Exercise 2.4 - Chapter 2 - Complex Numbers - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Ex 2.4

Question 1.
Write the following in the rectangular form:
(i) $\overline{(5+9 i)+(2-4 i)}$
(ii) $\frac{10-5 i}{6+2 i}$
(iii) $\overline{3 i}+\frac{1}{2-i}$
Solution:
(i) $z=\overline{(5+9 i)+(2-4 i)}=(\overline{5+9 i})+(\overline{2-4 i})=5-9 i+2+4 i=7-5 i$
(ii) $z=\frac{10-5 i}{6+2 i} \times \frac{6-2 i}{6-2 i}=\frac{60-20 i-30 i+10 i^2}{36+4}=\frac{60-50 i-10}{40}$ $=\frac{50-50 i}{40}=\frac{50(1-i)}{40}=\frac{5}{4}(1-i)$
(iii) $z=\overline{3 i}+\frac{1}{2-i}=-3 i+\frac{1}{2-i} \times \frac{2+i}{2+i}=-3 i+\frac{2+i}{4+1}$
$
=\frac{-15 i+2+i}{5}=\frac{2-14 i}{5}=\frac{2}{5}-\frac{14}{5} i
$
Question 2.
If $z=x+i y$, find the following in rectangular form.
(i) $\operatorname{Re}\left(\frac{1}{z}\right)$
(ii) $\operatorname{Re}(\mathrm{i} \bar{z})$
(iii) $\operatorname{Im}(3 z+4 \bar{z}-4$ i $)$
Solution:
(i) $\operatorname{Re}\left(\frac{1}{z}\right)=\operatorname{Re}\left(\frac{1}{x+i y} \times \frac{x-i y}{x-i y}\right)$
$=\operatorname{Re}\left(\frac{x-i y}{x^2+y^2}\right)$
$=\frac{x}{x^2+y^2}$
(ii) $\operatorname{Re}(\mathrm{i} \bar{z})=\operatorname{Re}[\mathrm{i}(\overline{x+i y})]$
$=\operatorname{Re}(\mathrm{ix}+\mathrm{y})$
$=y$
(iii) $\operatorname{Im}(3 z+4 \bar{z}-4 i)$
$=\operatorname{Im}(3(x+i y)+4(x-i y)-4 i)$
$=\operatorname{Im}(3 \mathrm{x}+3 \mathrm{iy}+4 \mathrm{x}-4 \mathrm{iy}-4 \mathrm{i})$
$=\operatorname{Im}(3 x+4+\mathrm{i}(3 \mathrm{y}-4 \mathrm{y}-4)$
$=\operatorname{Im}(3 x+4 x+i(-y-4))$

$
\begin{aligned}
& =\operatorname{Im}[7 x+i(-y-4)] \\
& =-y-4 \\
& =-(y+4)
\end{aligned}
$
Question 3.
If $z_1=2-i$ and $z_2=-4+3 i$, find the inverse of $z_1 z_2$ and $\frac{z_1}{z_2}$
Solution:
$
\begin{aligned}
& z_1=2-i, z_2=-4+3 i \\
& \text { (i) } z_1 z_2=(2-i)(-4+3 i) \\
& =\left(-8+6 i+4 i-3 i^2\right) \\
& =(-8+10 i+3) \\
& =(-5+10 i) \\
& \quad\left(z_1 z_2\right)^{-1}=\frac{1}{\left(z_1 z_2\right)}=\frac{1}{-5+10 i} \times \frac{-5-10 i}{-5-10 i}=\frac{-5-10 i}{25+100}=\frac{5(-1-2)}{125}=\frac{-1-2 i}{25}
\end{aligned}
$
(ii) $\frac{z_1}{z_2}=\frac{2-i}{-4+3 i} \times \frac{-4-3 i}{-4-3 i}=\frac{-8-6 i+4 i+3 i^2}{16+9}=\frac{-8-2 i-3}{25}=\frac{-11-2 i}{25}$
$
\begin{aligned}
\left(\frac{z_1}{z_2}\right)^{-1} & =\frac{25}{-11-2 i}=\frac{25}{-11-2 i} \times \frac{-11+2 i}{-11+2 i}=\frac{25(-11+2 i)}{121+4}=\frac{25(-11+2 i)}{125} \\
& =\frac{1}{5}(-11+2 i)
\end{aligned}
$

