Question

For every real number c$\ge 0$, find all the complex number z which satisfy the equation

$|z{|}^2$ - 2iz + 2c(1 + i) = 0

• A. c + i( -1) Where 0 - 1
• B. c + i( -1 Where 0
• C. c + i( -2) Where 0
• D. c + i( -2) Where c > -1

Answer 1

The correct option is B

c + i( -1 Where 0

Given equation $|z{|}^2$ - 2iz + 2c(1 + i) = 0

Let z = x + iy

Modulus of complex number

$|z{|}^2$ = $x^2$ + $y^2$

Substitute the value of z in the given equation

$x^2$ + $y^2$ - 2i(x + iy) + 2c(1 + i) = 0

($x^2$ + $y^2$ + 2c + 2y) + i (-2x + 2c) = 0

Comparing the real and imaginary part on both side

$x^2$ + $y^2$ + 2c + 2y = 0 ---------------(1)

-2x + 2c = 0 ------------(2)

x = c

Substitute x =c in equation (1)

$x^2$ + $y^2$ + 2y + 2c = 0

$y^2$ + 2y + $c^2$ + 2c = 0

y = $\frac{-2\underset{–}{+}\sqrt{4-4(c^2+2c)}}{2}$ = -1$\underset{–}{+}\sqrt{1-c^2-2c)}$

Since x and y are real

Part inside the square should be greater than equal to zero.

1 - $c^2$ - 2c $\ge 0$

$c^2$ + 2c + 1 $\le 2$

$(c+1{)}^2$ $\le 2$

-$\sqrt{2}$ -1 $\le c\le \sqrt{2}$ - 1

$\because$ Given c $\ge 0$

So, c lies 0 $\le c$ $\le \sqrt{2}$ - 1

z = x + iy = c + i( -1$\pm +\sqrt{1-c^2-2c}$) For 0 $\le c$ $\le \sqrt{2}$ - 1