Exercise 2.9 - Chapter 2 Complex Numbers 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $2.9$
Choose the correct or the most suitable answer from the given four alternatives:
Question 1.
$\mathrm{i}^{\mathrm{n}}+\mathrm{i}^{\mathrm{n}+1}+\mathrm{i}^{\mathrm{n}+2}+\mathrm{i}^{\mathrm{n}+3}$ is
(a) 0
(b) 1
(c) $-1$
(d) $\mathrm{i}$
Answer:
(a) 0
Question 2.
The value of $\sum_{i=1}^{13}\left(i^n+i^{n-1}\right)$ is
(a) $1+i$
(b) $\mathrm{i}$
(c) 1
(d) 0
Answer:
(a) $1+\mathrm{i}$
$
\text { Hint. } \begin{aligned}
\sum_{i=1}^{13}=\left(i+i^0\right)+\left(i^2+i^1\right) & +\left(i^3+i^2\right)+\left(i^4+i^3\right)+\ldots\left(i^{13}+i^{12}\right) \\
& =1+2\left(i+i^2+i^3+\ldots+i^{12}\right)+i^{13}=1+2(0)+i \\
& =1+i
\end{aligned}
$
Question 3.
The area of the triangle formed by the complex numbers $z, i z$, and $z+i z$ in the Argand's diagrams is
(a) $\frac{1}{2}|z|$
(b) $|z|^2$
(c) $\frac{3}{2}|z|^2$
(d) $2|z|^2$
Answer:

(a) $\frac{1}{2}|z|$
Hint: Area of triangle $=\frac{1}{2}$ bh
$
\begin{aligned}
& =\frac{1}{2}|z| \mid \text { iz } \mid \\
& =\frac{1}{2}|z|^2
\end{aligned}
$
Question 4.
The conjugate of a complex number is $\frac{1}{i-2}$. Then, the complex number is
(a) $\frac{1}{i+2}$
(b) $\frac{-1}{i+2}$
(c) $\frac{-1}{i-2}$
(d) $\frac{1}{i-2}$
Answer:
(b) $\frac{-1}{i+2}$
Hint:
$
\begin{aligned}
& \bar{z}=\frac{1}{i-2} \quad \Rightarrow(\bar{z})=\overline{\left(\frac{1}{i-2}\right)} \\
& z=\frac{1}{-i-2}=\frac{-1}{i+2}
\end{aligned}
$
Question 5.
If $z=\frac{(\sqrt{3}+i)^3(3 i+4)^2}{(8+6 i)^2}$ then $|z|$ is equal to
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(c) 2
Hint:
$
\begin{aligned}
|z| & =\left|\frac{(\sqrt{3}+i)^3(3 i+4)^2}{(8+6 i)^2}\right|=\frac{|\sqrt{3}+i|^3|4+3 i|^2}{|8+6 i|^2} \\
& =\frac{(\sqrt{4})^3(\sqrt{25})^2}{(\sqrt{100})^2}=\frac{2^3 \times 25}{100}=2
\end{aligned}
$

Question 6.
If $\mathrm{z}$ is a non zero complex number, such that $2 \mathrm{iz}^2=\bar{z}$ then $|\mathrm{z}|$ is
(a) $\frac{1}{2}$
(b) 1
(c) 2
(d) 3
Answer:
(a) $\frac{1}{2}$
Hint:
$
\begin{array}{rlrl}
2 i & z^2=(\bar{z}) & & \Rightarrow\left|2 i z^2\right|=|\bar{z}| \\
2|i||z|^2 & =|z| & & \Rightarrow 2|z|^2=|z| \\
|z| & =\frac{1}{2} &
\end{array}
$
Question 7.
If $|z-2+i| \leq 2$, then the greatest value of $|z|$ is
(a) $\sqrt{3}-2$
(b) $\sqrt{3}+2$
(c) $\sqrt{5}-2$
(d) $\sqrt{5}+2$
Answer:
(d) $\sqrt{5}+2$
Hint:
$
\begin{array}{rlrl}
|z-2+i| & \leq 2 \\
|| z_1|-| z_2|| & \leq\left|z_1-z_2\right| \leq 2 & \Rightarrow & || z_1|-| 2-i|\leq| z-2+i \mid \leq 2 \\
\left|z_1\right|-|\sqrt{5}| & \leq 2 & & \left|z_1\right|-\sqrt{5} \leq 2 \\
\left|z_1\right| & \leq 2+\sqrt{5} &
\end{array}
$

