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NDA Math Complex Numbers

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For. the next two (2) items that follow :

Let

z_1,z_2

and

z_3

be non zero complex numbers satisfying

z^2=\overset{-}{iz},

where

i=\sqrt{-1}

[NDA I 2016]

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Question : 40 of 66

Marks: +1, -0

What is

z_1 + z_2 + z_3

$i$
$-i$
$0$
$1$

Solution:

Given

z^2 = \overset{-}{iz}

Let us suppose that

z = x + iy

\Rightarrow (x+iy)^2 = i(x-iy)

\Rightarrow x^2 - y^2 + 2xyi = ix + y

Comparing real and imaginary part of both sides

x^2 - y^2 = y

and

2xy = x

Taking

2xy = x

(2y - 1)x = 0

\Rightarrow x = 0

and

y = \frac{1}{2}

if

x = 0

\therefore y + y^2 = 0

(\because x^2 - y^2 = y)

y(y+1) = 0

y = 0

and

-1

x = 0

y = 0

and

-1

if

y = \frac{1}{2}

x^2 - \frac{1}{4} = \frac{1}{2}

\Rightarrow x^2 = \frac{3}{4}

\Rightarrow x = \pm \frac{\sqrt{3}}{2}

Since given numbers are non zero complex numbers.

So,

z_1 = 0 + (-1)i = -i

z_2 = \frac{\sqrt{3}}{2} + \frac{1}{2} i

z_3 = \frac{-\sqrt{3}}{2} + \frac{1}{2} i

z_1 + z_2 + z_3 = (-i) + \left( \frac{\sqrt{3}}{2} + \frac{1}{2} i \right) + \left( -\frac{\sqrt{3}}{2} + \frac{1}{2} i \right) = 0

Hence

z_1 + z_2 + z_3 = 0

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