Exercise 2.8 - Chapter 2 Complex Numbers 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $2.8$
Question 1.
If to $\omega \neq 1$ is a cube root of unity, then show that $\frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2}+\frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}=1$
Solution:
Since $\omega$ is a cube root of unity, we have $\omega^3=1$ and $1+\omega+\omega^2=0$
$
\begin{aligned}
\mathrm{LHS} & =\frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2}+\frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}=\frac{\omega^2\left(a+b \omega+c \omega^2\right)}{\omega^2\left(b+c \omega+a \omega^2\right)}+\frac{\omega^2\left(a+b \omega+c \omega^2\right)}{\omega^2\left(c+a \omega+b \omega^2\right)} \\
& =\frac{a \omega^2+b+c \omega}{\omega^2\left(b+c \omega+a \omega^2\right)}+\frac{\omega^2\left(a+b \omega+c \omega^2\right)}{\left(c \omega^2+a+b \omega\right)} \quad\left(\therefore \omega^3=1, \omega^4=\omega\right) \\
& =\frac{1}{\omega^2}+\frac{\omega^2}{1}=\frac{1+\omega^4}{\omega^2}=\frac{1+\omega}{\omega^2}=\frac{-\omega^2}{\omega^2}=-1
\end{aligned}
$

Question 2.
Show that $\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^5+\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^5=-\sqrt{3}$
Solution:
$
\mathrm{LHS}=\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^5+\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^5
$
Polar form of $\frac{\sqrt{3}}{2}+\frac{i}{2}=r(\cos \theta+i \sin \theta)$
$
\begin{aligned}
& r=\sqrt{x^2+y^2}=\sqrt{\frac{3}{4}+\frac{1}{4}}=1 \\
& \alpha=\tan ^{-1}\left|\frac{y}{x}\right|=\tan ^{-1}\left|\frac{1}{\sqrt{3}}\right|=\frac{\pi}{6} \\
& \theta=\alpha=\frac{\pi}{6} \\
& \frac{\sqrt{3}}{2}+\frac{i}{2}=1\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right) \\
& \text { Similarly } \frac{\sqrt{3}}{2}-\frac{i}{2}=\left(\cos \frac{\pi}{6}-i \sin \frac{\pi}{6}\right) \\
& \text { (1) } \Rightarrow \text { LHS }=\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)^5+\left(\cos \frac{\pi}{6}-i \sin \frac{\pi}{6}\right)^5 \\
& =\cos \frac{5 \pi}{6}+i \sin \frac{5 \pi}{6}+\cos \frac{5 \pi}{6}-i \sin \frac{5 \pi}{6} \\
& =2 \cos \frac{5 \pi}{6}=2 \cos \left(\pi-\frac{\pi}{6}\right)=-2 \cos \frac{\pi}{6}=-2 \times \frac{\sqrt{3}}{2}=-\sqrt{3}=\text { RHS } \\
&
\end{aligned}
$

Question 3.
Find the value of $\left(\frac{1+\sin \frac{\pi}{10}+i \cos \frac{\pi}{10}}{1+\sin \frac{\pi}{10}-i \cos \frac{\pi}{10}}\right)^{10}$
Solution:
$
\begin{aligned}
& \left(\frac{1+\sin \frac{\pi}{10}+i \cos \frac{\pi}{10}}{1+\sin \frac{\pi}{10}-i \cos \frac{\pi}{10}}\right)^{10}=\left(\frac{1+z}{1+\frac{1}{z}}\right)^{10} \\
& \text { Let } z=\sin \frac{\pi}{10}+i \cos \frac{\pi}{10} \\
& =\left(\frac{1+z}{(z+1)} \times z\right)^{10}=z^{10}\left[\because|z|=1 \Rightarrow|z|^2=1 \Rightarrow z \bar{z}=1 \Rightarrow \bar{z}=\frac{1}{z}\right] \\
& =\left(\sin \frac{\pi}{10}+i \cos \frac{\pi}{10}\right)^{10}=i^{10}\left[\cos \frac{\pi}{10}-i \sin \frac{\pi}{10}\right]^{10} \\
& =i^8 \cdot i^2[\cos \pi-i \sin \pi]=-1[-1]=1 \\
&
\end{aligned}
$

Question 4.
If $2 \cos \alpha=x+\frac{1}{x}$ and $2 \cos \beta=y+\frac{1}{y}$, show that

