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UPSC NDA I 2016 Mathematics Solved Paper

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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 : 48 of 120

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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