Question 8 .
If $\left|z-\frac{3}{z}\right|$, then the least value of $|z|$ is
(a) 1
(b) 2
(c) 3
(d) 5
Answer:
(a) 1
Hint:
$
\begin{aligned}
\left|z-\frac{3}{z}\right| & =2 \\
|| z|-| \frac{3}{z}|| & \leq\left|z-\frac{3}{z}\right|=2 \\
|z|-\frac{3}{|z|} & \leq 2 \quad \text { Let } t=|z| \\
t-\frac{3}{t} & \leq 2 \quad \Rightarrow t^2-3 \leq 2 t \quad \Rightarrow t^2-2 t-3 \leq 0 \\
t & =\frac{2 \pm \sqrt{4+12}}{2} \quad \Rightarrow t=\frac{2 \pm 4}{2} \\
t & =\frac{2-4}{2}, t=\frac{2+4}{2} \\
t & =-1,3 . \text { The least value of }|z|=1
\end{aligned}
$

Question 9.
If $|z|=1$, then the value of $\frac{1+z}{1+\bar{z}}$ is
(a) $\mathrm{z}$
(b) $\bar{z}$
(c) $\frac{1}{z}$
(d) 1
Answer:
(a) $z$
Hint:
$
\begin{aligned}
& |z|=1 \quad \Rightarrow \bar{z}=\frac{1}{z} \\
& \frac{1+z}{1+\bar{z}}=\frac{1+z}{1+\frac{1}{z}}=\frac{(1+z)}{(z+1)} \times z=z
\end{aligned}
$
Question 10.
The solution of the equation $|z|-z=1+2 i$ is
(a) $\frac{3}{2}-2 \mathrm{i}$
(b) $-\frac{3}{2}+2 \mathrm{i}$
(c) $2-\frac{3}{2} \mathrm{i}$
(d) $2+\frac{3}{2} \mathrm{i}$
Answer:

(a) $\frac{3}{2}-21$
Hint.
$
\begin{aligned}
& \begin{aligned}
|z|-z & =1+2 i \\
\sqrt{x^2+y^2}-(x+i y) & =1+2 i
\end{aligned} \\
& \sqrt{x^2+y^2}-x-i y=1+2 i \\
& y=-2 \quad \Rightarrow \sqrt{x^2+y^2}-x=1 \\
& \sqrt{x^2+4}=(1+x) \\
&
\end{aligned}
$
Squaring on both sides
$
\begin{aligned}
& x^2+4=(1+x)^2 \\
& x^2+4=1+x^2+2 x \quad \Rightarrow 2 x=3 \quad \Rightarrow x=\frac{3}{2} \\
& z=\frac{3}{2}-2 i \\
&
\end{aligned}
$
Question 11.
If $\left|z_1\right|=1,\left|z_2\right|=2,\left|z_3\right|=3$ and $\left|9 z_1 z_2+4 z_1 z_3+z_2 z_3\right|=12$, then the value of $\left|z_1+z_2+z_3\right|$ is
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2
Hint: $\left|z_1+z_2+z_3\right|=2$
Question 12.
If $z$ is a complex number such that $z \in C / R$ and $z+\frac{1}{z} \in R$, then $|z|$ is
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(b) 1
Hint: We have
$
\begin{array}{ll}
z+\bar{z}=2 \operatorname{Re}(z) & \therefore \frac{1}{z}=\bar{z} \text { only when }|z|=1 \\
z+\frac{1}{z}=z+\bar{z}=2 \operatorname{Re}(z) \quad \therefore|z|=1
\end{array}
$

Question 13.
$\mathrm{z}_1, \mathrm{z}_2$ and $\mathrm{z}_3$ are complex numbers such that $\mathrm{z}_1+\mathrm{z}_2+\mathrm{z}_3=0$ and $\left|\mathrm{z}_1\right|=\left|\mathrm{z}_2\right|=\left|\mathrm{z}_3\right|=1$ then $z_1^2+z_2^2+z_3^2$ is
(a) 3
(b) 2
(c) 1
(d) 0
Answer:
(d) 0
Hint:
$
\begin{aligned}
& \left|z_1\right|=1 \quad \Rightarrow\left|z_1\right|^2=1 \\
& \Rightarrow \quad z_1 \bar{z}_1=1 \quad \Rightarrow \bar{z}_1=\frac{1}{z_1} \\
& z_1+z_2+z_3=0 \quad \Rightarrow \frac{1}{\bar{z}_1}+\frac{1}{\bar{z}_2}+\frac{1}{\bar{z}_3}=0 . \\
& \frac{\bar{z}_2 \bar{z}_3+\bar{z}_1 \bar{z}_3+\bar{z}_1 \bar{z}_2}{\bar{z}_1 \bar{z}_2 \bar{z}_3}=0 \quad \Rightarrow \bar{z}_2 \bar{z}_3+\bar{z}_1 \bar{z}_3+\bar{z}_1 \bar{z}_2=0 \\
& \Rightarrow \overline{z_2 z_3}+\overline{z_1 z_3}+\overline{z_1 z_2}=0 \\
&
\end{aligned}
$

