Exercise 2.7-Additional Problems - Chapter 2 - Complex Numbers - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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AdditionalProblems
Question 1.
Write the following complex numbers in the polar form:
(i) $-3 \sqrt{2}+3 \sqrt{2} i$
(ii) $1+i$
(iii) $-1-i$
(iv) $1-i$
Solution:
(i) Let $z=-3 \sqrt{2}+3 \sqrt{2} i$. Then $r=|z|=\sqrt{(-3 \sqrt{2})^2+(3 \sqrt{2})^2}=6$
Let $\alpha$ be the acute angle given by $\tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$. Then, $\tan \alpha=1 \Rightarrow \alpha=\frac{\pi}{4}$
The point representing $z$ lies in the second quadrant.
So, the argument $\theta$ of $z$ is given by $\theta=\pi-\alpha=\pi-\left(\frac{\pi}{4}\right)=3 \frac{\pi}{4}$.
Hence, the polar form of $z=-3 \sqrt{2}+3 \sqrt{2} i$ is
$
z=r(\cos \theta+\sin \theta)=6\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)
$

(ii) Let. $\quad z=1+i$. Then, $r=|z|=\sqrt{1^2+1^2}=\sqrt{2}$.
Let $\alpha$ be the acute angle given by $\tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$. Then, $\tan \alpha=\left|\frac{1}{1}\right|=1 \Rightarrow \alpha=\frac{\pi}{4}$
We find that the point $(1,1)$ representing $z$ lies in first quadrant. Therefore, the argument of $z$ is given by $\theta=\alpha=\frac{\pi}{4}$.
Hence, the polar form of $z=1+i$ is
$
z=r(\cos \theta+i \sin \theta)=\sqrt{2}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)
$

Let $z=-1-i$. Then, $r=|z|=\sqrt{(-1)^2+(-1)^2}=\sqrt{2}$.
Let $\alpha$ be the acute angle given by $\tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$. Then, $\tan \alpha=\left|\frac{-1}{-1}\right|=1 \Rightarrow \alpha=\frac{\pi}{4}$
Clearly, the point $(-1,-1)$ representing $z$ lies in third quadrant. Therefore, the argument of $z$ is given by $\theta=-(\pi-\alpha)=-\left(\pi-\frac{\pi}{4}\right)=-\frac{3 \pi}{4}$.
Hence, the polar form of $z=-1-i$ is
$
z=r(\cos \theta+i \sin \theta)=\sqrt{2}\left\{\cos \left(\frac{-3 \pi}{4}\right)+i \sin \left(\frac{-3 \pi}{4}\right)\right\}=\sqrt{2}\left(\cos \frac{3 \pi}{4}-i \sin \frac{3 \pi}{4}\right)
$

(iv) Let $z=1-i$. Then, $|z|=\sqrt{1^2+(-1)^2}=\sqrt{2}$.
Let $\alpha$ be the acute angle given by $\tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$. Then, $\tan \alpha=\left|\frac{-1}{1}\right|=1 \Rightarrow \alpha=\frac{\pi}{4}$.
We find that the point $(1,-1)$ representing $z$ lies in the fourth quadrant. Therefore, the argument of $z$ is given by $\theta=-\alpha=-\frac{\pi}{4}$.
Hence, the polar form of $z=1-i$ is
$
r(\cos \theta+i \sin \theta)=\sqrt{2}\left\{\cos \left(-\frac{\pi}{4}\right)+i \sin \left(-\frac{\pi}{4}\right)\right\}=\sqrt{2}\left\{\cos \frac{\pi}{4}-i \sin \frac{\pi}{4}\right\}
$

Question 2.
Find the modulus and principal argument of $(1+i)$ and hence express it in the polar form.
Solution:
Let $z=1+i$. Then $|z|=\sqrt{1^2+1^2}=\sqrt{2}$.
Let $\alpha$ be the acute angle given by $\tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$. Then, $\tan \alpha=\left|\frac{1}{1}\right|=1 \Rightarrow \alpha=\frac{\pi}{4}$
Clearly, the point $(1,1)$ representing $z=1+i$ lies in first quadrant.
Therefore, $\theta=\arg (z)=\frac{\pi}{4}$.
Hence, the polar form of $z=1+i$ is $z=|z|(\cos \theta+i \sin \theta)=\sqrt{2}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)$.

