Convert the complex numbers given in Exercises in the polar form:… - Vidyadip

Question

Convert the complex numbers given in Exercises in the polar form:-1 - i.

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Answer

-1 - iLet $\text{r}\cos\theta=-1$ and $\text{r}\sin\theta=-1$
On squaring and adding, we obtain
$\text{r}^2\cos^2\theta+\text{r}^2\sin^2\theta=(-1)^2+(-1)^2$
$\Rightarrow\ \text{r}^2(\cos^2\theta+\sin^2\theta)=1+1$
$\Rightarrow\ \text{r}^2=2$
$\Rightarrow\ \text{r}=\sqrt{2}$ [Conventionally, r > 0]
$\therefore\ \sqrt{2}\cos\theta=-1$ and $\sqrt{2}\sin\theta=-1$
$\Rightarrow\ \cos\theta=-\frac{1}{\sqrt{2}}$ and $\sin\theta=-\frac{1}{\sqrt{2}}$
$\therefore\ \theta=-\Big(\pi-\frac{\pi}{4}\Big)=-\frac{3\pi}{4}$ [As $\theta$ lies in the III quadrant]
$\therefore\ -1-\text{i}=\text{r}\cos\theta+\text{i r}\sin\theta$
$=\sqrt{2}\cos\frac{-3\pi}{4}+\text{i}\sqrt{2}\sin\frac{-3\pi}{4}$
$=\sqrt{2}\Big(\cos\frac{-3\pi}{4}+\text{i}\sin\frac{-3\pi}{4}\Big)$
This is the required polar form.

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