Exercise 2.7 - Chapter 2 Complex Numbers 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $2.7$
Question 1.
Write in polar form of the following complex numbers.
(i) $2+i 2 \sqrt{ } 3$
(ii) $3-\mathrm{i} \sqrt{3}$
(iii) $-2-\mathrm{i} 2$
(iv) $\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}$
Solution:
(i)
$
\begin{aligned}
2+i 2 \sqrt{3} & =r(\cos \theta+i \sin \theta) \\
r & =\sqrt{x^2+y^2}=\sqrt{4+12}=\sqrt{16}=4 \\
\alpha & =\tan ^{-1}\left|\frac{y}{x}\right|=\tan ^{-1}\left|\frac{2 \sqrt{3}}{2}\right|=\tan ^{-1}|\sqrt{3}|=\frac{\pi}{3}
\end{aligned}
$
$2+2 i \sqrt{3}$ lies in I quadrant
$
\theta=\alpha=\frac{\pi}{3}
$
So $\quad(2+i 2 \sqrt{3})=4\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)=4\left[\cos \left(2 k \pi+\frac{\pi}{3}\right)+i \sin \left(2 k \pi+\frac{\pi}{3}\right)\right], k \in z$
(ii)
$
\begin{aligned}
3-i \sqrt{3} & =r(\cos \theta+i \sin \theta) \\
r & =\sqrt{x^2+y^2}=\sqrt{9+3}=\sqrt{12}=2 \sqrt{3} \\
\alpha & =\tan ^{-1}\left|\frac{y}{x}\right|=\tan ^{-1}\left|\frac{-\sqrt{3}}{3}\right|=\tan ^{-1}\left|\frac{-1}{\sqrt{3}}\right|=\frac{\pi}{6}
\end{aligned}
$
$(3-i \sqrt{3})$ lies in IV quadrant
$
\begin{aligned}
\theta & =-\alpha=\frac{-\pi}{6} \\
(3-i \sqrt{3}) & =2 \sqrt{3}\left[\cos \frac{-\pi}{6}+i \sin \left(\frac{-\pi}{6}\right)\right]
\end{aligned}
$

$
=2 \sqrt{3}\left[\cos \left(2 k \pi-\frac{\pi}{6}\right)+i \sin \left(2 k \pi-\frac{\pi}{6}\right)\right], k \in z
$
(iii)
$
\begin{aligned}
-2-i 2 & =r(\cos \theta+i \sin \theta) \\
r & =\sqrt{x^2 \quad y^2}=\sqrt{4+4}=2 \sqrt{2} \\
\alpha & =\tan ^{-1}\left|\frac{y}{x}\right|=\tan ^{-1}\left|\frac{-2}{-2}\right|=\tan ^{-1}|1|=\frac{\pi}{4}
\end{aligned}
$
$(-2-i 2)$ lies in III quadrant
$
\begin{aligned}
\theta & =-(\pi-\alpha)=-\left(\pi-\frac{\pi}{4}\right)=-\frac{3 \pi}{4} \\
\therefore \quad-2-i 2 & =2 \sqrt{2}\left[\cos \left(\frac{-3 \pi}{4}\right)+i \sin \left(\frac{-3 \pi}{4}\right)\right] \\
& =2 \sqrt{2}\left[\cos 2 k \pi-\frac{3 \pi}{4}+i \sin 2 k \pi-\frac{3 \pi}{4}\right] k \in z
\end{aligned}
$

