Find zw and $\frac{z}{w}$. Leave the answers in polar form.

$$\begin{gathered} z=1-i \\ w=1-\sqrt{3}i \end{gathered}$$

Consider the given question,

$$\begin{gathered} z=1-i \\ w=1-\sqrt{3}i \end{gathered}$$

The multiplication of the polar form of complex numbers is given by,

$$\begin{gathered} z_1{z}_2=r_1{r}_2\left[\cos \left({\theta }_1+{\theta }_2\right)+i\sin \left({\theta }_1+{\theta }_2\right)\right]......\left(i\right) \\ \frac{z_1}{z_2}=\frac{r_1}{r_2}\left[\cos \left({\theta }_1-{\theta }_2\right)+i\sin \left({\theta }_1-{\theta }_2\right)\right]......\left(ii\right) \end{gathered}$$

We need to write the given numbers in polar form.

$$\begin{gathered} r=\sqrt{x^2+y^2} \\ \theta ={\tan }^{-1}\left(\frac{y}{x}\right) \end{gathered}$$

Determine the polar form of z,

$$\begin{gathered} r_z=\sqrt{{\left(1\right)}^2+{\left(-1\right)}^2} \\ {r}_z=\sqrt{2} \\ {\theta }_z={\tan }^{-1}\left(\frac{-1}{1}\right) \\ {\theta }_z=-45° \end{gathered}$$

Therefore, $z=\sqrt{2}\left[\cos \left(-45°\right)+i\sin \left(-45°\right)\right]$.

Determine the polar form of w,

$$\begin{gathered} r_w=\sqrt{1^2+{\left(-\sqrt{3}\right)}^2} \\ {r}_w=2 \\ {\theta }_w={\tan }^{-1}\left(\frac{-\sqrt{3}}{1}\right) \\ {\theta }_w=-60° \end{gathered}$$

Therefore,$w=2\left[\cos \left(-60°\right)+i\sin \left(-60°\right)\right]$

Substitute the values of z,w in equation (i),

$$\begin{gathered} z·w=\sqrt{2}\left[\cos \left(-45°\right)+i\sin \left(-45°\right)\right]·2\left[\cos \left(-60°\right)+i\sin \left(-60°\right)\right] \\ z·w=\left(2\right)\sqrt{2}\left[\cos \left(-45°+\left(-60°\right)\right)\right]+i\sin \left(-45°+\left(-60°\right)\right) \\ z·w=2\sqrt{2}\left[\cos \left(-105°\right)+i\sin \left(-105°\right)\right] \end{gathered}$$

Substitute the values of $\frac{z}{w}$in equation (ii),

localid="1646664280313" $$\begin{gathered} \frac{z}{w}=\frac{\sqrt{2}\left[\cos \left(-45°\right)+i\sin \left(-45°\right)\right]}{2\left[\cos \left(-60\right)°+i\sin \left(-60\right)°\right]} \\ \frac{z}{w}=\frac{1}{\sqrt{2}}\left[\cos \left(-45°+60°\right)+i\sin \left(-45°+60°\right)\right] \\ \frac{z}{w}=\frac{1}{\sqrt{2}}\left(\cos 15°+i\sin 15°\right) \end{gathered}$$