# COMPLEX NUMBERS WORKSHEET

Choose the correct or the most suitable answer from the given four alternatives :

Problem 1 :

in + in + 1 + in+ 2 + in + 3 is

(1)  0   (2)  1  (3)  -1  (4)  i

Solution

Problem 2 :

(1)  1 + i    (2)  i   (3)  1   (4)   0

Solution

Problem 3 :

The area of the triangle formed by the complex numbers z, iz, and z + iz in the Argand’s diagram is

(1)  1/2 |z|2   (2)  |z|2    (3)  3/2  |z|2   (4)  2|z|2

Solution

Problem 4 :

The conjugate of a complex numbers is 1/(I - 2). Then, the complex number is

 (1) 1i + 2 (3) -1i - 2 (2) -1i + 2 (4) 1i - 2

Solution

Problem 5 :

(1)   0   (2)  1  (3)  2  (4)  3

Solution

1)  0, option (1)

2)  i + 1, option (1)

3)   1/2 × |z|2, option (1)

4)  z = -1/(i + 2), option (2)

5)  2, option 3.

Problem 1 :

If z is a non zero complex number, such that 2iz2 = z̄ then |z| is

(1)   1/2    (2)   1   (3)  2  (4)  3

Solution

Problem 2 :

If |z – 2 + i| ≤ 2, then the greatest value of |z| is

(1)√3 - 2  (2)  √3 + 2  (3)  √5 - 2  (4) √5 + 2

Solution

Problem 3 :

If |z – 3/z| = 2, then the least value of |z| is

(1)1  (2)  2  (3)  3  (4)  5

Solution

Problem 4 :

If |z| = 1, then the value of (1 + z)/(1 + z̄) is

(1)  z   (2)  z̄   (3)  1/z  (4)  1

Solution

Problem 5 :

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

 (1) 32 - 2i (2) -32 + 2i (3) 2 -32i (4) 2 + 32i

Solution

1)  |z| = 1/2, option (1)

2)  2 + √5, So, option (4)

3)  The last value is 1, option (1)

4)  z, option (1)

5) z = 3/2 - 2i, option (1)

Problem 1 :

If |z1| = 1, |z2| = 2, |z3| = 3 and |9z1z2 + 4z1z3 + z2z3| = 12, then the value of |z1 + z2 + z3| is

(1)   1   (2)   2   (3)   3   (4)   4

Solution

Problem 2 :

If z is a complex number such that z ϵ ℂ \ ℝ and z + 1/z ϵ ℝ, then |z| is

(1)   0   (2)   1   (3)   2   (4)   3

Solution

Problem 3 :

z1, z2 and z3 are complex numbers such that z1 + z2 + z3 = 0 and |z1| = |z2| = |z3| = 1 then z12 + z22 + z32 is

(1)  3   (2)   2   (3)   1   (4)   0

Solution

Problem 4 :

If (z - 1)/(z + 1) is purely imaginary, then |z| is

(1)   1/2    (2)   1   (3)   2   (4)   3

Solution

Problem 5 :

If z = x + iy is a complex number such that |z + 2| = |z - 2|, then the locus of z is

(1)   real axis   (2)   imaginary axis   (3)   ellipse   (4)   circle

Solution

1) |z1 + z2 + z3| = 2, option (2)

2)  |z| = 1, option (2)

3)  z12 + z22 + z32 = 0, option (4)

4)  |z| = 1, option (2)

5)  x = 0, option (2)

Problem 1 :

The principal argument of 3/(-1 + i) is

 (1) -5𝜋6 (2) -2𝜋3 (3) -3𝜋4 (4) -𝜋2

Solution

Problem 2 :

The principal argument of (sin 40º + i cos 40º)5 is

(1)  -110º    (2)   -70º   (3)   70º   (4)   110º

Solution

Problem 3 :

If (1 + i) (1 + 2i) (1 + 3i) … (1 + ni) = x + iy, then 2 ⋅ 5 ⋅ 10 … (1 + n2) is

(1)1   (2)   i   (3)   x2 + y2    (4)   1 + n2

Solution

Problem 4 :

If ω ≠ 1 is a cubic root of unity (1 + ω)7 = A + Bω, then (A, B) equals

(1)   (1, 0)   (2)   (-1, 1)  (3)   (0,1)   (4)   (1, 1)

Solution

Problem 5 :

 (1) 2𝜋3 (2) 𝜋6 (3) 5𝜋6 (4) 𝜋2

Solution

1)  -3𝜋/4, option (3)

2)  θ = -110º, option (3)

3)  2 · 5 · 10 .... 1 + n2 = x2 + y2, option (3)

4)  (A, B) = (1, 1), option (4)

5)  𝜃 = 𝜋/2, option (4)

Problem 1 :

If α and β are the roots of x2 + x + 1 = 0, then α2020 + β2020 is

(1)   -2   (2)  -1    (3)    1   (4)   2

Solution

Problem 2 :

(1)   -2   (2)   -1    (3)   1    (4)  2

Solution

Problem 3 :

(1)   1    (2)  -1   (3)  √3i    (4)   -√3i

Solution

Problem 4 :

 (1) cis 2𝜋3 (2) cis 4𝜋3 (3) -cis 2𝜋3 (4) -cis 4𝜋3

Solution

Problem 5 :

Solution

1)  -1, option (2)

2)  1, option (2)

3)  -√3 i, option (4)

4)  cis 2𝜋/3, option (1)

5)  z = 0, option (1)

## Recent Articles

1. ### Finding Range of Values Inequality Problems

May 21, 24 08:51 PM

Finding Range of Values Inequality Problems

2. ### Solving Two Step Inequality Word Problems

May 21, 24 08:51 AM

Solving Two Step Inequality Word Problems