Exercise 2.5-Additional Problems - Chapter 2 - Complex Numbers - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Question 1.
Find the modulus and argument of the following complex numbers and convert them in polar form.
(i) $\frac{1+2 i}{1-3 i}$
(ii) $\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}$
(iii) $\frac{1+3 i}{1-2 i}$
Solution:
(i)
Let $z=\frac{1+2 i}{1-3 i}$.
Then, $z=\frac{1+2 i}{1-3 i} \times \frac{1+3 i}{1+3 i}=\frac{(1-6)+i(2+3)}{1+9}=-\frac{1}{2}+\frac{1}{2} i$
$\therefore \quad r=|z|=\sqrt{\left(-\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2}=\frac{1}{\sqrt{2}}$
Let $\alpha$ be the acute angle given by $\tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$. Then, $\tan \alpha=\left|\frac{-1 / 2}{1 / 2}\right|=1 \Rightarrow \alpha=\frac{\pi}{4}$

We find that $\operatorname{Re}(z)=-\frac{1}{2}<0$ and $\operatorname{Im}(z)=\frac{1}{2}>0$. So, the point representing $z$ lies in the second quadrant.
$
\therefore \quad \theta=\arg (z)=\pi-\alpha=\pi-\frac{\pi}{4}=\frac{3 \pi}{4}
$
Hence, the polar form of $z$ is $r(\cos \theta+i \sin \theta)=\frac{1}{\sqrt{2}}\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)$.

$
\begin{aligned}
& \text { (ii) Let } z=\frac{i-1}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}} \text {. Then, } \\
& z=\frac{i-1}{\frac{1}{2}+i \frac{\sqrt{3}}{2}}=\frac{2(-1+i)}{1+i \sqrt{3}} \times \frac{1-i \sqrt{3}}{1-i \sqrt{3}}=\frac{2\{(-1+\sqrt{3})+i(1+\sqrt{3})\}}{1+3}=\left(\frac{\sqrt{3}-1}{2}\right)+i\left(\frac{\sqrt{3}+1}{2}\right) \\
& \therefore|z|=\sqrt{\left(\frac{\sqrt{3}-1}{2}\right)^2+\left(\frac{\sqrt{3}+1}{2}\right)^2}=\sqrt{\frac{(\sqrt{3}-1)^2+(\sqrt{3}+1)^2}{4}}=\sqrt{\frac{2(3+1)}{4}}=\sqrt{2}
\end{aligned}
$
Let $\alpha$ be the acute angle given by $\tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$. Then,
$
\begin{aligned}
\tan \alpha & =\left|\frac{\frac{\sqrt{3}+1}{2}}{\frac{\sqrt{3}-1}{2}}\right|=\frac{\sqrt{3}+1}{\sqrt{3}-1}=\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}=\frac{\tan \frac{\pi}{4}+\tan \frac{\pi}{6}}{1-\tan \frac{\pi}{4} \tan \frac{\pi}{6}}=\tan \left(\frac{\pi}{4}+\frac{\pi}{6}\right)=\tan \frac{5 \pi}{12} \\
\therefore \alpha & =\frac{5 \pi}{12}
\end{aligned}
$
Clearly, $\operatorname{Re}(z)=\frac{\sqrt{3}-1}{2}>0$ and $\operatorname{Im}(z)=\frac{\sqrt{3}+1}{2}>0$.
So, the point representing $z$ lies in the first quadrant. Therefore, $\theta=\arg (z)=\frac{5 \pi}{12}$.
Therefore, $\theta=\arg (z)=\frac{5 \pi}{12}$.
Hence, the polar form of $z$ is $r(\cos \theta+i \sin \theta)=\sqrt{2}\left(\cos \frac{5 \pi}{12}+i \sin \frac{5 \pi}{12}\right)$

(iii) Let $z=\frac{1+3 i}{1-2 i}$ Then,
$
\begin{aligned}
& z=\frac{1+3 i}{1-2 i}=\frac{1+3 i}{1-2 i} \times \frac{1+2 i}{1+2 i}=\frac{(1-6)+i(3+2)}{1+4}=-1+i \\
\therefore \quad r & =|z|=\sqrt{\left(-1^2+1^2\right)}=\sqrt{2}
\end{aligned}
$

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 $\operatorname{Re}(z)<0$ and $\operatorname{Im}(z)>0$. So, the point representing $z$ lies in the second quadrant. $\therefore \arg (z)=\pi-\alpha=\pi-\frac{\pi}{4}=\frac{3 \pi}{4}$.
Hence the polar form of $z$ is $r(\cos \theta+i \sin \theta)=\sqrt{2}\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)$

Question 2 .
Find the square roots of $-15-8 \mathrm{i}$
Solution:
Let $\sqrt{-15-8 i}=x+i y$. Then, $\sqrt{-15-8 i}=x+i y$.

