Question

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

$$\begin{gathered} z=3\left(\cos {130}^{\circ }+i\sin {130}^{\circ }\right) \\ w=4\left(\cos {270}^{\circ }+i\sin {270}^{\circ }\right) \end{gathered}$$

Short answer

The polar form of $zw$is, role="math" localid="1646761395771" $12\left[\cos {40}^{\circ }+i\sin {40}^{\circ }\right]$.

The polar form of$\frac{z}{w}$is,$\frac{3}{4}\left[\cos {220}^{\circ }+i\sin {220}^{\circ }\right]$.

Step-by-step solution

Step 1 Given that,

$$\begin{gathered} z=3\left(\cos {130}^{\circ }+i\sin {130}^{\circ }\right) \\ w=4\left(\cos {270}^{\circ }+i\sin {270}^{\circ }\right) \end{gathered}$$

First find $zw$.

$$\begin{gathered} z.w=3\left(\cos {130}^{\circ }+i\sin {130}^{\circ }\right).4\left(\cos {270}^{\circ }+i\sin {270}^{\circ }\right) \\ =12\left[\cos \left({130}^{\circ }+{270}^{\circ }\right)+i\sin \left({130}^{\circ }+{270}^{\circ }\right)\right] \\ =12\left[\cos {400}^{\circ }+i\sin {400}^{\circ }\right] \end{gathered}$$

we know that ,$\cos \left({360}^{\circ }+\theta \right)=\cos \theta ,$so $\cos \left({360}^{\circ }+{40}^{\circ }\right)=\cos {40}^{\circ }$.

$z.w=12\left[\cos {40}^{\circ }+i\sin {40}^{\circ }\right]$.

Step 2 Next find zw.

$$\begin{gathered} \frac{z}{w}=\frac{3\left(\cos {130}^{\circ }+i\sin {130}^{\circ }\right)}{4\left(\cos {270}^{\circ }+i\sin {270}^{\circ }\right)} \\ =\frac{3}{4}[\cos \left({130}^{\circ }-{270}^{\circ }\right)+i\sin \left({130}^{\circ }-{270}^{\circ }\right)] \\ =\frac{3}{4}\left[\cos \left(-{140}^{\circ }\right)+i\sin \left(-{140}^{\circ }\right)\right] \end{gathered}$$

we know that,

$$\begin{gathered} \cos \left(-{140}^{\circ }\right)=\cos \left({360}^{\circ }-{220}^{\circ }\right) \\ =\cos {220}^{\circ } \end{gathered}$$

Thus, $\frac{z}{w}=\frac{3}{4}\left[\cos {220}^{\circ }+i\sin {220}^{\circ }\right]$.