Express the following complex in the form r(cos θ + i sin θ):
Question:

Express the following complex in the form r(cos θ + i sin θ):

(i) $1+i \tan a$

(ii) $\tan \alpha-i$

(iii) $1-\sin \alpha+i \cos \alpha$

(iv) $\frac{1-i}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}$

Solution:

(i) Let $\mathrm{z}=1+i \tan \alpha$

$\because \tan \alpha$ is periodic with period $\pi .$ So, let us take

$\alpha \in[0, \pi / 2) \cup(\pi / 2, \pi]$

Case I :

When $\alpha \in[0, \pi / 2)$

$z=1+i \tan \alpha$

$\Rightarrow|\mathrm{z}|=\sqrt{1+\tan ^{2} \alpha}$

$=|\sec \alpha|$   $\left[\because 0<\alpha<\frac{\pi}{2}\right]$

$=\sec \alpha$

Let $\beta$ be an acute angle given by $\tan \beta=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$

$\tan \beta=|\tan \alpha|$

$=\tan \alpha$

$\Rightarrow \beta=\alpha$

As $z$ lies in the first quadrant. Therefore, $\arg (z)=\beta=\alpha$Thus, $z$ in the polar form is given by

Thus, $z$ in the polar form is given by

$z=\sec \alpha(\cos \alpha+i \sin \alpha)$

Case II :

$z=1+i \tan \alpha$

$\Rightarrow|z|=\sqrt{1+\tan ^{2} \alpha}$

$=|\sec \alpha|$    $\left[\because \frac{\pi}{2}<\alpha<\pi\right]$

$=-\sec \alpha$

Let $\beta$ be an acute angle given by $\tan \beta=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$

$\tan \beta=|\tan \alpha|$

$=-\tan \alpha$

$\Rightarrow \tan \beta=\tan (\pi-\alpha)$

$\Rightarrow \beta=\pi-\alpha$

As, $z$ lies in the fourth quadrant.

$\therefore \arg (z)=-\beta=\alpha-\pi$

Thus, $z$ in the polar form is given by

$z=-\sec \alpha\{\cos (\alpha-\pi)+i \sin (\alpha-\pi)\}$

(ii) Let $\mathrm{z}=\tan \alpha-i$

$\because \tan \alpha$ is periodic with period $\pi$. So, let us take

$\alpha \in[0, \pi / 2) \cup(\pi / 2, \pi]$

Case I :

$z=\tan \alpha-\mathrm{i}$

$\Rightarrow|z|=\sqrt{\tan ^{2}+1}$

$=|\sec \alpha|$

$=\sec \alpha$

$\left[\because 0<\alpha<\frac{\pi}{2}\right]$

Let $\beta$ be an acute angle given by $\tan \beta=\left|\frac{\operatorname{Im}(\mathrm{z})}{\operatorname{Re}(\mathrm{z})}\right|$

$\tan \beta=\frac{1}{|\tan \alpha|}$

$=|\cot \alpha|$

$=\cot \alpha$

$=\tan \left(\frac{\pi}{2}-\alpha\right)$

$\Rightarrow \beta=\frac{\pi}{2}-\alpha$

We can see that $\operatorname{Re}(z)>0$ and $\operatorname{Im}(z)<0$. So, $z$ lies in the fourth quadrant.

$\therefore \arg (z)=-\beta=\alpha-\frac{\pi}{2}$

Thus, $z$ in the polar form is given by

$z=\sec \alpha\left\{\cos \left(\alpha-\frac{\pi}{2}\right)+i \sin \left(\alpha-\frac{\pi}{2}\right)\right\}$

Case II :

$z=\tan \alpha-i$

$\Rightarrow|z|=\sqrt{\tan ^{2}+1}$

$=|\sec \alpha|$ $\left[\because \frac{\pi}{2}<\alpha<\pi\right]$

$=-\sec \alpha$

Let $\beta$ be an acute angle given by $\tan \beta=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$

$\tan \beta=\frac{1}{|\tan \alpha|}$

$=|\cot \alpha|$

$=-\cot \alpha$

$=\tan \left(\alpha-\frac{\pi}{2}\right)$

$\Rightarrow \beta=\alpha-\frac{\pi}{2}$

We can see that $\operatorname{Re}(z)<0$ and $\operatorname{Im}(z)<0$. So, $z$ lies in the third quadrant.

