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Complex Numbers — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSComplex Numbers5 Marks
Question
Express the following complex numbers in the form $\text{r}(\cos\theta+\text{i}\sin\theta):$
$1-\sin\alpha+\text{i}\cos\alpha$
✓

Answer

Let $\text{z}=(1-\sin\alpha)+\text{i}\cos\alpha$
Since sine and cosine are periodic functions with periodic with peroid $2\pi.$
So, let us take $\alpha$ lying in the interval $[0,2\pi]$
Now, $\text{z}=(1-\sin\alpha)+\text{i}\cos\alpha$
$\Rightarrow|\text{z}|=\sqrt{(1-\sin\alpha)^2+\cos^2\alpha}\\=\sqrt{2-2\sin\alpha}=\sqrt{2}\sqrt{1-\sin\alpha}$
$\Rightarrow|\text{z}|=\sqrt{2}\sqrt{\Big(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big)^2}=\sqrt{2}\Big|\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big|$
Let $\beta$ be acute angle given by $\tan\beta=\frac{|\text{Im(z)}|}{|\text{Re(z)|}}.$
$\tan\beta=\frac{|\cos\alpha|}{|1-\sin\alpha|}=\frac{|\cos\alpha|}{|1-\sin\alpha|}=\Bigg|\frac{\cos^2\frac{\alpha}{2}-\sin^2\frac{\alpha}{2}}{\Big(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big)}\Bigg|=\Bigg|\frac{\cos\frac{\alpha}{2}+\sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}}\Bigg|$
$\Rightarrow\tan\beta=\Bigg|\frac{1+\tan\frac{\alpha}{2}}{1-\tan\frac{\alpha}{2}}\Bigg|=\Big|\tan\big(\frac{\pi}{4}+\frac{\alpha}{2}\big)\Big|$
Following cases arise:
Case I: when $0\leq\alpha<\frac{\pi}{2}$
$\cos\frac{\alpha}{2}>\sin\frac{\alpha}{2}$ and $\frac{\pi}{4}+\frac{\alpha}{2}\in\Big[\frac{\pi}{4},\frac{\pi}{2}\Big)$
$\therefore\text{arg(z)}=\frac{\pi}{2}+\frac{\alpha}{2}$
So polar form of z is 
$\sqrt{2}\Big(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big)\Big(\cos\Big(\frac{\pi}{4}+\frac{\alpha}{2}\Big)+\text{i}\sin\Big(\frac{\pi}{4}+\frac{\alpha}{2}\Big)\Big)$
Case II: when $\frac{\pi}{2}<\alpha<\frac{3\pi}{2}$
$\cos\frac{\alpha}{2}<\sin\frac{\alpha}{2}$ and $\frac{\pi}{4}+\frac{\alpha}{2}\in\Big(\frac{\pi}{2},\pi\Big)$
$\therefore|\text{z}|=\sqrt{2}\Big|\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big|=\sqrt{2}\Big(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big)$
and, $\tan\beta=\Bigg|\tan\Big(\frac{\pi}{4}+\frac{\alpha}{2}\Big)\Bigg|=-\tan\Big(\frac{\pi}{4}+\frac{\alpha}{2}\Big)\\=\tan\Big\{\pi\Big(\frac{\pi}{4}+\frac{\alpha}{2}\Big)\Big\}=\tan\Big(\frac{3\pi}{4}-\frac{\alpha}{2}\Big)$
$\Rightarrow\beta=\frac{3\pi}{4}-\frac{\alpha}{2}$
Since $1-\sin\alpha>0$ and $\cos\alpha<0.$
Clearly, z lies in the fourth quadrant.
$\therefore\text{arg(z)}=-\beta=\frac{\alpha}{2}-\frac{3\pi}{4}$
So polar form of z is $\sqrt{2}\Big(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big)\Big(\cos\Big(\frac{\alpha}{2}-\frac{3\pi}{4}\Big)+\text{i}\sin\Big(\frac{\alpha}{2}-\frac{3\pi}{4}\Big)\Big)$
Case III: when $\frac{3\pi}{2}<\alpha<2\pi$
$\cos\frac{\alpha}{2}<\sin\frac{\alpha}{2}$ and $\frac{\pi}{4}+\frac{\alpha}{2}\in\Big(\pi,\frac{5\pi}{4}\Big)$
$\therefore|\text{z}|=\sqrt{2}\Big|\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big|=-\sqrt{2}\Big(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big)$
and, $\tan\beta=\Bigg|\tan\Big(\frac{\pi}{4}+\frac{\alpha}{2}\Big)\Bigg|=\tan\Big(\frac{\pi}{4}+\frac{\alpha}{2}\Big)\\=-\tan\Big\{\pi-\Big(\frac{\pi}{4}+\frac{\alpha}{2}\Big)\Big\}=\tan\Big(\frac{\alpha}{2}-\frac{3\pi}{4}\Big)$
$\Rightarrow\beta=\frac{\alpha}{2}-\frac{3\pi}{4}$
Clearly, $\text{Re(z)}<0$ and $\text{Im(z)}>0.$
$\therefore\text{arg(z)}=\beta=\frac{\alpha}{2}-\frac{3\pi}{4}$
So polar form of z is $-\sqrt{2}\Big(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big)\Big(\cos\Big(\frac{\alpha}{2}-\frac{3\pi}{4}\Big)+\text{i}\sin\Big(\frac{\alpha}{2}-\frac{3\pi}{4}\Big)\Big).$

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