Question

If $z$ is a complex number such that $|z|=1$, and maximum value of $|z^4+z^3-2z^{2}i+z+1|$ is $\alpha$, then ${\alpha }^2$ is -

Answer 1

$|z|=1$
$u=|z^2||z^2+\frac{1}{z^2}+z+\frac{1}{z}-2i|$
$=|z^2+{\overline{z}}^2+(z+\overline{z})-2i|$
$z=x+iy$
$u=|(2x{)}^2-2+2x-2i|$
$u=2|2x^2+x-1-i|$
$u^2=4\left(\right(2x^2+x-1{)}^2+1)$
$|z|=1\Rightarrow x^2+y^2=1$ So, $-1\le x\le 1$
$t=2x^2+x-1=2({(x+\frac{1}{4})}^2-\frac{9}{16})$

So, $-\frac{9}{8}\le t\le 2$
So, $u_{\text{max}}^2=20={\alpha }^2$

${\alpha }^2=20$