Thank you for reporting, we will resolve it shortly
Q.
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
KEAMKEAM 2013Complex Numbers and Quadratic Equations
Solution:
Let $z=x+i y$
Then, we have $z+|z|=8+12 i$
$\Rightarrow \, (x+i y)+|x+i y|=8+12 i$
$\Rightarrow \, \left(x+\sqrt{x^{2}+y^{2}}\right)+i y=8+12 i$
On comparing the real and imaginary part, we get
and $ y =12$
and $ x+\sqrt{x^{2}+y^{2}} =8 $
$\Rightarrow \, \sqrt{x^{2}+144}=8-x$
On squaring both sides, we get
$x^{2}+144=64+x^{2}-16 \,x$
$\Rightarrow \, 16 x=-80$
$\Rightarrow \, x=-5$
$\therefore \, z=x+i y=-5+i \cdot 12$
Then, $|z|=\sqrt{25+144}=\sqrt{169}=13$
$\Rightarrow \, |z|^{2}=169 $
$ \Rightarrow \, \left|z^{2}\right|=169$
$\left(\because\left|z^{n}\right|=|z|^{n}\right)$