(iv) Consider $-1+i=r(\cos \theta+i \sin \theta)$
$
\begin{aligned}
& r=\sqrt{x^2+y^2}=\sqrt{1+1}=\sqrt{2} \\
& \alpha=\tan ^{-1}\left|\frac{1}{-1}\right|=\tan ^{-1}|1|=\frac{\pi}{4}
\end{aligned}
$
$-1+i$ lies in II quadrant
$
\begin{aligned}
\theta & =\pi-\alpha=\pi-\frac{\pi}{4}=\frac{3 \pi}{4} \\
-1+i & =\sqrt{2} \cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4} \\
\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}} & =\frac{\sqrt{2}\left[\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right]}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}} \quad\left[\arg \left(\frac{z_1}{z_2}\right)=\arg \left(z_1\right)-\arg \left(z_2\right)\right] \\
& =\sqrt{2}\left[\cos \left(\frac{3 \pi}{4}-\frac{\pi}{3}\right)+i \sin \left(\frac{3 \pi}{4}-\frac{\pi}{3}\right)\right] \\
& =\sqrt{2}\left[\cos \left(\frac{9 \pi-4 \pi}{12}\right)+i \sin \left(\frac{9 \pi-4 \pi}{12}\right)\right] \\
& =\sqrt{2}\left[\cos \left(\frac{5 \pi}{12}\right)+i \sin \left(\frac{5 \pi}{12}\right)\right]=\sqrt{2}\left[\operatorname{cis} 2 k \pi+\frac{5 \pi}{12}\right], k \in z
\end{aligned}
$

Question 2.
Find the rectangular form of the following complex numbers.
(i) $\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)\left(\cos \frac{\pi}{12}+i \sin \frac{\pi}{12}\right)$
(ii) $\frac{\cos \frac{\pi}{6}-i \sin \frac{\pi}{6}}{2\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)}$
Solution:
(i) $\arg \left(z_1 z_2\right)=\operatorname{Arg}\left(z_1\right)+\operatorname{Arg}\left(z_2\right)$
$
\begin{aligned}
& =\cos \left(\frac{\pi}{6}+\frac{\pi}{12}\right)+i \sin \left(\frac{\pi}{6}+\frac{\pi}{12}\right)=\cos \left(\frac{2 \pi+\pi}{12}\right)+i \sin \left(\frac{2 \pi+\pi}{12}\right) \\
& =\cos \left(\frac{3 \pi}{12}\right)+i \sin \left(\frac{3 \pi}{12}\right)=\cos \frac{\pi}{4}+i \sin \frac{\pi}{4} \\
& =\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}=\frac{1}{\sqrt{2}}(1+i)
\end{aligned}
$
(ii)
$
\begin{aligned}
\frac{\cos \frac{\pi}{6}-i \sin \frac{\pi}{6}}{2\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)} & =\frac{1}{2} \frac{\cos \left(\frac{-\pi}{6}\right)+i \sin \left(\frac{-\pi}{6}\right)}{\cos \frac{\pi}{3}+i \sin \left(\frac{\pi}{3}\right)} \\
& =\frac{1}{2} \cos \left(\frac{-\pi}{6}-\frac{\pi}{3}\right)+i \sin \left(\frac{-\pi}{6}-\frac{\pi}{3}\right)=\frac{1}{2}\left[\cos \left(\frac{-\pi}{2}\right)+i \sin \left(\frac{-\pi}{2}\right)\right] \\
& =\frac{1}{2}\left[\cos \left(\frac{\pi}{2}\right)+i \sin \left(\frac{\pi}{2}\right)\right]=\frac{1}{2}[-i]=0-\frac{i}{2}=-\frac{i}{2}
\end{aligned}
$