(i) $\frac{x}{y}+\frac{y}{x}=2 \cos (\alpha-\beta)$
(ii) $x y-\frac{1}{x y}=2 i \sin (\alpha+\beta)$
(iii) $\frac{x^m}{y^n}-\frac{y^n}{x^m}=2 i \sin (m \alpha-n \beta)$
(iv) $x^m y^n+\frac{1}{x^m y^n}=2 \cos (m \alpha+n \beta)$
Solution:
$
\text { (i) } \begin{aligned}
& 2 \cos \alpha=\mathrm{x}+\frac{1}{x} \\
& \Rightarrow 2 \cos \alpha=\frac{x^2+1}{x} \\
& \Rightarrow 2 \mathrm{x} \cos \alpha=\mathrm{x}^2+1 \\
& \Rightarrow \mathrm{x}^2-2 \mathrm{x} \cos \alpha+1=0 \\
& x=\frac{2 \cos \alpha \pm \sqrt{4 \cos ^2 x-4}}{2} \\
& x=\frac{2 \cos \alpha \pm 2 \sqrt{-\left(1-\cos ^2 \alpha\right)}}{2} \\
& \text { Let }=\frac{2 \cos \alpha \pm 2 i \sin \alpha}{2} \\
& x=\cos \alpha \pm i \sin \alpha \\
& x=e^{i \alpha}, y=e^{i \beta} \\
& \frac{x}{y}=\frac{e^{i \alpha}}{e^{i \beta}}=e^{i \alpha} e^{-i \beta}=e^{i(\alpha-\beta)} \\
& \frac{x}{y}=\cos (\alpha-\beta)+i \sin (\alpha-\beta) \\
& \text { Similarly } \frac{y}{x}=\cos (\alpha-\beta)-i \sin (\alpha-\beta) \\
& \frac{x}{y}+\frac{y}{x}=2 \cos (\alpha-\beta)
\end{aligned}
$

$\begin{aligned}
& \Rightarrow z=3(-1)^{1 / 3} \\
& z=3[\cos (2 k \pi+\pi)+i \sin 2 k \pi+\pi]^{\frac{1}{3}} \\
&=3\left[\cos \left(\frac{2 k \pi+\pi}{3}\right)+i \sin \left(\frac{2 k \pi+\pi}{3}\right)\right] \\
& k=0,1,2 \\
& k=0, z=3\left[\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right] \\
& k=1, z=3\left[\cos \frac{3 \pi}{3}+i \sin \frac{3 \pi}{3}\right]=-3 \\
& k=2, z=3\left[\cos \frac{5 \pi}{3}+i \sin \frac{5 \pi}{3}\right]
\end{aligned}$

Question 6.
If $\omega \neq 1$ is a cube root of unity, show that the roots of the equation $(z-1)^3+8=0$ are $-1,1-2 \omega, 1-2 \omega^2$

Solution:
$
\begin{aligned}
(z-1)^3+8 & =0 \\
\Rightarrow(z-1)^3 & =-8 \\
(z-1)^3 & =(-2)^3 \times 1 \quad \Rightarrow(z-1)=\left[(-2)^3\right]^{\frac{1}{3}}[1]^{\frac{1}{3}} \\
z-1 & =-2[1]^{\frac{1}{3}}=-2[\cos 2 k \pi+i \sin 2 k \pi]^{\frac{1}{3}} \\
z-1 & =-2\left[\cos \frac{2 k \pi}{3}+i \sin \frac{2 k \pi}{3}\right] ; k=0,1,2 \\
k=0, z-1 & =-2(1) \quad \Rightarrow z=-2+1=-1 \\
k=1, z-1 & =-2\left[\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right]=-2 \omega \\
z & =1-2 \omega \\
k=2, z-1 & =-2\left[\cos \frac{4 \pi}{3}+i \sin \frac{4 \pi}{3}\right] \\
z-1 & =-2 \omega^2 \quad \Rightarrow z=1-2 \omega^2
\end{aligned}
$
The roots are $-1,1-2 \omega, 1-2 \omega^2$

Question 7.
Find the value of $\sum_{k=1}^8\left(\cos \frac{2 k \pi}{9}+i \sin \frac{2 k \pi}{9}\right)$
Solution:
$
\sum_{k=1}^8\left(\cos \frac{2 k \pi}{9}+i \sin \frac{2 k \pi}{9}\right)
$
We know that 9 th roots of unit are $1, \omega, \omega^2, \ldots \ldots, \omega^8$ Sum of the roots:
$
1+\omega+\omega^2+\ldots+\omega^8=0 \Rightarrow \omega+\omega^2+\omega^3+\ldots \ldots+\omega^8=-1
$
If $k=1$
$
\begin{array}{ll}
\text { If } k=1 & \Rightarrow \cos \frac{2 \pi}{9}+i \sin \frac{2 \pi}{9}=\omega \\
\text { If } k=2 & \Rightarrow \cos \frac{4 \pi}{9}+i \sin \frac{4 \pi}{9}=\omega^2
\end{array}
$
If $k=8 \quad \Rightarrow \cos \frac{16 \pi}{9}+i \sin \frac{16 \pi}{9}=\omega^8$
The sum of all the terms $\sum_{k=1}^8\left(\cos \frac{2 k \pi}{9}+i \sin \frac{2 k \pi}{9}\right)=-1$