$\therefore \arg (z)=\pi+\beta=\frac{\pi}{2}+\alpha$

Thus, $z$ in the polar form is given by

$z=-\sec \alpha\left\{\cos \left(\frac{\pi}{2}+\alpha\right)+i \sin \left(\frac{\pi}{2}+\alpha\right)\right\}$

(iii) Let $\mathrm{z}=(1-\sin \alpha)+i \cos \alpha$.

$\because$ sine and cosine functions are periodic functions with period $2 \pi$.

So, let us take $\alpha \in[0,2 \pi]$

Now, $z=1-\sin \alpha+i \cos \alpha$

$\Rightarrow|z|=\sqrt{(1-\sin \alpha)^{2}+\cos ^{2} \alpha}=\sqrt{2-\sin \alpha}=\sqrt{2} \sqrt{1-\sin \alpha}$

$\Rightarrow|z|=\sqrt{2} \sqrt{\left(\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}\right)^{2}}=\sqrt{2}\left|\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}\right|$

Let $\beta$ be an acute angle given by $\tan \beta=\frac{|\operatorname{Im}(z)|}{|\operatorname{Re}(z)|}$. Then,

$\tan \beta=\frac{|\cos \alpha|}{|1-\sin \alpha|}=\left|\frac{\cos ^{2} \frac{\alpha}{2}-\sin ^{2} \frac{\alpha}{2}}{\left(\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}\right)^{2}}\right|=\left|\frac{\cos \frac{\alpha}{2}+\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}}\right|$

$\Rightarrow \tan \beta=\left|\frac{1+\tan \frac{\alpha}{2}}{1-\tan \frac{\alpha}{2}}\right|=\left|\tan \left(\frac{\pi}{4}+\frac{\alpha}{2}\right)\right|$

Case I: When $0 \leq \alpha<\frac{\pi}{2}$

In this case, we have,

$\cos \frac{\alpha}{2}>\sin \frac{\alpha}{2}$ and $\frac{\pi}{4}+\frac{\alpha}{2} \in\left[\frac{\pi}{4}, \frac{\pi}{2}\right)$

$\Rightarrow|z|=\sqrt{2}\left(\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}\right)$

and $\tan \beta=\left|\tan \left(\frac{\pi}{4}+\frac{\alpha}{2}\right)\right|=\tan \left(\frac{\pi}{4}+\frac{\alpha}{2}\right)$

$\Rightarrow \beta=\frac{\pi}{4}+\frac{\alpha}{2}$

Clearly, $z$ lies in the first quadrant. Therefore, $\arg (z)=\frac{\pi}{4}+\frac{\alpha}{2}$

Hence, the polar form of $z$ is

$\sqrt{2}\left(\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}\right)\left\{\cos \left(\frac{\pi}{4}+\frac{\alpha}{2}\right)+i \sin \left(\frac{\pi}{4}+\frac{\alpha}{2}\right)\right\}$

Case II : When $\frac{\pi}{2}<\alpha<\frac{3 \pi}{2}$

In this case, we have,

$\cos \frac{\alpha}{2}<\sin \frac{\alpha}{2}$ and $\frac{\pi}{4}+\frac{\alpha}{2} \in\left(\frac{\pi}{2}, \pi\right)$

$\Rightarrow|z|=\sqrt{2}\left(\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}\right)=-\sqrt{2}\left(\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}\right)$

and $\tan \beta=\left|\tan \left(\frac{\pi}{4}+\frac{\alpha}{2}\right)\right|=-\tan \left(\frac{\pi}{4}+\frac{\alpha}{2}\right)=\tan \left\{\pi-\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)\right\}=\tan \left(\frac{3 \pi}{4}-\frac{\alpha}{2}\right)$

$\Rightarrow \beta=\frac{3 \pi}{4}-\frac{\alpha}{2}$

Clearly, $z$ lies in the fourth quadrant. Therefore, $\arg (z)=-\beta=-\left(\frac{3 \pi}{4}-\frac{\alpha}{2}\right)=\frac{\alpha}{2}-\frac{3 \pi}{4}$

