Gujarat BoardEnglish MediumSTD 11 ScienceMATHSCOMPLEX NUMBERS AND QUADRATIC EQUATIONS4 Marks
Question
Convert the following in the polar form: $\frac{1+7\text{i}}{(2-\text{i})^2}$
✓
Answer
Here, $\text{z}=\frac{1+7\text{i}}{(2-\text{i})^2}$ $=\frac{1+7\text{i}}{(2-\text{i})^2}=\frac{1+7\text{i}}{4+\text{i}^2-4\text{i}}=\frac{1+7\text{i}}{4-1-4\text{i}}$ $=\frac{1+7\text{i}}{3-4\text{i}}\times\frac{3+4\text{i}}{3+4\text{i}}=\frac{3+4\text{i}+21\text{i}+28\text{i}^2}{3^2+4^2}$ $=\frac{3+4\text{i}+21\text{i}-28}{3^2+4^2}=\frac{-25+25\text{i}}{25}$ $=-1+\text{i}$ 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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