Gujarat BoardEnglish MediumSTD 11 ScienceMATHSCOMPLEX NUMBERS AND QUADRATIC EQUATIONS4 Marks
Question
Convert the following in the polar form: $\frac{1+3\text{i}}{1-2\text{i}}$
✓
Answer
Here, $\text{z}=\frac{1+3\text{i}}{1-2\text{i}}$$=\frac{1+3\text{i}}{1-2\text{i}}\times\frac{1+2\text{i}}{1+2\text{i}}$
$=\frac{1+2\text{i}+3\text{i}-6}{1+4}$
$=\frac{-5+5\text{i}}{5}=-1+\text{i}$
Let $\text{r}\cos\theta=-1$ and $\text{r}\sin\theta=1$
On squaring and adding, we obtain
$\text{r}^2(\cos^2\theta+\sin^2\theta)=1+1$
$\Rightarrow\ \text{r}^2(\cos^2\theta+\sin^2\theta)=2$
$\Rightarrow\ \text{r}^2=2$ $[\cos^2\theta+\sin^2\theta=1]$
$\Rightarrow\ \text{r}=\sqrt{2}$ [Conventionally, r > 0]
$\therefore\ \sqrt{2}\cos\theta=-1$ and $\sqrt{2}\sin\theta=1$
$\Rightarrow\ \cos\theta=\frac{-1}{\sqrt{2}}$ and $\sin\theta=\frac{1}{\sqrt{2}}$
$\therefore\ \theta=\pi-\frac{\pi}{4}=\frac{3\pi}{4}$ [As $\theta$ lies in II quadrant]
$\therefore\ \text{z}=\text{r}\cos\theta+\text{ir}\sin\theta$
$=\sqrt{2}\cos\frac{3\pi}{4}+\text{i}\sqrt{2}\sin\frac{3\pi}{4}$
$=\sqrt{2}\Big(\cos\frac{3\pi}{4}+\text{i}\sin\frac{3\pi}{4}\Big)$
This is the required polar form.
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