Question 3.
If $\left(x_1+i y_1\right)\left(x_2+i y_2\right)\left(x_3+i y_3\right) \cdots\left(x_n+i y_n\right)=a+i b$, show that
(i) $\left(x_1^2+y_1^2\right)\left(x_2^2+y_2^2\right)\left(x_3^2+y_3^2\right) \cdots\left(x_n^2+y_n^2\right)=a^2+b^2$
(ii) $\sum_{r=1}^n \tan ^{-1}\left(\frac{y_r}{x_r}\right)=\tan ^{-1}\left(\frac{b}{a}\right)+2 k \pi, k \in \mathbb{Z}$
Solution:
(i) $\left(x_1+\mathrm{iy}_1\right)\left(\mathrm{x}_2+\mathrm{iy}_2\right)\left(\mathrm{x}_3+\mathrm{iy}_3\right) \ldots \ldots .\left(\mathrm{x}_{\mathrm{n}}+\mathrm{iy}_{\mathrm{n}}\right)=\mathrm{a}+\mathrm{ib}$
Taking modulus on both sides,
$
\begin{aligned}
& \left|\left(x_1+i y_1\right)\left(x_2+i y_2\right)\left(x_3+i y_3\right) \ldots \ldots .\left(x_n+i y_n\right)\right|=|a+i b| \\
& \left|x_1+i y_1\right|\left|x_2+i y_2\right|\left|x_3+i y_3\right| \ldots . .\left|x_n+i y_n\right|=|a+i b|
\end{aligned}
$

$
\sqrt{x_1^2+y_1^2} \sqrt{x_2^2+y_2^2} \sqrt{x_3^2+y_3^2} \ldots \sqrt{x_n^2+y_n^2}=\sqrt{a^2+b^2}
$
Squaring on both sides
$
\left(x_1^2+y_1^2\right)\left(x_2^2+y_2^2\right)\left(x_3^2+y_3^2\right) \ldots\left(x_n^2+y_n^2\right)=a^2+b^2
$
(ii) Taking Argument on both sides of (1)
$
\begin{aligned}
& \arg \left[\left(x_1+i y_1\right)\left(x_2+i y_2\right)\left(x_3+i y_3\right) \ldots\left(x_n+i y_n\right)\right]=\arg (a+i b) \\
& \arg \left(x_1+i y_1\right)+\arg \left(x_2+i y_2\right)+\arg \left(x_3+i y_3\right)+\ldots \arg \left(x_n+i y_n\right)=\arg (a+i b) \\
& \tan ^{-1}\left(\frac{y_1}{x_1}\right)+\tan ^{-1}\left(\frac{y_2}{x_2}\right)+\tan ^{-1}\left(\frac{y_3}{x_3}\right)+\ldots+\tan ^{-1}\left(\frac{y_n}{x_n}\right)=2 k \pi+\tan ^{-1}\left(\frac{b}{a}\right) \\
& \sum_{r=1}^n \tan ^{-1}\left(\frac{y_r}{x_r}\right)=2 k \pi+\tan ^{-1}\left(\frac{b}{a}\right)(k \in z)
\end{aligned}
$

Question 4.
If $\frac{1+z}{1-z}=\cos 2 \theta+i \sin 2 \theta$, show that $z=i \tan \theta$.
Solution:
$
\begin{aligned}
& \frac{1+z-1+z}{1+z+1-z}=\frac{\cos 2 \theta+i \sin 2 \theta-1}{\cos 2 \theta+i \sin 2 \theta+1} \\
& \frac{2 z}{2}=\frac{1-2 \sin ^2 \theta+2 i \sin \theta \cos \theta-1}{2 \cos ^2 \theta-1+i 2 \sin \theta \cos \theta+1} \\
& {[(x+i y)=i(y-x i)]} \\
& z=\frac{2 \sin \theta[i \cos \theta-\sin \theta]}{2 \cos \theta[\cos \theta+i \sin \theta]}=\frac{i \sin \theta[\cos \theta+i \sin \theta]}{\cos \theta[\cos \theta+i \sin \theta]} \\
& z=i \tan \theta \\
&
\end{aligned}
$

