Exercise 2.5 - Chapter 2 Complex Numbers 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $2.5$
Question 1.
Find the modulus of the following complex numbers.
(i) $\frac{2 i}{3+4 i}$
(ii) $\frac{2-i}{1+i}+\frac{1-2 i}{1-i}$
(iii) $(1-i)^{10}$
(iv) $2 \mathrm{i}(3-4 \mathrm{i})(4-3 \mathrm{i})$
Solution:
(i) $\left|\frac{2 i}{3+4 i}\right|=\frac{|2 i|}{|3+4 i|}=\frac{2}{\sqrt{9+16}}=\frac{2}{5}$

$\text { (ii) } \begin{aligned}
\left|\frac{2-i}{1+i}+\frac{1-2 i}{1-i}\right| & =\left|\frac{(2-i)(1-i)+(1-2 i)(1+i)}{(1+i)(1-i)}\right|=\left|\frac{2-2 i-i-1+1+i-2 i+2}{1+1}\right| \\
& =\left|\frac{4-4 i}{2}\right|=|2-2 i|=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}
\end{aligned}$
$\begin{aligned}
& \text { (iii) }\left|(1-\mathrm{i})^{10}\right|=(|1-\mathrm{i}|)^{10} \\
& =(\sqrt{1+1})^{10}=(\sqrt{2})^{10}=2^5=32
\end{aligned}$

$\begin{aligned}
& \text { (iv) }|2 i(3-4 \mathrm{i})(4-3 \mathrm{i})| \\
& =|2 \mathrm{i}||3-4 \mathrm{i}||4-3 \mathrm{i}| \\
& =2 \sqrt{9+16} \sqrt{16+9} \\
& =2 \times 5 \times 5 \\
& =50
\end{aligned}$
Question 2.
For any two complex numbers $z_1$ and $z_2$, such that $\left|z_1\right|=\left|z_2\right|=1$ and $z_1 z_2 \neq-1$, then show that $\frac{z_1+z_2}{1+z_1 z_2}$ is a real number.
Solution:
$
\begin{aligned}
& \left|z_1\right|^2=1 \\
& \Rightarrow z_1 \bar{z}_1=1 \\
& \Rightarrow z_1=\frac{1}{\bar{z}_1}
\end{aligned}
$

$
\begin{aligned}
& \text { Similarly } z_2=\frac{1}{\bar{z}_2} \\
& \qquad \frac{z_1+z_2}{1+z_1 z_2}=\frac{\frac{1}{\bar{z}_1}+\frac{1}{\bar{z}_2}}{1+\frac{1}{\bar{z}_1} \frac{1}{\bar{z}_2}}=\frac{\frac{\bar{z}_2+\bar{z}_1}{\bar{z}_1 \bar{z}_2}}{\frac{\bar{z}_1 \bar{z}_2+1}{\bar{z}_1 \bar{z}_2}}=\frac{\bar{z}_1+\bar{z}_2}{\bar{z}_1 \bar{z}_2+\overline{1}}
\end{aligned}
$
$\Rightarrow$ We have if $z=\bar{z}$ only when $z$ is real.
$
\begin{aligned}
& =\frac{\overline{z_1+z_2}}{\overline{1+z_1 z_2}} \\
\left(\frac{z_1+z_2}{1+z_1 z_2}\right) & =\overline{\left(\frac{z_1+z_2}{1+z_1 z_2}\right)} \quad \therefore \frac{z_1+z_2}{1+z_1 z_2} \text { is real. }
\end{aligned}
$