Question 4.
The complex numbers $\mathrm{u}, \mathrm{v}$, and $\mathrm{w}$ are related by $\frac{1}{u}=\frac{1}{v}+\frac{1}{w}$. If $\mathrm{v}=3-4 \mathrm{i}$ and $\mathrm{w}=4+3 \mathrm{i}$, find $\mathrm{u}$ in rectangular form.
Solution:
$
\begin{aligned}
& v=3-4 i, w=4+3 i=i(3-4 i) \\
& \frac{1}{u}=\frac{1}{v}+\frac{1}{w}=\frac{1}{3-4 i}+\frac{1}{i(3-4 i)}=\frac{1}{3-4 i}\left[1+\frac{1 \times-i}{i \times-i}\right] \\
& =\frac{1}{3-4 i}[1-i]=\frac{1-i}{3-4 i} \times \frac{3+4 i}{3+4 i}=\frac{3+4 i-3 i+4}{9+16} \quad \Rightarrow \frac{1}{u}=\frac{7+i}{25} \\
& u=\frac{25}{7+i} \times \frac{7-i}{7-i}=\frac{25(7-i)}{49+1}=\frac{25(7-i)}{50}=\frac{1(7-i)}{2}=\frac{7}{2}-\frac{i}{2} \\
&
\end{aligned}
$
Question 5.
Prove the following properties:
(i) $z$ is real if and only if $\mathrm{z}=\bar{z}$
(ii) $\operatorname{Re}(\mathrm{z})=\frac{z+\bar{z}}{2}$ and $\operatorname{Im}(\mathrm{z})=\frac{z-\bar{z}}{2 i}$
Solution:
(i) $z$ is real iff $z=\bar{z}$
Let $z=x+i y$
$\mathrm{z}=\bar{z}$
$\Rightarrow \mathrm{x}+\mathrm{iy}=\mathrm{x}-\mathrm{iy}$
$
\begin{aligned}
& \Rightarrow 2 \mathrm{iy}=0 \\
& \Rightarrow \mathrm{y}=0 \\
& \Rightarrow \mathrm{z} \text { is real. }
\end{aligned}
$
$\mathrm{z}$ is real iff $\mathrm{z}=\bar{z}$
(ii) $\frac{z+\bar{z}}{2 i}=\frac{x+i y+x-i y}{2}=\frac{2 x}{2}=x$
Real part of $\mathrm{z}=\mathrm{x}$
(iii) $\frac{z-\bar{z}}{2 i}=\frac{(x+i y)-(x-i y)}{2 i}=\frac{x+i y-x+i y}{2 i}=\frac{2 i y}{2 i}=y$
Im part of $z=y$.

