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

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

AdditionalProblems
Question 1.
Prove that: $(1+i)^n+(1-i)^n=2^{\frac{n+2}{2}} \cos \frac{n \pi}{4}$.
Solution:
Let $z=1+i$
$
\begin{aligned}
|z| & =\sqrt{1+1}=\sqrt{2}=2^{\frac{1}{2}}=r \\
\tan \theta & =\frac{1}{1}=1 \Rightarrow \theta=\frac{\pi}{4} \\
\therefore 1+i & =2^{\frac{1}{2}}\left[\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right] \\
\text { So, } 1-i & =2^{\frac{1}{2}}\left[\cos \frac{\pi}{4}-i \sin \frac{\pi}{4}\right] \quad \text { (The point is in the 1st quadrant) } \\
\text { LHS } & =(1+i)^n+(1-i)^n \\
& =\left\{2^{\frac{1}{2}}\left[\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right]\right\}+\left\{2^{\frac{1}{2}}\left[\cos \frac{\pi}{4}-i \sin \frac{\pi}{4}\right]\right\} \\
& =2^{\frac{n}{2}}\left[\cos \frac{n \pi}{4}+i \sin \frac{n \pi}{4}\right]+2^{\frac{n}{2}}\left[\cos \frac{n \pi}{4}-i \sin \frac{n \pi}{4}\right] \\
& =2^{\frac{n}{2}}\left[\cos \frac{n \pi}{4}+i \sin \frac{n \pi}{4}+\cos \frac{n \pi}{4}-i \sin \frac{n \pi}{4}\right] \\
& \left.=2^{\frac{n}{2}}\left[2 \cos \frac{n \pi}{4}\right]=2^{\frac{n}{2}+1} \cos \frac{n \pi}{4}=2^{\frac{n+2}{2}} \cos \frac{n \pi}{4}=\text { RHS } i \text { to }-i\right]
\end{aligned}
$

Question 2.
Prove that: $(1+\cos \theta+i \sin \theta)^n+(1+\cos \theta-i \sin \theta)^n=2^{n+1} \cos ^n\left(\frac{\theta}{2}\right) \cos \left(\frac{n \theta}{2}\right)$
Solution:

$\begin{aligned}
& \text { Let } z=1+\cos \theta+i \sin \theta ; \mathrm{RP} \text { of } z=1+\cos \theta \text {; IP of } z=\sin \theta \\
& |z|=\sqrt{(1+\cos \theta)^2+\sin ^2 \theta}=\sqrt{1+\cos ^2 \theta+2 \cos \theta+\sin ^2 \theta} \\
& =\sqrt{2+2 \cos \theta}=\sqrt{2(1+\cos \theta)} \\
& =\sqrt{2 \times 2 \cos ^2 \frac{\theta}{2}}=\sqrt{4 \cos ^2 \frac{\theta}{2}}=2 \cos \frac{\theta}{2}=r \\
& \tan \theta=\frac{\mathrm{IP}}{\mathrm{RP}}=\frac{\sin \theta}{1+\cos \theta}=\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^2 \frac{\theta}{2}}=\tan \frac{\theta}{2} \Rightarrow \arg z=\frac{\theta}{2} \\
& \therefore 1+\cos \theta+i \sin \theta=2 \cos \frac{\theta}{2}\left[\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}\right] \\
& \text { Similarly, } 1+\cos \theta-i \sin \theta=2 \cos \frac{\theta}{2}\left[\cos \frac{\theta}{2}-i \sin \frac{\theta}{2}\right] \\
& \text { LHS: }(1+\cos \theta+i \sin \theta)^n+(1+\cos \theta-i \sin \theta)^n \\
& =\left[2 \cos \left(\frac{\theta}{2}\right)\left(\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}\right)\right]^n+\left\{2 \cos \left(\frac{\theta}{2}\right)\left[\cos \frac{\theta}{2}-i \sin \frac{\theta}{2}\right]\right\}^n \\
& =2^n \cos ^n\left(\frac{\theta}{2}\right)\left[\cos \frac{n \theta}{2}+i \sin \frac{n \theta}{2}\right]+2^n \cos ^n\left(\frac{\theta}{2}\right)\left[\cos \frac{n \theta}{2}-i \sin \frac{n \theta}{2}\right] \\
& =2^n \cos ^n\left(\frac{\theta}{2}\right)\left[\cos \frac{n \theta}{2}+i \sin \frac{n \theta}{2}+\cos \frac{n \theta}{2}-i \sin \frac{n \theta}{2}\right] \\
& =2^n \cos ^n\left(\frac{\theta}{2}\right)\left(2 \cos \frac{n \theta}{2}\right)=2^{n+1} \cos ^n\left(\frac{\theta}{2}\right) \cos \frac{n \theta}{2}=\text { RHS } \\
&
\end{aligned}$

