The mod-amplitude form of $\frac{1+7\mathrm{i}}{{\left(2-\mathrm{i}\right)}^2}$ is

The given complex number is $\frac{1+7\mathrm{i}}{{\left(2-\mathrm{i}\right)}^2}$

Simplify the complex number $\frac{1+7\mathrm{i}}{{\left(2-\mathrm{i}\right)}^2}$ as 

    $\begin{array}{rcl}\frac{1+7\mathrm{i}}{4-1-4i}& =& \frac{1+7i}{3-4i}\\ & =& \frac{\left(1+7i\right)\left(3+4i\right)}{25}\\ & =& \frac{1}{25}\left(3+4i+21i-28\right)\\ & =& \frac{-\left(25\right)\left(1-i\right)}{25}\\ & =& -1+i\end{array}$

$\left|\frac{1+7\mathrm{i}}{{\left(2-\mathrm{i}\right)}^2}\right|=\left|-1+i\right|=\sqrt{2}$

If $\theta$ is amplitude of a complex number $a+ib$ then $\theta ={\tan }^{-1}\left(\frac{b}{a}\right)$, if both $a,b$ are positive. 

The complex number $-1+i$ lies in  the second quadrant in argand plane

Hence, amplitude is $\pi -{\tan }^{-1}1=\frac{3\pi }{4}$

Therefore, the polar form of  $\frac{1+7\mathrm{i}}{{\left(2-\mathrm{i}\right)}^2}$is $\boxed{\sqrt{\mathbf{2}}\mathbf{c}\mathbf{i}\mathbf{s}\left(\frac{3\pi }{4}\right)}$