Question 8.
If $\omega \neq 1$ is a cube root of unity, show that
(i) $\left(1-\omega+\omega^2\right)^6+\left(1+\omega-\omega^2\right)^6=128$
(ii) $(1+\omega)\left(1+\omega^2\right)\left(1+\omega^4\right)\left(1+\omega^8\right) \ldots\left(1+\omega^{2 n}\right)=1$
Solution:
(i) $\omega$ is a cube root of unity $\omega^3=1 ; 1+\omega+\omega^2=0$
$
\begin{aligned}
& \left(1-\omega+\omega^2\right)^6+\left(1+\omega-\omega^2\right)^6 \\
& =(-\omega-\omega)^6+\left(-\omega^2-\omega^2\right)^6 \\
& =(-2 \omega)^6+\left(-2 \omega^2\right)^6 \\
& =(-2)^6\left(\omega^6+\omega^{12}\right) \\
& =(64)(1+1) \\
& =128
\end{aligned}
$
(ii) $(1+\omega)\left(1+\omega^2\right)\left(1+\omega^4\right)\left(1+\omega^8\right) \ldots \ldots\left(1+\omega^{2 n}\right)$
$=(1+\omega)\left(1+\omega^2\right)\left(1+\omega^4\right)\left(1+\omega^8\right) \ldots . .2 \mathrm{n}$ factors
$=\left(-\omega^2\right)(-\omega)\left(-\omega^2\right)(-\omega) \ldots . .2 \mathrm{n}$ factors
$=\omega^3 \cdot \omega^3$
$=1$

Question 9.
If $z=2-2 i$, find the rotation of $z$ by $\theta$ radians in the counter clockwise direction about the origin when
(i) $\theta=\frac{\pi}{3}$
(ii) $\theta=\frac{2 \pi}{3}$
(iii) $\theta=\frac{3 \pi}{2}$
Solution:
(i) $z=2-2 i=2(1-i)=r(\cos \theta+i \sin \theta)$
$r=\sqrt{x^2+y^2}=2 \sqrt{1+1}=2 \sqrt{2}$
$\alpha=\tan ^{-1}=\left|\frac{y}{x}\right|=\tan ^{-1}|1|=\frac{\pi}{4}$
(1-i) lies in IV quadrant
$\theta=-\alpha=-\frac{\pi}{4}$
$\Rightarrow z=2 \sqrt{2}\left[\cos \left(\frac{-\pi}{4}\right)+i \sin \left(\frac{-\pi}{4}\right)\right]$
$\mathrm{z}$ is rotated by $\theta=\frac{\pi}{3}$ in the counter clock wise direction.
$
z=2 \sqrt{2}\left[\cos \left(\frac{\pi}{3}-\frac{\pi}{4}\right)+i \sin \left(\frac{\pi}{3}-\frac{\pi}{4}\right)\right]=2 \sqrt{2}\left[\cos \frac{\pi}{12}+i \sin \frac{\pi}{12}\right]
$
(ii) $\mathrm{z}$ is rotated by $\theta=\frac{2 \pi}{3}$ in the counter clockwise direction.
$
\begin{aligned}
z & =2 \sqrt{2}\left[\cos \left(\frac{2 \pi}{3}-\frac{\pi}{4}\right)+i \sin \left(\frac{2 \pi}{3}-\frac{\pi}{4}\right)\right] \\
& =2 \sqrt{2}\left[\cos \frac{5 \pi}{12}+i \sin \frac{5 \pi}{12}\right]
\end{aligned}
$
(iii) $\mathrm{z}$ is rotated by $\theta=\frac{3 \pi}{2}$ in the counter clockwise direction.
$
\begin{aligned}
z & =2 \sqrt{2}\left[\cos \left(\frac{3 \pi}{2}-\frac{\pi}{4}\right)+i \sin \left(\frac{3 \pi}{2}-\frac{\pi}{4}\right)\right] \\
& =2 \sqrt{2}\left[\cos \left(\frac{5 \pi}{4}\right)+i \sin \left(\frac{5 \pi}{4}\right)\right]
\end{aligned}
$

Question 10.
Prove that the values of $\sqrt[4]{-1}$ are $\pm \frac{1}{\sqrt{2}}(1 \pm i)$
Solution:
$
\begin{aligned}
\text { Let } z & =(-1)^{\frac{1}{4}} \\
z & =[\cos (2 k \pi+\pi)+i \sin (2 k \pi+\pi)]^{\frac{1}{4}} \\
z & =\left[\cos \left(\frac{2 k \pi+\pi}{4}\right)+i \sin \left(\frac{2 k \pi+\pi}{4}\right)\right] \\
k & =0,1,2,3 \\
k=0, z & =\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}} \\
k=1, z & =\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}=-\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}} \\
k=2, z & =\cos \frac{5 \pi}{4}+i \sin \frac{5 \pi}{4}=-\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}} \\
k=3, z & =\cos \frac{7 \pi}{4}+i \sin \frac{7 \pi}{4}=\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}} \\
\text { The roots } & \operatorname{are} \pm \frac{1}{\sqrt{2}}(1 \pm i)
\end{aligned}
$