Question

Convert the complex number $\frac{1+7i}{{\left(2-i\right)}^2}$ into polar form

Answer 1

Step 1: Simplify the given complex number into $x+iy$ form

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

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

$=\frac{1+7i}{4+i^2-4i}$

$=\frac{1+7i}{4-1-4i}$

$=\frac{1+7i}{3-4i}\times \frac{3+4i}{3+4i}$

$=\frac{3+4i+21i-28}{9+16}$

$=\frac{-25+25i}{25}$

$\Rightarrow$ $z=-1+i$

Step 2: Solve for the required polar form

We know that the general polar form is $z=r\cos \left(\theta \right)+ir\sin \left(\theta \right)$

Let $r\cos \left(\theta \right)=-1$ and $r\sin \left(\theta \right)=1$

On squaring and adding we get

$\Rightarrow$ ${\left(r\cos \left(\theta \right)\right)}^2+{\left(r\sin \left(\theta \right)\right)}^2=1+1$

$\Rightarrow$$r^2\left(\left({\cos }^2\left(\theta \right)\right)+\left({\sin }^2\left(\theta \right)\right)\right)=2$

$\Rightarrow$ $r^2=2$

$\Rightarrow$ $r=\sqrt{2}$

$\Rightarrow$ $\cos \left(\theta \right)=\frac{-1}{\sqrt{2}}$ and $\sin \left(\theta \right)=\frac{1}{\sqrt{2}}$

As sine is positive and cosine is negative $\theta$ belongs to Quadrant $II$.

$\Rightarrow$$\theta =\mathrm{\pi }-\frac{\mathrm{\pi }}{4}=\frac{3\mathrm{\pi }}{4}$

Thus as $r=\sqrt{2}$ and $\theta =\frac{3\mathrm{\pi }}{4}$

$z=r\cos \left(\theta \right)+ir\sin \left(\theta \right)$

$\Rightarrow$$z=\sqrt{2}\cos \left(\frac{3\mathrm{\pi }}{4}\right)+i\sqrt{2}\sin \left(\frac{3\mathrm{\pi }}{4}\right)$

Hence, $\sqrt{2}\cos \left(\frac{3\mathrm{\pi }}{4}\right)+i\sqrt{2}\sin \left(\frac{3\mathrm{\pi }}{4}\right)$ is the polar form of the complex number $\frac{1+7i}{{\left(2-i\right)}^2}$