Question 5.
If $\cos \alpha+\cos \beta+\cos \gamma=\sin \alpha+\sin \beta+\sin \gamma=0$, then show that
(i) $\cos 3 \alpha+\cos 3 \beta+\cos 3 \gamma=3 \cos (\alpha+\beta+\gamma)$
(ii) $\sin 3 \alpha+\sin 3 \beta+\sin 3 \gamma=3 \sin (\alpha+\beta+\gamma)$
Solution:
Let $\mathrm{a}=\cos \alpha+\mathrm{i} \sin \alpha=\mathrm{e}^{\mathrm{i} \alpha}$
$\mathrm{b}=\cos \beta+\mathrm{i} \sin \beta=\mathrm{e}^{\mathrm{i} \beta}$
$c=\cos \gamma+i \sin \gamma=e^{i \gamma}$
$a+b+c=(\cos \alpha+\cos \beta+\cos \gamma)+i(\sin \alpha+\sin \beta+\sin \gamma)$
$\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=0+\mathrm{i} 0$
$\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=0$
If $a+b+c=0$ then $a^3+b^3+c^3=3 a b c$
$
\begin{aligned}
\left(e^{i \alpha}\right)^3+\left(e^{i \beta}\right)^3+\left(e^{\gamma \gamma}\right)^3 & =3 e^{i \alpha} e^{i \beta} e^{i \gamma} \\
e^{i 3 \alpha}+e^{i 3 \beta}+e^{i 3 \gamma} & =3 e^{i(\alpha+\beta+\gamma)}
\end{aligned}
$
$(\cos 3 \alpha+i \sin 3 \alpha+\cos 3 \beta+i \sin 3 \beta+\cos 3 \gamma+i \sin 3 \gamma)=3[\cos (\alpha+\beta+\gamma)+i \sin (\alpha+\beta+\gamma)]$ $(\cos 3 \alpha+\cos 3 \beta+\cos 3 \gamma)+i(\sin 3 \alpha+\sin 3 \beta+\sin 3 \gamma)=3 \cos (\alpha+\beta+\gamma)+i 3 \sin (\alpha+\beta+\gamma)$
Equating real and Imaginary parts
(i) $\cos 3 \alpha+\cos 3 \beta+\cos 3 \gamma=3 \cos (\alpha+\beta+\gamma)$
(ii) $\sin 3 \alpha+\sin 3 \beta+\sin 3 \gamma=3 \sin (\alpha+\beta+\gamma)$

Question 6.
If $z=x+i y$ and $\arg \left(\frac{z-i}{z+2}\right)=\frac{\pi}{4}$, then show that $x^2+y^2+3 x-3 y+2=0$.
Solution:
$\arg \left(\frac{z-i}{z+2}\right)=\frac{\pi}{4}$
We have $\arg \left(\frac{z_1}{z_2}\right)=\arg \left(\mathrm{z}_1\right)-\arg \left(\mathrm{z}_2\right)$
$\arg (z-i)-\arg (z+2)=\frac{\pi}{4}$
Let $z=x+i y$
$\arg (\mathrm{x}+\mathrm{iy}-\mathrm{i})-\arg (\mathrm{x}+\mathrm{iy}+2)=\frac{\pi}{4}$
$\arg (x+i(y-1))-\arg (x+2+i y)=\frac{\pi}{4}$

$
\begin{aligned}
& \tan ^{-1}\left(\frac{y-1}{x}\right)-\tan ^{-1}\left(\frac{y}{x+2}\right)=\frac{\pi}{4} \Rightarrow \tan ^{-1}\left(\frac{\frac{y-1}{x}-\frac{y}{x+2}}{1+\frac{y-1}{x} \times \frac{y}{x+2}}\right)=\frac{\pi}{4} \\
& \tan ^{-1}\left(\frac{\frac{(y-1)(x+2)-x y}{x(x+2)}}{\frac{x(x+2)+y(y-1)}{x(x+2)}}\right)=\frac{\pi}{4} \Rightarrow \tan ^{-1}\left(\frac{x y+2 y-x-2-x y}{x^2+2 x+y^2-y}\right)=\frac{\pi}{4} \\
&\left(\frac{2 y-x-2}{x^2+y^2+2 x-y}\right)=\tan \frac{\pi}{4}=1 \\
& 2 y-x-2=x^2+y^2+2 x-y \\
& x^2+y^2+3 x-3 y+2=0
\end{aligned}
$
Hence proved.