PROBLEMS ON OPERATIONS OF COMPLEX NUMBERS

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 :

Given, z is a non zero complex number.

2iz² = z̄

|2iz²| =  |z̄|

|2| |i| |z|² = |z|

(2) (1) |z|² = |z|

|z| = 1/2

So, option (1) is correct.

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 :

Given, |z – 2 + i| ≤ 2

||z₁| - |z₂|| ≤ |z₁ - z₂| ≤ 2

||z| - |2 - i|| ≤  |z – 2 + i| ≤ 2

|z| - |√5| ≤ 2

|z| - √5 ≤ 2

|z| ≤ 2 + √5

Greatest value of |z| is 2 + √5.

So, option (4) is correct.

Problem 3 :

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

(1)1  (2)  2  (3)  3  (4)  5

Solution : 

t = |-1| and t = 1

t = |3| and t = 3

So, the least value of |z| is 1.

Hence, option (1) is correct.

Problem 4 :

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

(1)  z   (2)  z̄   (3)  1/z  (4)  1

Solution :

Given, |z| = 1

z̄ = 1/z

Hence, option (1) is correct.

Problem 5 :

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

Solution :

Given, |z| - z = 1 + 2i

z = x + iy

|x + iy| - (x + iy) = 1 + 2i

√(x² + y²) - x - iy = 1 + 2i

Equating real and imaginary part.

√(x² + y²) - x = 1

-iy = 2

y = -2

√(x² + (-2)²) - x = 1

√(x² + 4) - x = 1

√(x² + 4)  = 1 + x

Square on both sides.

x² + 4 = (1 + x)²

x² + 4 = 1 + x² + 2x

x² + 4 - 1 - x² - 2x = 0

3 - 2x = 0

-2x = -3

x = 3/2

z = x + iy

z = 3/2 - 2i

Hence, option (1) is correct.