Express the following complex numbers in the form r( +i ): 1+i

Question

Express the following complex numbers in the form $\text{r}(\cos\theta+\text{i}\sin\theta):$
$1+\text{i}\tan\alpha$

✓

Answer

Let $\text{z}=1+\text{i}\tan\alpha$$\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 - 1: when $\alpha\in\Big[0,\frac{\pi}{2}\Big)$
$|\text{z}|=\sqrt{1+\tan^2\alpha}=\sqrt{\sec^2\alpha}=|\sec\alpha|=\sec\alpha$
Let $\beta$ be acute angle given by $\tan\beta=\frac{|\text{Im(z)}|}{|\text{Re(z)|}}.$
$\tan\beta=|\tan\alpha|=\tan\alpha$
$\Rightarrow\beta=\alpha$
As z represented by a point in first quadrant.
$\therefore \ \text{arg(z)}=\beta=\alpha$
So polar form of z is $\sec\alpha\big(\cos\alpha+\text{i}\sin\alpha\big)$
Case - 2: when $\alpha\in\Big(\frac{\pi}{2},\pi\Big]$
$|\text{z}|=\sqrt{1+\tan^2\alpha}=\sqrt{\sec^2\alpha}=|\sec\alpha|=-\sec\alpha$
Let $\beta$ be acute angle given by $\tan\beta=\frac{|\text{Im(z)}|}{|\text{Re(z)|}}.$
$\tan\beta=|\tan\alpha|=-\tan\alpha=\tan(\pi-\alpha)$
$\Rightarrow\beta=\pi-\alpha$
As z represented by a point in fourth quadarnt.
$\therefore \ \text{arg(z)}=-\beta=\alpha-\pi$
So polar form of z is $-\sec\alpha\big(\cos(\alpha-\pi)+\text{i}\sin(\alpha-\pi)\big).$

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Similar questions

Match the statements of Column $A$ and Column $B.$

	Column $A$		Column $B$
$a.$	The polar form of $\text{i}+\sqrt{3}$ is	$i.$	Perpendicular bisector of segment joining $(– 2, 0)$ and $(2, 0).$
$b.$	The amplitude of $-1+\sqrt{-3}$ is	$ii.$	On or outside the circle having centre at $(0, – 4)$ and radius $3.$
$c.$	If $\|z + 2\| = \|z - 2\|,$ then locus of $z$ is	$iii.$	$\frac{2\pi}{3}$
$d.$	If $\|z + 2i\| = \|z - 2i\|,$ then locus of $z$ is	$iv.$	Perpendicular bisector of segment joining $(0, – 2)$ and $(0, 2).$
$e.$	Region represented by $\|\text{z}+4\text{i}\|\geq3$ is	$v.$	$2\Big(\cos\frac{\pi}{6}+\text{i}\sin\frac{\pi}{6}\Big)$
$f.$	Region represented by $\|\text{z}+4\|\leq3$ is	$vi.$	On or inside the circle having centre $(– 4, 0)$ and radius $3$ units.
$g.$	Conjugate of $\frac{1+2\text{i}}{1-\text{i}}$ lies in	$vii.$	First quadrant.
$h.$	Reciprocal of $1 - i$ lies in	$viii.$	Third quadrant.