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Complex Numbers — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsComplex Numbers4 Marks
Question
Express the following complex numbers in the form $\text{r}(\cos\theta+\text{i}\sin\theta):$
$\tan\alpha-\text{i}$
✓

Answer

Let $\text{z}=\tan\alpha-\text{i}$$\tan\alpha$ is periodic function with period $\pi$
So, let us take $\alpha$ lying in the interval $\Big[0,\frac{\pi}{2}\Big)\cup\Big(\frac{\pi}{2},\pi\Big].$
Case I: when $\alpha\in\Big[0,\frac{\pi}{2}\Big)$
$|\text{z}|=\sqrt{\tan^2\alpha+1}=\sqrt{\sec^2\alpha}=|\sec\alpha|=\sec\alpha$
Let $\beta$ be acute angle given by $\tan\beta=\frac{|\text{Im(z)}|}{|\text{Re(z)|}}.$
$\tan\beta=\frac{1}{|\tan\alpha|}=|\cot\alpha|=\cot\alpha=\tan\Big(\frac{\pi}{2}-\alpha\Big)$
$\Rightarrow\beta={\frac{\pi}{2}}-\alpha$
As z represented by a point in first quadrant.
$\therefore \ \text{arg(z)}=\beta=\alpha-\frac{\pi}{2}.$
So polar form of z is $\sec\alpha\Big(\cos\Big(\alpha-\frac{\pi}{2}\Big)+\text{i}\sin\Big(\alpha-\frac{\pi}{2}\Big)\Big)$
Case II: when $\alpha\in\Big(\frac{\pi}{2},\pi\Big]$
$|\text{z}|=\sqrt{1+\tan^2\alpha+1}=\sqrt{\sec^2\alpha}=|\sec\alpha|=-\sec\alpha$
Let $\beta$ be acute angle given by $\tan\beta=\frac{|\text{Im(z)}|}{|\text{Re(z)|}}.$
$\tan\beta=|\tan\alpha|=-\tan\alpha=\tan(\pi-\alpha)$
$\Rightarrow\beta=\alpha-\frac{\pi}{2}$
As z represented by a point in fourth quadrant.
$\therefore \ \text{arg(z)}=\pi+\beta=\frac{\pi}{2}+\alpha.$
So polar form of z is $-\sec\alpha\Big(\cos\Big(\frac{\pi}{2}+\alpha\Big)+\text{i}\sin\Big(\frac{\pi}{2}+\alpha\Big)\Big).$

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