$\begin{aligned}
& \Rightarrow \quad-15-8 i=(x+i y)^2 \quad \Rightarrow-15-8 i=\left(x^2-y^2\right)+2 i x y \\
& \Rightarrow \quad-15=x^2-y^2 \quad \ldots(i) \text { and, } 2 x y=-8 \\
& \text { Now, } \quad\left(x^2+y^2\right)^2=\left(x^2-y^2\right)^2+4 x^2 y^2 \Rightarrow\left(x^2+y^2\right)^2=(-15)^2+64=289 \\
& \Rightarrow \quad x^2+y^2=17 \\
&
\end{aligned}$

On solving (i) and (iii), we get $x^2=1$ and $y^2=16 \Rightarrow x=\pm 1$ and $y=\pm 4$ From (ii), we observe that $2 x y$ is negative. So, $x$ and $y$ are of opposite signs.
Hence, $\quad \sqrt{-15-8 i}=\pm(1-4 i)$
Question 3.
Find the square roots of $i$.
Solution:
$
\begin{aligned}
& \text { Let } \sqrt{i}=x+i y . \quad \text { Then, } \sqrt{i}=x+i y \\
& \Rightarrow \quad i=(x+i y)^2 \\
& \Rightarrow \quad\left(x^2-y^2\right)+2 i x y=0+i \\
& x^2-y^2=0 \quad \text {...(i) and } 2 x y=1 \\
& \quad \text { Now, } \quad\left(x^2+y^2\right)^2=\left(x^2-y^2\right)^2+4 x^2 y^2 \\
& \Rightarrow \quad\left(x^2+y^2\right)^2=0+1=1 \\
& \Rightarrow \quad x^2+y^2=1 \quad\left[\because x^2+y^2>0\right] \\
&
\end{aligned}
$
Solving (i) and (iii), we get $x^2=1 / 2$ and $y^2=1 / 2 \Rightarrow x=\pm 1 / \sqrt{2}$ and $y=\pm 1 / \sqrt{2}$

From (ii) we observe that we find that $2 x y$ is positive. So, $x$ and $y$ are of same sign.
$\therefore\left(x=\frac{1}{\sqrt{2}}\right.$ and $\left.y=\frac{1}{\sqrt{2}}\right)$ or, $\left(x=-\frac{1}{\sqrt{2}}\right.$ and $\left.y=-\frac{1}{\sqrt{2}}\right)$
Hence, $\sqrt{i}=\pm\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} i\right)=\pm \frac{1}{\sqrt{2}}(1+i)$
Aliter:
Let $z=i$. Then, $\operatorname{Re}(z)=0$ and $|z|=1$.
$
\begin{aligned}
& \therefore \quad \sqrt{i}=\pm\left\{\sqrt{\frac{|z|+\operatorname{Re}(z)}{2}}+i \sqrt{\frac{|z|-\operatorname{Re}(z)}{2}}\right\} \quad[\because \operatorname{Im}(z)>0] \\
& \Rightarrow \quad \sqrt{i}=\pm\left\{\sqrt{\frac{1+0}{2}}+i \sqrt{\frac{1-0}{2}}\right\}=\pm\left(\frac{1}{\sqrt{2}}+i \frac{1}{\sqrt{2}}\right)=\pm \frac{1}{\sqrt{2}}(1+i) \\
&
\end{aligned}
$

Question 4.
Find the modulus or the absolute value of $
\frac{(1+3 i)(1-2 i)}{(3+4 i)}
$

Solution:
$
\begin{aligned}
\left|\frac{(1+3 i)(1-2 i)}{(3+4 i)}\right| & =\frac{|1+3 i||1-2 i|}{|3+4 i|} \\
& =\frac{\sqrt{1^2+3^2} \sqrt{1^2+(2)^2}}{\sqrt{3^2+4^2}}=\frac{\sqrt{10} \sqrt{5}}{\sqrt{25}}=\frac{\sqrt{10} \sqrt{5}}{5}=\sqrt{2}
\end{aligned}
$
Question 5.
Find the modulus and argument of the following complex numbers:

