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