Question 3.
Which one of the points $10-8 i, 11+6 i$ is closest to $1+i$.
Solution:
$
\begin{aligned}
& \mathrm{A}(1+\mathrm{i}), \mathrm{B}(10-8 \mathrm{i}), \mathrm{C}(11+6 \mathrm{i}) \\
& |\mathrm{AB}|=|(10-8 \mathrm{i})-(1+\mathrm{i})| \\
& =|10-8 \mathrm{i}-1-\mathrm{i}| \\
& =|9-9 \mathrm{i}| \\
& =\sqrt{81+81} \\
& =\sqrt{162} \\
& =9(1.414) \\
& =12.726 \\
& \mathrm{CA}=|(11+6 \mathrm{i})-(1+\mathrm{i})| \\
& =|11+6 \mathrm{i}-1-\mathrm{i}| \\
& =|10+5 \mathrm{i}| \\
& =\sqrt{100+25} \\
& =\sqrt{125} \\
& \mathrm{C}(11+6 \mathrm{i}) \text { is closest to the point } \mathrm{A}(1+\mathrm{i})
\end{aligned}
$
Question 4.
If $|z|=3$, show that $7 \leq|z+6-8 i| \leq 13$.
Solution:
$|z|=3$, To find the lower bound and upper bound we have
$
\begin{aligned}
& || z_1|-| z_2|| \leq\left|z_1+z_2\right| \leq\left|z_1\right|+\left|z_2\right| \\
& || z|-| 6-8 i|| \leq|z+6-8 i| \leq|z|+|6-8 i| \\
& |3-\sqrt{36+64}| \leq|z+6-8 i| \leq 3+\sqrt{36+64} \\
& |3-10| \leq|z+6-8 i| \leq 3+10 \\
& 7 \leq|z+6-8 i| \leq 13
\end{aligned}
$
Question 5.
If $|z|=1$, show that $2 \leq\left|z^2-3\right| \leq 4$.
Solution:
$
\begin{aligned}
& |z|=1 \Rightarrow|z|^2=1 \\
& || z_1|-| z_2|\leq| z_1+z_2|\leq| z_1|+| z_2 \mid
\end{aligned}
$
$
\begin{aligned}
& \left.|| z\right|^2-|-3||\leq| z^2-\left.3|\leq| z\right|^2+|-3| \\
& |1-3| \leq\left|z^2-3\right| \leq 1+3 \\
& 2 \leq\left|z^2-3\right| \leq 4
\end{aligned}
$

Question 6.
If $\left|z-\frac{2}{z}\right|=2$, show that the greatest and least value of $|z|$ are $\sqrt{3}+1$ and $\sqrt{3}-1$ respectively.
Solution:
$
\left|z-\frac{2}{z}\right|=2
$
We know that
$
\begin{aligned}
|| z|-| \frac{2}{z}|| & \leq\left|z-\frac{2}{z}\right|=2 \\
|z|-\left|\frac{2}{z}\right| & \leq 2
\end{aligned}
$
Case (i) Let
$
t=|z| \quad \Rightarrow\left|t-\frac{2}{t}\right| \leq 2
$

$
\begin{aligned}
-2<t-\frac{2}{t}<2 & \Rightarrow t-\frac{2}{t}>-2 \text { (or) } t-\frac{2}{t}<2 \\
t^2-2<2 t & \Rightarrow t^2-2 t-2<0
\end{aligned}
$
Case (ii)
$
t^2-2 t-2<0 \quad \Rightarrow t=\frac{2 \pm \sqrt{4+8}}{2}=1 \pm \sqrt{3}
$
The minimum value of $|z|$ is $|1-\sqrt{3}|=\sqrt{ } 3-1$
The greatest value of $|z|$ is $\sqrt{3}+1$
Question 7.
If $z_1, z_2$ and $z_3$ are three complex numbers such that $\left|z_1\right|=1,\left|z_2\right|=2,\left|z_3\right|=3$ and $\left|z_1+z_2+z_3\right|=1$, show that $\left|9 z_1 z_2+4 z_1 z_3+z_2 z_3\right|=6$.
Solution:
$
\begin{aligned}
\left|z_1\right|=1 & \Rightarrow\left|z_1\right|^2=1 \\
z_1 \bar{z}_1=1 & \Rightarrow z_1=\frac{1}{\bar{z}_1} \\
\left|z_2\right|=2 & \Rightarrow\left|z_2\right|^2=4 \\
z_2 \bar{z}_2=4 & \Rightarrow z_2=\frac{4}{\bar{z}_2}
\end{aligned}
$

