Re(z)>3

Solution :
|z-4| < |z-2|
z =x+iy
squaring both sides (does not affect the sign of inequality )
(x-4)² + y² < (x-2)² + y²
x² + 16 -8x + y² < x² + 4 - 4x + y²
12 < 4x
3 < x

Question no. 40

The locus of the center of a circle which touches the circle |z-z₁| = a and |z - z₂|=b externally (z, z₁, and z₂ are complex numbers) will be

looks_one ellipse

looks_two hyperbola

looks_3 circle

looks_4 none of these

Option looks_two hyperbola

Solution :
|z_o - z₁| = r + a
|z_o - z₂| = r + b
|z_o - z₁| - |z_o - z₂| = a - b

Question no. 41

Let z₁ and z₂ be two roots of the equation z² + az + b = 0, z being complex. Further, assume that the origin, z₁ and z₂ form an equilateral triangle, then

looks_one a²=b

looks_two a²=2b

looks_3 a²=3b

looks_4 a²=4b

Option looks_3 a²=3b

Solution :
For equilateral triangle
(z₁)² + (z₂)² + (z₃)² - z₁.z₂ - z₂.z₃ - z₂.z₃ = 0
one point is in origin so z₃ = 0
(z₁)² + (z₂)² - z₁.z₂ = 0
(z₁ + z₂)² -3z₁z₂ = 0
z₁ + z₂ = sum of zeros = -a
z₁z₂ = b= product of zeros
(z₁ + z₂)² -3z₁z₂ = a² - 3b =0

Question no. 42

If z and w are two non-zero complex number such that |zw|=1, and Arg (z)-Arg(w)=π/2 , then z¯w is equal to

looks_one 1

looks_two -1

looks_3 -i

looks_4 i

Option looks_3 -i

Solution :
|zw|=1, and Arg (z)-Arg(w)=π/2
arg(z) = π/2 + arg(w)
z = iw
|zw| = 1
|iw.w|= |w²| = 1
|w|= 1
z¯w = -iw .w = -i|w²| = -i

Question no. 43

Let z, w be complex numbers such that z + iw=0 and arg zw =π . Then arg z equals

looks_one π/4

looks_two 5π/4

looks_3 3π/4

looks_4 π/2

Option looks_3 3π/4

Solution :
z + iw=0
arg zw =π
z = -iw
z= iw
arg(z) = arg(w) + π/2
arg(z) +arg(w) = π
arg(w) = π - arg(z)
arg(z) = π - arg(z) + π/2
arg(z) = 3π/4

Question no. 44

If |z²-1|=|z|²+1, then z lies on

looks_one the real axis

looks_two a circle

looks_3 an ellipse

looks_4 the imaginary axis.

Option looks_4 the imaginary axis.

Solution :
|z²-1|=|z|²+1
Squaring both the sides
|z²-1|²=(|z|²+1)²
|z²-1|² = (z² - 1)(z² -1) = z².z² + 1 - z² - z² (1)
(|z|²+1)² = |z|⁴ + 2 |z|² + 1 (2)
Equate (1) and (2)
z².z² + 1 - z² - z² = |z|⁴ + 2 |z|² + 1
z² + z² + 2z.z = 0
z + z =0
if z = x+iy
z + z = x+iy +x-iy = 0
x = 0

Question no. 45

If the cube roots of unity are 1,ω,ω² then the roots of the equation (x – 1)³ + 8 = 0, are

looks_one -1,-1+2ω , -1 -2ω²

looks_two -1,1-2ω , 1 -2ω²

looks_3 -1,-1,-1

looks_4 -1,1+2ω , 1 +2ω²

Option looks_two -1,1-2ω , 1 -2ω²

Solution:
1,ω,ω² are the roots of unity
(x – 1)³ + 8 = 0
let x -1 = Y
Y³ + 8 = 0
Y = -2 , -2ω , -2ω²
x- 1 = -2 , -2ω , -2ω²
x = -1 , 1-2ω , 1 -2ω²

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