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KEAM 2013 Math Paper

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Question : 9 of 120

Marks: +1, -0

If z is a complex number such that

z+\left|z\right|=8+12i,

then the value of

\left|z^{2}\right|

is equal to

228
144
121
169
189

Solution:

Let

z=x+iy

Then, we have

z+\left|z\right|=8+12i

\Rightarrow \;\; (x+iy)+\left|x+iy\right|=8+12i

\Rightarrow \;\; (x+\sqrt{x^{2}+y^{2}})+iy=8+12i

On comparing the real and imaginary part, we get

y=12

and

x+\sqrt{x^{2}+y^{2}}=8

\Rightarrow \;\; \sqrt{x^{2}+144}=8-x

On squaring both sides, we get

x^{2}+144=64+x^{2}-16x

\Rightarrow \;\; 16x=-80

\Rightarrow \;\; x=-5

\therefore \;\; z=x+iy=-5+i\cdot 12

Then,

\left|z\right|=\sqrt{25+144}=\sqrt{169}=13

\Rightarrow \;\; \left|z\right|^{2}=169

\Rightarrow \;\; \left|z^{2}\right|=169

(\because \left|z^{n}\right|=\left|z\right|^{n})

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