Question

In Problems $33-40$, find $zw$and $\frac{z}{w}$.Leave your answers in polar form.

$$\begin{gathered} z=4\left(\cos \frac{3\mathrm{\pi }}{8}+i\sin \frac{3\mathrm{\pi }}{8}\right) \\ w=2\left(\cos \frac{9\mathrm{\pi }}{16}+i\sin \frac{9\mathrm{\pi }}{16}\right) \end{gathered}$$

Short answer

The polar form of $zw$is, $8\left(\cos \frac{15\mathrm{\pi }}{16}+i\sin \frac{15\mathrm{\pi }}{16}\right)$.

The polar form of$\frac{z}{w}$is,$2\left(\cos \frac{3\mathrm{\pi }}{16}-i\sin \frac{3\mathrm{\pi }}{16}\right)$.

Step-by-step solution

Step 1 Given that,

$$\begin{gathered} z=4\left(\cos \frac{3\mathrm{\pi }}{8}+i\sin \frac{3\mathrm{\pi }}{8}\right) \\ w=2\left(\cos \frac{9\mathrm{\pi }}{16}+i\sin \frac{9\mathrm{\pi }}{16}\right) \end{gathered}$$

First find $zw$.

$$\begin{gathered} z.w=4\left(\cos \frac{3\mathrm{\pi }}{8}+i\sin \frac{3\mathrm{\pi }}{8}\right).2\left(\cos \frac{9\mathrm{\pi }}{16}+i\sin \frac{9\mathrm{\pi }}{16}\right) \\ =8\left(\cos \left(\frac{3\mathrm{\pi }}{8}+\frac{9\mathrm{\pi }}{16}\right)+i\sin \left(\frac{3\mathrm{\pi }}{8}+\frac{9\mathrm{\pi }}{16}\right)\right) \\ =8\left(\cos \left(\frac{6\mathrm{\pi }+9\mathrm{\pi }}{16}\right)+i\sin \left(\frac{6\mathrm{\pi }+9\mathrm{\pi }}{16}\right)\right) \\ =8\left(\cos \frac{15\mathrm{\pi }}{16}+i\sin \frac{15\mathrm{\pi }}{16}\right) \end{gathered}$$

Step 2 Next find zw.

$$\begin{gathered} \frac{z}{w}=\frac{4\left(\cos \frac{3\mathrm{\pi }}{8}+i\sin \frac{3\mathrm{\pi }}{8}\right)}{2\left(\cos \frac{9\mathrm{\pi }}{16}+i\sin \frac{9\mathrm{\pi }}{16}\right)} \\ =2\left[\cos \left(\frac{3\mathrm{\pi }}{8}-\frac{9\mathrm{\pi }}{16}\right)+i\sin \left(\frac{3\mathrm{\pi }}{8}-\frac{9\mathrm{\pi }}{16}\right)\right] \\ =2\left[\cos \left(\frac{6\mathrm{\pi }-9\mathrm{\pi }}{16}\right)+i\sin \left(\frac{6\mathrm{\pi }-9\mathrm{\pi }}{16}\right)\right] \\ =2\left[\cos \left(\frac{-3\mathrm{\pi }}{16}\right)+i\sin \left(\frac{-3\mathrm{\pi }}{16}\right)\right] \end{gathered}$$

We know that,

$$\begin{gathered} \cos \left(-\theta \right)=\cos \theta \\ \sin \left(-\theta \right)=-\sin \theta \end{gathered}$$

Thus,

$\frac{z}{w}=2\left[\cos \left(\frac{3\mathrm{\pi }}{16}\right)-i\sin \left(\frac{3\mathrm{\pi }}{16}\right)\right]$