$\frac{1+7i}{(2-i{)}^2}$=$\frac{(1+7i)(3+4i)}{(3-4i)(3+4i)}$=$\frac{-25+25i}{25}$= -1+i

Let z=x+iy = -1+i

$\therefore$ rcos$\theta$ =-1 And rsin$\theta$ =1 $\therefore$ $\theta$=$\frac{3\pi }{4}$ and r=$\sqrt{2}$

Thus $\frac{1+7i}{(2-i{)}^2}$=$\sqrt{2}$ $[cos\frac{3\pi }{4}+isin\frac{3\pi }{4}]$

Alter: $\left|\frac{1+7i}{(2-i{)}^2}\right|$=$\left|\frac{1+7i}{3-4i}\right|$=$\sqrt{2}$

And arg $\left(\frac{1+7i}{(2-i{)}^2}\right)$=${\tan }^{-1}7-{\tan }^{-1}\left(\frac{-4}{3}\right)$

                                                                    =${\tan }^{-1}7+{\tan }^{-1}\left(\frac{4}{3}\right)$=$\frac{3\pi }{4}$

$\therefore$ $\frac{1+7i}{(2-i{)}^2}$=$\sqrt{2}$ $[cos\frac{3\pi }{4}+isin\frac{3\pi }{4}]$

z = 3+5i , $\overline{z}$ = 3-5i

⇒ $z^3=(3+5i{)}^3=3^3+(5i{)}^3+\mathrm{3.3.5}i(3+5i)$

              = -198+10i

Hence, $z^3+\overline{z}$ + 198 = 10i-198+3-5i+198 =  3+5i . 

${(\frac{i-1}{i+1}\times \frac{i-1}{i-1})}^n={\left(\frac{-2i}{-2}\right)}^n=i^n$  

Hence, to make the real number the least positive integar is 2.

We have a $=\sqrt{2i}=\sqrt{2}{(cos\frac{\pi }{2}+isin\frac{\pi }{2})}^{\frac{1}{2}}=\sqrt{2}{\left(e^{i^{\frac{\pi }{2}}}\right)}^{\frac{1}{2}}=\sqrt{2}{e}^{i^{\frac{\pi }{4}}}=\sqrt{2}(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}i)=1+i$
Trick:Check with options.

$z=x+iyiz=-y+ix$

$Given,|z|=Re\left(iz\right)+1\Rightarrow \sqrt{x^2+y^2}=-y+1\Rightarrow x^2+y^2=y^2-2y+1\Rightarrow x^2=1-2y$