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Complex Numbers — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsComplex Numbers5 Marks
Question
Express the following complex numbers in the form $\text{r}(\cos\theta+\text{i}\sin\theta):$
$\frac{1-\text{i}}{\cos\frac{\pi}{3}+\text{i}\sin\frac{\pi}{3}}$
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Answer

Let $\text{z}=\frac{1-\text{i}}{\cos\frac{\pi}{3}+\text{i}\sin\frac{\pi}{3}}=\frac{1-\text{i}}{\frac{1}{2}+\text{i}\frac{\sqrt{3}}{2}}=\frac{2-2\text{i}}{1+\text{i}\sqrt{3}}=\frac{(2-2\text{i})(1-\text{i}\sqrt{3})}{(1+\text{i}\sqrt{3})(1-\text{i}\sqrt{3})}=\\\frac{(2-2\sqrt{3})-\text{i}(2\sqrt{3}+2)}{4}=\frac{(1-\sqrt{3})}{2}-\text{i}\frac{({\sqrt{3}+1)}}{2}$
$|\text{z}|=\sqrt{\frac{(1-\sqrt{3})^2}{4}+\frac{(1+\sqrt{3})^2}{4}}=\sqrt{\frac{8}{4}}=\sqrt{2}$
Let $\beta$ be acute angle by $\tan\beta=\frac{|\text{Im(z)}|}{|\text{Re(z)|}}.$
$\tan\beta=\frac{\Big|-\frac{(\sqrt{3}+1)}{2}\Big|}{\Big|\frac{1-\sqrt{3}}{2}\Big|}=\Big|\frac{-(\sqrt{3}+1)}{(1-\sqrt{3})}\Big|=\big|2+\sqrt{3}\big|=\tan\Big(\frac{7\pi}{12}\Big)$
$\Rightarrow\beta=\frac{7\pi}{12}$
Z is represented by a poiny in second quadrant.
So polar form of z is $\sqrt{2}\Big(\cos\frac{7\pi}{12}-\text{i}\sin\frac{7\pi}{12}\Big).$

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