Question 14.
If $\frac{z-1}{z+1}$ is purely imaginary, then $|z|$ is
(a) $\frac{1}{2}$
(b) 1
(c) 2
(d) 3
Answer:
(b) 1
Hint. $\frac{z-1}{z+1}$ is purely imaginary. $\quad \Rightarrow \operatorname{Re}\left(\frac{z-1}{z+1}\right)=0$
$
\operatorname{Re}\left(\frac{x+i y-1}{x+i y+1}\right)=0 \quad \Rightarrow \operatorname{Re}\left(\frac{(x-1)+i y}{(x+1)+i y} \times \frac{(x+1)-i y}{x+1-i y}\right)=0
$
Considering only the real part
$
\begin{aligned}
& \frac{(x-1)(x+1)+y^2}{(x+1)^2+y^2}=0 \\
& \Rightarrow|z|=1 \\
&
\end{aligned}
$
Question 15 .
If $z=x+i y$ is a complex number such that $|z+2|=|z-2|$, then the locus of $z$ is
(a) real axis
(b) imaginary axis
(c) ellipse
(d) circle
Answer:
(b) imaginary axis
Hint:
$
|z+2|=|z-2|
$

$
\begin{aligned}
& \Rightarrow|x+i y+2|=|x+i y-2| \\
& \Rightarrow|x+2+i y|^2=|x-2+i y|^2 \\
& \Rightarrow(x+2)^2+y^2=(x-2)^2+y^2 \\
& \Rightarrow x^2+4+4 x=x^2+4-4 x \\
& \Rightarrow 8 x=0 \\
& \Rightarrow x=0
\end{aligned}
$
Question 16.
The principal argument of $\frac{3}{-1+i}$ is
(a) $-\frac{5 \pi}{6}$
(b) $-\frac{2 \pi}{3}$
(c) $-\frac{3 \pi}{4}$
(d) $\frac{-\pi}{2}$
Answer:
(c) $-\frac{3 \pi}{4}$
Hint:
$
\begin{aligned}
\frac{3}{-1+i} & =\frac{3(-1-i)}{(-1+i)(-1-i)}=\frac{3(-1-i)}{2} \\
\alpha & =\tan ^{-1}\left|\frac{y}{x}\right|=\tan ^{-1}|1|=\frac{\pi}{4}
\end{aligned}
$
The complex number lies in III quadrant. $\theta=-(\pi-\alpha)=-\left(\pi-\frac{\pi}{4}\right)=\frac{-3 \pi}{4}$
Question 17.
The principal argument of $\left(\sin 40^{\circ}+i \cos 40^{\circ}\right)^5$ is
(a) $-110^{\circ}$
(b) $-70^{\circ}$
(c) $70^{\circ}$
(d) $110^{\circ}$
Answer:
(a) $-110^{\circ}$
Hint:
$\left(\sin 40^{\circ}+i \cos 40^{\circ}\right)^5$
$=\left(\cos 50^{\circ}+i \sin 50^{\circ}\right)^5$
$=\left(\cos 250^{\circ}+i \sin 250^{\circ}\right)$
$250^{\circ}$ lies in III quadrant.
To find the principal argument the rotation must be in a clockwise direction which coincides with $250^{\circ}$ $\theta=-110^{\circ}$