Hence, the polar form of $z$ is

$-\sqrt{2}\left(\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}\right)\left\{\cos \left(\frac{\alpha}{2}-\frac{3 \pi}{4}\right)+i \sin \left(\frac{\alpha}{2}-\frac{3 \pi}{4}\right)\right\}$

Case III : When $\frac{3 \pi}{2}<\alpha<2 \pi$

In this case, we have,

$\cos \frac{\alpha}{2}<\sin \frac{\alpha}{2}$ and $\frac{\pi}{4}+\frac{\alpha}{2} \in\left(\pi, \frac{5 \pi}{4}\right)$

$\Rightarrow|z|=\sqrt{2}\left|\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}\right|=-\sqrt{2}\left(\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}\right)$

and $\tan \beta=\left|\tan \left(\frac{\pi}{4}+\frac{\alpha}{2}\right)\right|=\tan \left(\frac{\pi}{4}+\frac{\alpha}{2}\right)=-\tan \left\{\pi-\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)\right\}=\tan \left(\frac{\alpha}{2}-\frac{3 \pi}{4}\right)$

$\Rightarrow \beta=\frac{\alpha}{2}-\frac{3 \pi}{4}$

Clearly, $z$ lies in the first quadrant. Therefore, $\arg (z)=\beta=\frac{\alpha}{2}-\frac{3 \pi}{4}$

Hence, the polar form of $z$ is

$-\sqrt{2}\left(\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2}\right)\left\{\cos \left(\frac{\alpha}{2}-\frac{3 \pi}{4}\right)+i \sin \left(\frac{\alpha}{2}-\frac{3 \pi}{4}\right)\right\}$

$($ iv $)$ Let $z=\frac{1-i}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}$

$=\frac{1-i}{\frac{1}{2}+i \frac{\sqrt{3}}{3}}$

$=\frac{2-2 i}{1+i \sqrt{3}} \times \frac{1-i \sqrt{3}}{1-i \sqrt{3}}$

$=\frac{2-2 i-2 \sqrt{3} i+2 \sqrt{3} i^{2}}{1+3}$

$=\frac{2-2 \sqrt{3}-2 i(1+\sqrt{3})}{4}$

$=\frac{(1-\sqrt{3})+i(-1-\sqrt{3})}{2}$

$=\frac{(1-\sqrt{3})}{2}+i \frac{(-1-\sqrt{3})}{2}$

Now, $z=\frac{(1-\sqrt{3})}{2}+i \frac{(-1-\sqrt{3})}{2}$

$\Rightarrow|z|=\sqrt{\left(\frac{1-\sqrt{3}}{2}\right)^{2}+\left(\frac{-1-\sqrt{3}}{2}\right)^{2}}$

$=\sqrt{\left(\frac{1+3-2 \sqrt{3}}{4}\right)+\left(\frac{1+3+2 \sqrt{3}}{4}\right)}$

$=\sqrt{\frac{8}{4}}$

$=\sqrt{2}$

Let $\beta$ be an acute angle given by $\tan \beta=\frac{|\operatorname{Im}(z)|}{|\operatorname{Re}(z)|}$. Then,

$\tan \beta=\frac{\left|\frac{1+\sqrt{3}}{2}\right|}{\left|\frac{1-\sqrt{3}}{2}\right|}=\left|\frac{1+\sqrt{3}}{1-\sqrt{3}}\right|=\left|\frac{\tan \frac{\pi}{4}+\tan \frac{\pi}{3}}{1-\tan \frac{\pi}{4} \tan \frac{\pi}{3}}\right|$

$\Rightarrow \tan \beta=\left|\tan \left(\frac{\pi}{4}+\frac{\pi}{3}\right)\right|=\left|\tan \frac{7 \pi}{12}\right|$

$\Rightarrow \beta=\frac{7 \pi}{12}$

Clearly, $z$ lies in the fourth quadrant. Therefore, $\arg (z)=-\frac{7 \pi}{12}$

Hence, the polar form of $z$ is

$\sqrt{2}\left(\cos \frac{7 \pi}{12}-\sin \frac{7 \pi}{12}\right)$

 

Administrator

Leave a comment

Please enter comment.
Please enter your name.