(i) $-\sqrt{2}+i \sqrt{2}$
(ii) $1+i \sqrt{3}$
(iii) $-1-i \sqrt{3}$
Solution:
(i) Let $\quad-\sqrt{2}+i \sqrt{2}=r(\cos \theta+i \sin \theta)$
Equating the real and imaginary parts separately.
$
\begin{array}{c|l}
r \cos \theta=-\sqrt{2} & r \sin \theta=-\sqrt{2} \\
r^2 \cos ^2 \theta=2 & r^2 \sin ^2 \theta=2 \\
r^2\left(\cos ^2 \theta+\sin ^2 \theta\right) & =4 \\
r & =\sqrt{4}=2 .
\end{array}
$
$
\begin{gathered}
\cos \theta=\frac{-\sqrt{2}}{2}=\frac{-1}{\sqrt{2}} \\
\sin \theta=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}} \\
\theta=\pi-\frac{\pi}{4}=\frac{3 \pi}{4}
\end{gathered}
$
modulus $r=2$, argument $\theta=\frac{3 \pi}{4}$
Hence $-\sqrt{2}+i \sqrt{2}=2\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)$
(ii) Let $\quad 1+i \sqrt{3}=r(\cos \theta+i \sin \theta)$
Equating the real and imaginary parts separately.
$
\begin{aligned}
& \begin{array}{c|c}
r \cos \theta=1 & r \sin \theta=\sqrt{3} \\
r^2 \cos ^2 \theta=1 & r^2 \sin ^2 \theta=3
\end{array} \\
& r^2\left(\cos ^2 \theta+\sin ^2 \theta\right)=4 \Rightarrow r=2 \\
& \left.\begin{array}{l}
\cos \theta=\frac{1}{2} \\
\sin \theta=\frac{\sqrt{3}}{2}
\end{array}\right\} \Rightarrow \theta \text { lies in the } 1^{\text {st }} \text { quadrant } \\
&
\end{aligned}
$

$
\theta=\frac{\pi}{3}
$
modulus $r=2$, argument $\theta=\frac{\pi}{3}$
Hence $1+i \sqrt{3}=2\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)$
(iii) Let $\quad-1-i \sqrt{3}=r(\cos \theta+i \sin \theta)$
Equating the real and imaginary parts separately
$
\begin{aligned}
& \begin{array}{r|l}
r \cos \theta=-1 & r \sin \theta=-\sqrt{3} \\
r^2 \cos ^2 \theta=1 & r^2 \sin ^2 \theta=3
\end{array} \\
& r^2\left(\cos ^2 \theta+\sin ^2 \theta\right)=4 \Rightarrow r=2 \\
& \left.\begin{array}{l}
\cos \theta=\frac{-1}{2} \\
\sin \theta=\frac{-\sqrt{3}}{2}
\end{array}\right\} \Rightarrow \theta \text { in the } 3^{\text {rd }} \text { quadrant } \\
& \theta=-\pi+\frac{\pi}{3}=\frac{-2 \pi}{3} \\
&
\end{aligned}
$
modulus $r=2$, argument $\theta=\frac{-2 \pi}{3}$
Hence $-1-i \sqrt{3}=2\left[\cos \left(\frac{-2 \pi}{3}\right)+i \sin \left(\frac{-2 \pi}{3}\right)\right]=2\left[\cos \frac{2 \pi}{3}-i \sin \frac{2 \pi}{3}\right]$

Question 6 .
Show that the points representing the complex numbers $7+9 i,-3+7 i, 3+3 i$ form a right angled triangle on the Argand diagram.
Solution:
Let $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ represent the complex numbers $7+9 i,-3+7 i$ and $3+3 i$ in the Argand diagram respectively.

$
\begin{aligned}
\mathrm{AB} & =|(7+9 i)-(-3+7 i)| \\
& =|10+2 i|=\sqrt{10^2+2^2}=\sqrt{104} \\
\mathrm{BC} & =|(-3+7 i)-(3+3 i)| \\
& =|-6+4 i| \\
& =\sqrt{(-6)^2+4^2}=\sqrt{36+16}=\sqrt{52} \\
\mathrm{CA} & =|(3+3 i)-(7+9 i)|=|-4-6 i| \\
& =\sqrt{(-4)^2+(-6)^2}=\sqrt{16+36}=\sqrt{52} \\
\mathrm{AB}^2 & =\mathrm{BC}^2+\mathrm{CA}^2 \quad \Rightarrow \mid \mathrm{BCA}=90^{\circ}
\end{aligned}
$
Hence $\triangle \mathrm{ABC}$ is a right angled isosceles triangle.
Question 7.
Find the square root of $(-7+24 \mathrm{i})$.
Solution:
Let $\sqrt{-7+24 i}=x+i y$
On squaring, $\quad-7+24 i=\left(x^2-y^2\right)+2 i x y$
Equating the real and imaginary parts
$
\begin{aligned}
& x^2-y^2=-7 \text { and } 2 x y=24 \\
& x^2+y^2=\sqrt{\left(x^2-y^2\right)^2+4 x^2 y^2}=\sqrt{(-7)^2+(24)^2}=25
\end{aligned}
$
Solving, $x^2-y^2=-7$ and $x^2+y^2=25$
we get $x^2=9$ and $y^2=16 \Rightarrow x=\pm 3$ and $y=\pm 4$
$
\begin{aligned}
x & =3, y=4 & & \text { or } x=-3, y=-4 \\
\therefore \quad \sqrt{-7+24 i} & =(3+4 i) & & \text { or }(-3-4 i)
\end{aligned}
$