Question

Findzwand $\frac{z}{w}$. Leave the answers in polar form.

role="math" localid="1646661399572" $$\begin{gathered} z=2+2i \\ w=\sqrt{3}-i \end{gathered}$$

Short answer

zwin polar form is $4\sqrt{2}\left(\cos 15°+i\sin 15°\right)$.

$\frac{z}{w}$in polar form is $\sqrt{2}\left(\cos 75°+i\sin 75°\right)$.

Step-by-step solution

Step 1. Given information.

Consider the given question,

$$\begin{gathered} z=2+2i \\ w=\sqrt{3}-i \end{gathered}$$

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.

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

Step 2. Find the polar form of w.

Determine the polar form of z,

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

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

Determine the polar form of w,

$$\begin{gathered} r_w=\sqrt{{\left(\sqrt{3}\right)}^2+{\left(-1\right)}^2} \\ {r}_w=2 \\ {\theta }_w={\tan }^{-1}\left(\frac{-1}{\sqrt{3}}\right) \\ {\theta }_w=-30° \end{gathered}$$

Therefore,$w=2\left(\cos \left(-30°\right)+i\sin \left(-30°\right)\right)$

Step 3. Find zw and zw.

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

$$\begin{gathered} z·w=2\sqrt{2}\left[\cos \left(45°+i\sin 45°\right)\right]·2\left[\cos \left(-30°\right)+i\sin \left(-30°\right)\right] \\ z·w=\left(2\right)2\sqrt{2}\left[\cos \left(45°+\left(-30°\right)\right)+i\sin \left(45°+\left(-30°\right)\right)\right] \\ z·w=4\sqrt{2}\left(\cos 15°+i\sin 15°\right) \end{gathered}$$

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

$$\begin{gathered} \frac{z}{w}=\frac{2\sqrt{2}\left(\cos 45°+i\sin 45°\right)}{2\left(\cos \left(-30°\right)+i\sin \left(-30°\right)\right)} \\ \frac{z}{w}=\frac{2\sqrt{2}}{2}\left[\cos \left(45°-\left(-30°\right)\right)+i\sin \left(45°-\left(-30°\right)\right)\right] \\ \frac{z}{w}=\sqrt{2}\left(\cos 75°+i\sin 75°\right) \end{gathered}$$