Similarly
$
z_3=\frac{9}{\bar{z}_3}
$
$
\begin{aligned}
\left|z_1+z_2+z_3\right| & =1 \\
\left|\frac{\bar{z}_2 \bar{z}_3+4 \bar{z}_1 \bar{z}_3+9 \bar{z}_2 \bar{z}_1}{\bar{z}_1 \bar{z}_2 \bar{z}_3}\right| & =1 \\
\frac{\left|z_2 z_3+4 z_1 z_3+9 z_1 z_2\right|}{\left|\overline{z_1 z_2 z_3}\right|} & =1 \\
\left|z_2 z_3+4 z_1 z_3+9 z_1 z_2\right| & =1 \times 2 \times 3=6
\end{aligned}
$
$
\Rightarrow\left|z_2 z_3+4 z_1 z_3+9 z_1 z_2\right|=\left|z_1\right|\left|z_2\right|\left|z_3\right|
$
Question 8.
If the area of the triangle formed by the vertices $z, i z$, and $z+i z$ is 50 square units, find the value of $|z|$.
Solution:
The given vertices are $z, i z, z+i z \Rightarrow z$, iz are $\perp r$ to each other.
$
\begin{aligned}
& \text { Area of triangle }=\frac{1}{2} \text { bh }=50 \\
& \Rightarrow \frac{1}{2}|\mathrm{z}||\mathrm{iz}|=50 \\
& \Rightarrow \frac{1}{2}|\mathrm{z}||\mathrm{z}|=50 \\
& \Rightarrow|\mathrm{z}|^2=100 \\
& \Rightarrow|\mathrm{z}|=10
\end{aligned}
$
Question 9.
Show that the equation $z^3+2 \bar{z}=0$ has five solutions.
Solution:
Given that $z^3+2 \bar{z}=0$
$
z^3=-2 \bar{z}
$
Taking modulus on both sides,

$
\begin{aligned}
& \left|z^3\right|=|-2 \bar{z}| \quad \Rightarrow|z|^3=2|\bar{z}| \\
& |z|^3=2|z| \quad \Rightarrow|z|^3-2|z|=0 \\
& |z|\left(|z|^2-2\right)=0 \\
& |z|=0 \text { or }|z|^2-2=0 \\
& |z|=0 \Rightarrow z=0 \text { is a solution. } \\
& |z|^2-2=0 \Rightarrow|z|^2=2 \\
& z \bar{z}=2 \\
& \text { From (1) } \Rightarrow \quad \frac{-z^3}{2}=\bar{z} \therefore(2) \Rightarrow z\left(\frac{-z^3}{2}\right)=2 \quad \Rightarrow-z^4=4 \\
& z^4=-4 \\
&
\end{aligned}
$
$z$ has four non-zero solution.
Hence including zero solution. There are five solutions.
Question 10 .
Find the square roots of
(i) $4+3 \mathrm{i}$
(ii) $-6+8 i$
(iii) $-5-12 i$
Solution:
$
\begin{aligned}
& \text { (i) } z=4+3 i \\
& |z|=|4+3 i|=\sqrt{16+9}=5
\end{aligned}
$
We have
(ii)
$
\begin{aligned}
\sqrt{z} & =\pm\left(\sqrt{\frac{|z|+a}{2}}+i \frac{b}{|b|} \sqrt{\frac{|z|-a}{2}}\right) \\
\sqrt{4+3 i} & =\pm\left(\sqrt{\frac{5+4}{2}}+i \frac{3}{3} \sqrt{\frac{5-4}{2}}\right)=\pm\left(\frac{3}{\sqrt{2}}+i \frac{1}{\sqrt{2}}\right) \\
z & =-6+8 i \\
\sqrt{z} & =\pm\left(\sqrt{\frac{|z|+a}{2}}+i \frac{b}{|b|} \sqrt{\frac{|z|-a}{2}}\right) \\
\sqrt{-6+8 i} & =\pm\left(\sqrt{\frac{10-6}{2}}+i \frac{8}{8} \sqrt{\frac{10+6}{2}}\right)=\pm(\sqrt{2}+i 2 \sqrt{2})=\pm \sqrt{2}(1+2 i)
\end{aligned}
$

(iii) $z=-5-12 i$
$
\begin{aligned}
|z| & =\sqrt{25+144}=\sqrt{169}=13 \\
z & =\pm\left(\sqrt{\frac{|z|+a}{2}}+i \frac{b}{|b|} \sqrt{\frac{|z|-a}{2}}\right) \\
\sqrt{-5-12 i} & =\pm\left(\sqrt{\frac{13-5}{2}}-i \frac{12}{12} \sqrt{\frac{13+5}{2}}\right)=\sqrt{\frac{8}{2}}-i \sqrt{\frac{18}{2}}=\pm \sqrt{4}-i \sqrt{9} \\
& =\pm(2-i 3)=\pm(2-3 i)
\end{aligned}
$