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