Question 3.
Express the following complex numbers in the polar form.
(i) $\frac{1+i}{1-i}$
(ii) $\frac{2+6 \sqrt{3} i}{5+\sqrt{3} i}$
Solution:

(i) Let $z=\frac{1+i}{1-i}$. and, let $r(\cos \theta+i \sin \theta)$ be the polar form of $z$. Then, $r=|z|$ and $\theta=\arg (z)$. Now, $z=\frac{1+i}{1-i}=\frac{(1+i)(1+i)}{(1-i)(1+i)}=\frac{1-2 i+i^2}{1-i^2}=\frac{1-2 i-1}{1+1}=i=0+1 i$
$
\therefore r=|z|=\sqrt{0+1}=1 \text {. }
$
Clearly, the point $(0,1)$ representing $z=0+i$ lies on positive direction of imaginary axis Therefore, $\arg (z)=\frac{\pi}{2}$.
Hence, the polar form of $z$ is $z=1\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)=\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}$

(ii) Let $z=\frac{2+6 \sqrt{3} i}{5+\sqrt{3} i}$ and, let $r(\cos \theta+i \sin \theta)$ be the polar form of $z$.
Then, $r=|z|$ and $\theta=\arg (z)$
Clearly, $z=\frac{2+6 \sqrt{3} i}{5+\sqrt{3} i}=\frac{2+6 \sqrt{3} i}{5+\sqrt{3} i} \cdot \frac{(5-\sqrt{3} i)}{(5-\sqrt{3} i)}=\frac{28+28 \sqrt{3} i}{28}=1+i \sqrt{3}$
$
\therefore r=|z|=\sqrt{1+3}=2
$
Let $\alpha$ be the acute angle given by $\tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$.
Then, $\tan \alpha=\frac{\sqrt{3}}{1}=\sqrt{3} \Rightarrow \alpha=\frac{\pi}{3}$
Clearly, the point $(1, \sqrt{3})$ representing $z$ lies in first quadrant. Therefore, $\theta \arg (z)=\alpha=\frac{\pi}{3}$.
Hence, the polar form of $z$ is $2\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)$.

Question 4.
Express the following complex numbers in the polar form: $2+2 \sqrt{3} i$
Solution:
$
\begin{aligned}
z & =2+2 \sqrt{3} i \\
|z| & =\sqrt{2^2+(2 \sqrt{3})^2}=\sqrt{4+12}=4=r \\
\tan \theta & =\frac{2 \sqrt{3}}{2}=\sqrt{3} \Rightarrow \theta=\frac{\pi}{3} \\
\text { So, } z & =r[\cos \theta+i \sin \theta]=4\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)=4 \operatorname{cis} \frac{\pi}{3}
\end{aligned}
$

Question 5.
Express the following complex numbers in the polar form: $-1+i \sqrt{3}$

Solution:
Let $z=-1+i \sqrt{3} \quad \Rightarrow|z|=\sqrt{1+3}=2 ; \tan \theta=\frac{\sqrt{3}}{-1}=-\sqrt{3}$
Here the number $-1+i \sqrt{3}$ lies in II quadrant.
$\therefore \quad \theta=\pi-\frac{\pi}{3}=\frac{2 \pi}{3}$
So, $z=2\left[\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right]=2 \operatorname{cis} \frac{2 \pi}{3}$

Question 6.
Express the following complex numbers in the polar form: $-1-1$

Solution:
Let $z=-1-i$
$
|z|=\sqrt{1+1}=\sqrt{2} ; \tan \theta=\frac{-1}{-1}=1
$
Here the number $-1-i$ lies in III quadrant.
So,
$
\begin{array}{ll}
\text { So, } & \theta=\pi+\frac{\pi}{4}=\frac{5 \pi}{4} \text { or }-\frac{3 \pi}{4} \\
\therefore & z=\sqrt{2} \operatorname{cis}\left(\frac{5 \pi}{4}\right) \text { or } \sqrt{2} \operatorname{cis}\left(\frac{-3 \pi}{4}\right)
\end{array}
$
Question 7.
Express the following complex numbers in the polar form: $1-1$
Solution:
Let $z=1-i$
$
|z|=\sqrt{1+1}=\sqrt{2} ; \tan \theta=\frac{-1}{1}=-1
$
$1-i$ lies IV quadrant. So $\theta \frac{-\pi}{4} ; \therefore z=\sqrt{2}$ cis $\left(\frac{-\pi}{4}\right)$