Question 6.
Find the least value of the positive integer $n$ for which $(\sqrt{3}+i)^n$
(i) real
(ii) purely imaginary
Solution:
$(\sqrt{3}+i)^{\mathrm{n}}$
$(\sqrt{3}+i)^2$
$=3-1+2 \sqrt{3} \mathrm{i}$
$=(2+2 \sqrt{3}$ i $)$
$(\sqrt{3}+i)^3=(\sqrt{3}+i)^2(\sqrt{3}+i)$
$=(2+2 \sqrt{3} i)(\sqrt{3}+i)$
$=2 \sqrt{3}+2 i+6 i-2 \sqrt{3}$
$(\sqrt{3}+\mathrm{i})=8 \mathrm{i} \Rightarrow$ purely Imaginary when $\mathrm{n}=3$
$(\sqrt{3}+i)^4=(\sqrt{3}+i)^3(\sqrt{3}+i)$
$=8 \mathrm{i}(\sqrt{3}+\mathrm{i})$
$=(-8+8 \sqrt{3} i)$
$(\sqrt{3}+i)^5=(\sqrt{3}+i)^4(\sqrt{3}+i)$
$=(-8+8 \sqrt{3} i)(\sqrt{3}+i)$
$=-8 \sqrt{3}-8 i+24 i-8 \sqrt{3}$
$=-16 \sqrt{3}+16 i$
$(\sqrt{3}+i)^6=(\sqrt{3}+i)^5(\sqrt{3}+i)$
$=(\sqrt{3}+\mathrm{i})(-16 \sqrt{3}+16 \mathrm{i})$
$=16(\sqrt{3}+i)(-\sqrt{3}+i)$
$=16(-3+\mathrm{i} \sqrt{3}-\mathrm{i} \sqrt{3}-1)$
$=-64$ purely real when $n=6$
Another Method:
$(\sqrt{3}+i)=r(\cos \theta+i \sin \theta), \quad r=\sqrt{x^2+y^2}=\sqrt{4}=2$.
$(\sqrt{3}+i)=2\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right) \quad \alpha=\tan ^{-1}\left|\frac{y}{x}\right|=\tan ^{-1}\left|\frac{1}{\sqrt{3}}\right|=\frac{\pi}{6}$
$(\sqrt{3}+i)^n=2^n\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)^n=2^n\left[\cos \frac{n \pi}{6}+i \sin \frac{n \pi}{6}\right]$
when $n=3,(\sqrt{3}+i)^3=2^3\left[\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right]=+8(0+i)=8 i$ purely imaginary
when $n=6,(\sqrt{3}+i)^6=2^6[\cos \pi+i \sin \pi]=-2^6=-64$ purely real.

Question 7.
Show that
(i) $(2+\mathrm{i} \sqrt{ } 3)^{10}-(2-\mathrm{i} \sqrt{ } 3)^{10}$ is purely imaginary
(ii) $\left(\frac{19-7 i}{9+i}\right)^{12}+\left(\frac{20-5 i}{7-6 i}\right)^{12}$
Solution:
(i) $(2+i \sqrt{3})^{10}-(2-i \sqrt{ } 3)^{10}$
$
\begin{aligned}
& \frac{20-5 i}{7-6 i}=\frac{20-5 i}{7-6 i} \times \frac{7+6 i}{7+6 i} \\
& =\frac{140+120 i-35 i+30}{49+36}=\frac{170+85 i}{85}=\frac{85(2+i)}{85}=(2+i) \\
& z=\left(\frac{19-7 i}{9+i}\right)^{12}+\left(\frac{20-5 i}{7-6 i}\right)^{12}=(2-i)^{12}+(2+i)^{12} \\
& \bar{z}=\overline{(2-i)^{12}+(2+i)^{12}}=\overline{(2-i)^{12}}+\overline{(2+i)^{12}}=(2+i)^{12}+(2-i)^{12} \\
&
\end{aligned}
$
$\bar{z}=z \quad \Rightarrow z$ is purely real.
Let
$
\begin{aligned}
z & =(2+i \sqrt{3})^{10}-(2-i \sqrt{3})^{10} \\
\bar{z} & =\overline{(2+i \sqrt{3})^{10}-(2-i \sqrt{3})^{10}}=\overline{(2+i \sqrt{3})^{10}}-\overline{(2-i \sqrt{3})^{10}} \\
& =(2-i \sqrt{3})^{10}-(2+i \sqrt{3})^{10}=\left[(2+i \sqrt{3})^{10}-(2-i \sqrt{3})^{10}\right]
\end{aligned}
$
$\bar{z}=-z \quad \Rightarrow z$ is purely imaginary.
(ii) $\left(\frac{19-7 i}{9+i}\right)^{12}+\left(\frac{20-5 i}{7-6 i}\right)^{12}$
$
\begin{aligned}
\frac{19-7 i}{9+i} & =\frac{19-7 i}{9+i} \times \frac{9-i}{9-i}=\frac{171-19 i-63 i-7}{81+1} \\
& =\frac{164-82 i}{82}=\frac{82(2-i)}{82}=(2-i)
\end{aligned}
$