Answer
The division of the complex numbers in the polar form is $2\left( \cos 0{}^\circ +i\sin 0{}^\circ \right)$.
Work Step by Step
Consider the given complex number to write in the polar form,
$\begin{align}
& {{z}_{1}}=2-2i \\
& {{z}_{2}}=1-i \\
\end{align}$ (I)
For a complex number $z=x+iy$, the polar form is given by,
$z=r\left( \cos \theta +i\sin \theta \right)$ (II)
Here, $x=r\cos \theta \ \text{ and }\ y=r\sin \theta $, dividing the value of y by x, to get
$\tan \theta =\frac{y}{x}$ (III)
Also, the value of r is called the moduli of the complex number, given by,
For ${{z}_{1}}=2-2i$
$\begin{align}
& r=\sqrt{{{x}^{2}}+{{y}^{2}}} \\
& r=\sqrt{{{\left( 2 \right)}^{2}}+{{\left( -2 \right)}^{2}}} \\
& r=\sqrt{4+4} \\
& r=2\sqrt{2} \\
\end{align}$
$r=2\sqrt{2}$ (IV)
For any complex number $z=x+iy$, the sign of the value of x and y determine in which quadrant the value of $z=x+iy$ would lie,
If the value of x is positive and the value of y is positive, then the angle $\theta $ lies in the first quadrant having the value of $\theta $ as $\theta $ (V)
Also, if the value of x is negative and the value of y is positive, then the angle $\theta $ lies in the second quadrant having the value of $\theta $ as $\pi -\theta $ (VI)
Also, if the value of x is negative and the value of y is negative, then the angle $\theta $ lies in the third quadrant having the value of $\theta $ as $\pi +\theta $ (VII)
Also, if the value of x is positive and the value of y is negative, then the angle $\theta $ lies in the fourth quadrant having the value of $\theta $ as $2\pi -\theta $ (VIII)
For the given complex number, using (I), (III) to get,
$\begin{align}
& \tan \theta =\frac{y}{x} \\
& \tan \theta =\frac{-2}{2} \\
& \tan \theta =-1 \\
\end{align}$
For the given complex number, Using (V),
$\theta =\frac{7\pi }{4}$ (IX)
Using (II), (IV), and (IX), to get,
${{z}_{1}}=2\sqrt{2}\left( \cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4} \right)$
The polar form of the complex number is ${{z}_{1}}=2\sqrt{2}\left( \cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4} \right)$
For ${{z}_{2}}=1-i$
$\begin{align}
& r=\sqrt{{{x}^{2}}+{{y}^{2}}} \\
& r=\sqrt{{{\left( 1 \right)}^{2}}+{{\left( -1 \right)}^{2}}} \\
& r=\sqrt{1+1} \\
& r=\sqrt{2} \\
\end{align}$
$r=\sqrt{2}$ (X)
For the given complex number, using (I), (III) to get,
$\begin{align}
& \tan \theta =\frac{y}{x} \\
& \tan \theta =\frac{-1}{1} \\
& \tan \theta =-1 \\
\end{align}$
For the given complex number, Using (VIII),
$\theta =\frac{7\pi }{4}$ (XI)
Using (II), (IV), and (IX), to get,
${{z}_{2}}=\sqrt{2}\left( \cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4} \right)$
The polar form of the complex number is ${{z}_{2}}=\sqrt{2}\left( \cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4} \right)$
Consider any complex number, given by,
$\begin{align}
& {{z}_{1}}={{r}_{1}}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right) \\
& {{z}_{2}}={{r}_{2}}\left( \cos {{\theta }_{2}}+i\sin {{\theta }_{2}} \right) \\
\end{align}$
For a complex number in polar form, the division is calculated as,
$\frac{{{z}_{1}}}{{{z}_{2}}}=\frac{{{r}_{1}}}{{{r}_{2}}}\left( \cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)+i\sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right) \right)$ (XII)
The polar form after the division of the complex numbers,
$\begin{align}
& {{z}_{1}}=2\sqrt{2}\left( \cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4} \right) \\
& {{z}_{2}}=\sqrt{2}\left( \cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4} \right) \\
\end{align}$ (XIII)
Divide it using (XII) and (XIII), to get,
$\begin{align}
& \frac{{{z}_{1}}}{{{z}_{2}}}=\frac{{{r}_{1}}}{{{r}_{2}}}\left( \cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)+i\sin \left( {{\theta }_{1}}-{{\theta }_{2}} \right) \right) \\
& \frac{{{z}_{1}}}{{{z}_{2}}}=\frac{2\sqrt{2}}{\sqrt{2}}\left( \cos \left( \frac{7\pi }{4}-\frac{7\pi }{4} \right)+i\sin \left( \frac{7\pi }{4}-\frac{7\pi }{4} \right) \right) \\
& =2\left( \cos 0{}^\circ +i\sin 0{}^\circ \right)
\end{align}$
The division of the complex numbers in the polar form is $2\left( \cos 0{}^\circ +i\sin 0{}^\circ \right)$