Question 3.
Prove that: $(1+i)^{4 n}$ and $(1+i)^{4 n+2}$ are real and purely imaginary respectively.
Solution:
Let $z=1+i$
So,
$
\begin{aligned}
& \therefore & (1+i)^4 & =\left[\sqrt{2}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)\right]^4=2^2(\cos \pi+i \sin \pi)=4(-1) \\
& \therefore & (1+i)^4 & =-4 \\
\Rightarrow & & (1+i)^{4 n} & =(-4)^n \text { which is a real number. } \\
& \text { Now } & (1+i)^{4 n+2} & =(1+i)^{4 n} \cdot(1+i)^2=(-4)^n(1+i)^2=(-4)^n\left(1+i^2+2 i\right)
\end{aligned}
$
$
\begin{aligned}
|z| & =\sqrt{1+1}=\sqrt{2} \\
\tan \theta=\frac{1}{1} & =1 \quad \Rightarrow \theta=\frac{\pi}{4}\left(1^{\text {st }} \text { quadrant }\right) \\
1+i & =\sqrt{2}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)
\end{aligned}
$
$=2 \mathrm{i}(-1)^{\mathrm{n}}$ which is purely imaginary.

Question 4.
If $x=\cos \alpha+i \sin \alpha ; y=\cos \beta+i \sin \beta$, prove that $x^m y^n+\frac{1}{x^m y^n}=2 \cos (m \alpha+n \beta)$.
Solution:
$
\begin{aligned}
x & =\cos \alpha+i \sin \alpha ;=e^{i \alpha} ; y=\cos \beta+i \sin \beta=e^{i \beta} \\
x^m y^n & =\left(e^{i \alpha}\right)^m\left(e^{i \beta}\right)^n=e^{i m \alpha} \cdot e^{i n \beta} \\
& =e^{i(m \alpha+n \beta)}=\cos (m \alpha+n \beta)+i \sin (m \alpha+n \beta) \\
\frac{1}{x^m y^n} & =\frac{1}{e^{i(m \alpha+n \beta)}}=e^{-i(m \alpha+n \beta)}=\cos (m \alpha+n \beta)-i \sin (m \alpha+n \beta) \\
\text { LHS } & =x^m y^n+\frac{1}{x^m y^n} \\
& =\cos (m \alpha+n \beta)+i \sin (m \alpha+n \beta)+\cos (m \alpha+n \beta)-i \sin (m \alpha+n \beta) \\
& =2 \cos (m \alpha+n \beta)=\text { RHS }
\end{aligned}
$

Question 5.
If $\mathrm{a}=\cos 2 \alpha+\mathrm{i} \sin 2 \alpha, \mathrm{b}=\cos 2 \beta+\mathrm{i} \sin 2 \beta$ and $\mathrm{c}=\cos 2 \gamma+\mathrm{i} \sin 2 \gamma$, prove that
(i) $\sqrt{a b c}+\frac{1}{\sqrt{a b c}}=2 \cos (\alpha+\beta+\gamma)$
(ii) $\frac{a^2 b^2+c^2}{a b c}=2 \cos 2(\alpha+\beta-\gamma)$

