Question

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

$$\begin{gathered} z=2\left(\cos {80}^{\circ }+i\sin {80}^{\circ }\right) \\ w=6\left(\cos {200}^{\circ }+i\sin {200}^{\circ }\right) \end{gathered}$$

Short answer

The polar form of $zw$is, localid="1646762477334" $zw=12\left[\cos {280}^{\circ }+i\sin {280}^{\circ }\right]$.

The polar form of$\frac{z}{w}$is,$\frac{z}{w}=\frac{1}{3}\left[\cos {120}^{\circ }-i\sin {120}^{\circ }\right]$

Step-by-step solution

Step 1 Given that,

$$\begin{gathered} z=2\left(\cos {80}^{\circ }+i\sin {80}^{\circ }\right) \\ w=6\left(\cos {200}^{\circ }+i\sin {200}^{\circ }\right) \end{gathered}$$

First find $zw$.

$$\begin{gathered} zw=2\left(\cos {80}^{\circ }+i\sin {80}^{\circ }\right).6\left(\cos {200}^{\circ }+i\sin {200}^{\circ }\right) \\ =12\left[\cos \left({80}^{\circ }+{200}^{\circ }\right)+i\sin \left({80}^{\circ }+{200}^{\circ }\right)\right] \\ =12\left[\cos {280}^{\circ }+i\sin {280}^{\circ }\right] \end{gathered}$$

Step 2 Next find zw.

$$\begin{gathered} \frac{z}{w}=\frac{2\left(\cos {80}^{\circ }+i\sin {80}^{\circ }\right)}{6\left(\cos {200}^{\circ }+i\sin {200}^{\circ }\right)} \\ =\frac{1}{3}\left[\cos \left({80}^{\circ }-{200}^{\circ }\right)+i\sin \left({80}^{\circ }-{200}^{\circ }\right)\right] \\ =\frac{1}{3}\left[\cos \left(-{120}^{\circ }\right)+i\sin \left(-{120}^{\circ }\right)\right] \end{gathered}$$

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

Thus,$\frac{z}{w}=\frac{1}{3}\left[\cos {120}^{\circ }-i\sin {120}^{\circ }\right]$