Question 18.
If $(1+i)(1+2 \mathrm{i})(1+3 \mathrm{i}) \ldots(1+\mathrm{ni})=\mathrm{x}+\mathrm{iy}$, then $2.5 .10 \ldots \ldots\left(1+\mathrm{n}^2\right)$ is
(a) 1
(b) $\mathrm{i}$
(c) $x^2+y^2$
(d) $1+n^2$
Answer:
(c) $x^2+y^2$
Question 19.
If $\omega \neq 1$ is a cubic root of unity and $(1+\omega)^7=A+B \omega$, then $(A, B)$ equal to
(a) $(1,0)$
(b) $(-1,1)$
(c) $(0,1)$
(d) $(1,1)$
Answer:
(d) $(1,1)$
Hint:
$
\begin{aligned}
(1+\omega)^7 & =\mathrm{A}+\mathrm{B} \omega \\
\left(-\omega^2\right)^7 & =\mathrm{A}+\mathrm{B} \omega \\
-\omega^{14} & =\mathrm{A}+\mathrm{B} \omega \\
-\omega^2 & =\mathrm{A}+\mathrm{B} \omega \\
-[-(1+\omega)] & =\mathrm{A}+\mathrm{B} \omega \\
1+\omega & =\mathrm{A}+\mathrm{B} \omega \\
(\mathrm{A}, \mathrm{B}) & =(1,1) \quad\left[\text { Since } 1+\omega+\omega^2=0 ; \omega^2=-(1+\omega)\right]
\end{aligned}
$
Question 20.
The principal argument of the complex number $\frac{(1+i \sqrt{3})^2}{4 i(1-i \sqrt{3})}$ is
(a) $\frac{2 \pi}{3}$
(b) $\frac{\pi}{6}$
(c) $\frac{5 \pi}{6}$
(d) $\frac{\pi}{2}$
Answer:
(d) $\frac{\pi}{2}$
Hint:

$
\begin{aligned}
\frac{(1+i \sqrt{3})^2}{4 i(1-i \sqrt{3})} & =\frac{1-3+2 i \sqrt{3}}{4(i+\sqrt{3})}=\frac{-2+2 i \sqrt{3}}{4(\sqrt{3}+i)}=\frac{2(-1+i \sqrt{3})}{4(\sqrt{3}+i)} \\
& =\frac{i(\sqrt{3}+i)}{2(\sqrt{3}+i)}=\frac{i}{2} \quad \Rightarrow \frac{\pi}{2}
\end{aligned}
$
Question 21.
If $\alpha$ and $\beta$ are the roots of $x^2+x+1=0$, then $\alpha^{2020}+\beta^{2020}$ is
(a) $-2$
(b) $-1$
(c) 1
(d) 2
Answer:
(b) $-1$
Hint:
$
x^2+x+1=0
$
$\alpha$ and $\beta$ are the roots of the equation.
There are the two roots of cube roots of unity except 1 .
$
\begin{aligned}
\alpha & =\omega, \beta=\omega^2 \\
\alpha^{2020}+\beta^{2020} & =\omega^{2020}+\left(\omega^2\right)^{2020}=\omega^{3(673)+1}+\left(\omega^{3(673)+1}\right)^2 \\
& =\omega+\omega^2=-1
\end{aligned}
$
Question 22.
The product of all four values of $\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)^{\frac{3}{4}}$ is
(a) $-2$
(b) $-1$
(c) 1
(d) 2
Answer:
(c) 1

Hint. $\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)^{\frac{3}{4}}=(\cos \pi+i \sin \pi)^{\frac{1}{4}}=\operatorname{cis}\left(\frac{2 k \pi+\pi}{4}\right)$
The roots are $\operatorname{cis}\left(\frac{\pi}{4}\right) \operatorname{cis}\left(\frac{3 \pi}{4}\right) \operatorname{cis}\left(\frac{5 \pi}{4}\right) \operatorname{cis}\left(\frac{7 \pi}{4}\right)$
$
\begin{aligned}
\text { Product of roots } & =\operatorname{cis}\left(\frac{\pi}{4}+\frac{3 \pi}{4}+\frac{5 \pi}{4}+\frac{7 \pi}{4}\right)=\operatorname{cis}(4 \pi) \\
& =\cos 0+i \sin 0=1
\end{aligned}
$
Question 23.
If $\omega \neq 1$ is a cubic root of unity and $\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & -\omega^2-1 & \omega^2 \\ 1 & \omega^2 & \omega^7\end{array}\right|=3 \mathrm{k}$, then $\mathrm{k}$ is equal to
(a) 1
(b) $-1$
(c) $\mathrm{i} \sqrt{3}$
(d) $-1 \sqrt{3}$
Answer:
(d) $-1 \sqrt{3}$