Solution:
$
\begin{aligned}
& \text { (i) } \quad a b c=\left(e^{i \alpha} \cdot e^{i 2 \beta} \cdot e^{i 2 \gamma}\right)=e^{i 2(\alpha+\beta+\gamma)} \\
& \therefore \quad \sqrt{a b c}=e^{i(\alpha+\beta+\gamma)}=\cos (\alpha+\beta+\gamma)+i \sin (\alpha+\beta+\gamma) \text {. } \\
& \text { Similarly, } \quad \frac{1}{\sqrt{a b c}}=\cos (\alpha+\beta+\gamma)-i \sin (\alpha+\beta+\gamma) \\
& \therefore \sqrt{a b c}+\frac{1}{\sqrt{a b c}}=\cos (\alpha+\beta+\gamma)+i \sin (\alpha+\beta+\gamma)+\cos (\alpha+\beta+\gamma)-i \sin (\alpha+\beta+\gamma) \\
& =2 \cos (\alpha+\beta+\gamma) \\
&
\end{aligned}
$
(ii) $\frac{a^2 b^2+c^2}{a b c}=\frac{a b}{c}+\frac{c}{a b}$
$
\begin{aligned}
\frac{a b}{c} & =\frac{e^{i 2 \alpha} e^{i 2 \beta}}{e^{i 2 \gamma}}=e^{i 2(\alpha+\beta-\gamma)}=\cos 2(\alpha+\beta-\gamma)+i \sin 2(\alpha+\beta-\gamma) \\
\frac{c}{a b} & =\frac{1}{\left(\frac{a b}{c}\right)}=\cos 2(\alpha+\beta-\gamma)-i \sin 2(\alpha+\beta-\gamma) \\
\therefore \quad \frac{a b}{c}+\frac{c}{a b} & =\cos 2(\alpha+\beta-\gamma)+i \sin 2(\alpha+\beta-\gamma)+\cos 2(\alpha+\beta-\gamma) \\
& =2 \cos 2(\alpha+\beta-\gamma) \quad-i \sin 2(\alpha+\beta-\gamma)
\end{aligned}
$
Question 6.
Solve: $x^4+4=0$

Solution:
$
\begin{aligned}
x^4 & =(-4) \\
x & =(-4)^{\frac{1}{4}}=[(4)(-1)]^{\frac{1}{4}}=4^{\frac{1}{4}}(-1)^{\frac{1}{4}} \\
& =\left(2^2\right)^{\frac{1}{4}}(-1)^{\frac{1}{4}}=\sqrt{2}(-1)^{\frac{1}{4}} \\
& =\sqrt{2}(\cos \pi+i \sin \pi)^{\frac{1}{4}}
\end{aligned}
$
$[-1=\cos \pi+i \sin \pi]$
Adding $2 k \pi$ to $\arg , \sqrt{2}[\cos (2 k \pi+\pi)+i \sin (2 k \pi+\pi)]^{\frac{1}{4}}$
Applying Demoivre's theorem,
$
\begin{aligned}
& \sqrt{2}\left[\cos \frac{1}{4}(2 k+1) \pi+i \sin \frac{1}{4}(2 k+1) \pi\right] \\
& k=0,1,2,3
\end{aligned}
$
When $k=0, \sqrt{2} \operatorname{cis}\left(\frac{\pi}{4}\right) ; \quad$ When $k=1, \sqrt{2} \operatorname{cis} \frac{3 \pi}{4}$
When $k=2, \sqrt{2} \operatorname{cis} \frac{5 \pi}{4} ; \quad$ When $k=3, \sqrt{2} \operatorname{cis} \frac{7 \pi}{4}$
Question 7.
If $x=a+b, y=a \omega+b \omega^2, z=a \omega^2+b \omega$, show that
(i) $x y z=a^3+b^3$
(ii) $x^3+y^3+z^3=3\left(a^3+b^3\right)$
Solution:
(i) $x=a+A ; y=a \omega+b \omega^2, z=a \omega^2+b \omega$
Now $x y z=(a+b)\left(a \omega+b \omega^2\right)\left(a b \omega^2+b \omega\right)=(a+A)\left[a \omega^3+a b \omega^2+a b \omega+b^2 \omega^3\right]$
$
=(a+b)\left(a^2-a b+b^2\right)=a^3+b^3
$
$
x y z=a^3+b^3
$
(ii) $\mathrm{x}=\mathrm{a}+\mathrm{b}, \mathrm{y}=\mathrm{a} \omega+\mathrm{b} \omega^2, z=a \omega^2+\mathrm{b} \omega$
$x+y+z=\left(a+a \omega+a \omega^2\right)+\left(b+b \omega^2+b \omega\right)$
$=a\left(1+\omega+\omega^2\right)+b\left(1+\omega+\omega^2\right)=a(0)+b(0)=0$
Now $x+y+z=0 \Rightarrow x^3+y^3+z^3=3 x y z$
Here $x y z=a^3+b^3$
$
\therefore \mathrm{x}^3+\mathrm{y}^3+\mathrm{z}^3=3\left(\mathrm{a}^3+\mathrm{b}^3\right)
$