Hint:
$
\begin{aligned}
& \left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & -\omega^2-1 & \omega^2 \\
1 & \omega^2 & \omega^7
\end{array}\right|=3 k \quad \Rightarrow\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & -\omega^2-1 & \omega^2 \\
1 & \omega^2 & \omega
\end{array}\right|=3 k \\
& \left(1+\omega+\omega^2=0\right) \\
& \omega=\frac{-1+i \sqrt{3}}{2} \\
& \left|\begin{array}{ccc}
3 & 0 & 0 \\
1 & -1-\omega^2 & \omega^2 \\
1 & \omega^2 & \omega
\end{array}\right|=3 k\left(\mathrm{R}_1 \rightarrow \mathrm{R}_1+\mathrm{R}_2+\mathrm{R}_3\right) \\
& 3\left(-\omega-\omega^3-\omega^4\right)=3 k \quad \Rightarrow 3(-2 \omega-1)=3 k \\
& -2\left(\frac{-1}{2}+\frac{i \sqrt{3}}{2}\right)-1=k \\
& 1-i \sqrt{3}-1=k \quad \Rightarrow k=-i \sqrt{3} \\
&
\end{aligned}
$
Question 24.
The value of $\left(\frac{1+\sqrt{3} i}{1-\sqrt{3} i}\right)^{10}$ is
(a) $\operatorname{cis} \frac{2 \pi}{3}$
(b) $\operatorname{cis} \frac{4 \pi}{3}$
(c) $-\operatorname{cis} \frac{2 \pi}{3}$
(d) $-\operatorname{cis} \frac{4 \pi}{3}$
Answer:
(a) $\operatorname{cis} \frac{2 \pi}{3}$

Hint:
$
\begin{aligned}
1+i \sqrt{3} & =r(\cos \theta+i \sin \theta) \\
r & =\sqrt{x^2+y^2}=\sqrt{4}=2 \\
\alpha & =\theta=\tan ^{-1}\left|\frac{y}{x}\right|=\tan ^{-1}|| \sqrt{3} \mid=\frac{\pi}{3} \\
(1+i \sqrt{3}) & =2\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right) \\
(1-i \sqrt{3}) & =2\left(\cos \frac{-\pi}{3}+i \sin \frac{-\pi}{3}\right) \\
\left(\frac{1+\sqrt{3} i}{1-\sqrt{3} i}\right)^{10} & =\left[\frac{2 \operatorname{cis}\left(\frac{\pi}{3}\right)}{2 \operatorname{cis}\left(\frac{-\pi}{3}\right)}\right]^{10}=\left[\operatorname{cis}\left(\frac{\pi}{3}+\frac{\pi}{3}\right)\right]^{10}=\left[\operatorname{cis}\left(\frac{2 \pi}{3}\right)\right]^{10}=\operatorname{cis}\left(\frac{20 \pi}{3}\right) \\
& =\operatorname{cis}\left(6 \pi+\frac{2 \pi}{3}\right)=\operatorname{cis}\left(\frac{2 \pi}{3}\right)
\end{aligned}
$
Question 25.
If $\omega=\operatorname{cis} \frac{2 \pi}{3}$, then the number of distinct roots of $\left|\begin{array}{ccc}z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega\end{array}\right|=0$ are
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(a) 1
Hint:

$\begin{aligned}
& \left|\begin{array}{ccc}
z+1 & \omega & \omega^2 \\
\omega & z+\omega & 1 \\
\omega^2 & 1 & z+\omega
\end{array}\right|=0 \\
& \mathrm{R}_1 \rightarrow \mathrm{R}_1+\mathrm{R}_2+\mathrm{R}_3 \text { and } 1+\omega+\omega^2=0 \\
& \left|\begin{array}{ccc}
z & z & z \\
\omega & z+\omega & 1 \\
\omega^2 & 1 & z+\omega
\end{array}\right|=0 \\
& z\left|\begin{array}{ccc}
1 & 1 & 1 \\
\omega & z+\omega^2 & 1 \\
\omega^2 & 1 & z+\omega
\end{array}\right|=0 \quad \Rightarrow z\left|\begin{array}{ccc}
1 & 0 & 0 \\
\omega & z+\omega^2-\omega & 1-\omega \\
\omega^2 & 1-\omega^2 & z+\omega-\omega^2
\end{array}\right|=0 \\
& z\left[\left(z+\omega^2-\omega\right)\left(z+\omega-\omega^2\right)-\left(1-\omega^2\right)(1-\omega)\right]=0 \\
& z\left[\left(z^2-\left(\omega-\omega^2\right)^2\right)\right]-\left[1-\omega-\omega^2+\omega^3\right] \\
& z\left\{z^2-\left(\omega^2+\omega-2\right)-3\right\} \quad \Rightarrow z\left\{z^2\right\}=z^3=0 \\
& z=0 \text { one distinct root } \\
&
\end{aligned}$