Exercise 2.9-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

Additional Problems
Question 1.
$\left(\frac{1-i}{1+i}\right)^{106}=a+i b$ then $(a, b)$ is
(a) $(2,-1)$
(b) $(1,0)$
(c) $(0,1)$
(d) $(-1,2)$
Solution:
(b) $(1,0)$
Question 2.
If $\left(\frac{1+i}{1-i}\right)^3-\left(\frac{1-i}{1+i}\right)^3=x+i y$ the $(x, y)=$
(a) $(0,2)$
(b) $(-2,0)$
(c) $(0,-2)$
(d) $(-2,0)$
Solution:
(c) $(0,-2)$

Question 3.
The point represented by the complex number $2-i$ is rotated about origin through an angle of $\frac{\pi}{2}$. in the clockwise direction, the new position of point is ......
(a) $1+2 \mathrm{i}$
(b) $-1-2 \mathrm{i}$
(c) $2+\mathrm{i}$
(d) $-1+2 i$
Solution:
(b) $-1-2 i$
Hint.
Let $z=2-i$. Let the new position of point when the point represented by the complex number $z=2-i$ is rotated about origin through an angle of $\frac{\pi}{2}$ in the clockwise direction be denoted by $z_1$.
Now,
$
\begin{aligned}
\left|z_1\right| & =|z|=\sqrt{(2)^2+(-1)^2}=\sqrt{4+1}=\sqrt{5} \\
\frac{z_1}{z} & =\frac{\left|z_1\right|}{|z|} e^{-\frac{\pi}{2}} \Rightarrow z_1=\frac{z \sqrt{5}}{\sqrt{5}} e^{-\frac{\pi}{2}} \\
z_1 & =\frac{(2-i) \sqrt{5}}{\sqrt{5}}\left\{\cos \left(-\frac{\pi}{2}\right)+i \sin \left(-\frac{\pi}{2}\right)\right\} \\
& =(2-i)\left(\cos \frac{\pi}{2}-i \sin \frac{\pi}{2}\right)=(2-i)(0-i)=-2 i+i^2=-2 i-1
\end{aligned}
$
Fill in the blanks of the following:

Question 4.
For any two complex numbers $z_1, z_2$ and any real number $a, b$,
Solution:
$
\begin{aligned}
& \text { Since } \quad\left|z_1+z_2\right|^2=\left|z_1\right|^2+\left|z_2\right|^2+2 \operatorname{Re}\left(z_1 \bar{z}_2\right) \\
& \text { and } \quad\left|z_1-z_2\right|^2=\left|z_1\right|^2+\left|z_2\right|^2-2 \operatorname{Re}\left(z_1 \bar{z}_2\right) \\
& \left|a z_1-b z_2\right|^2+\left|b z_1+a z_2\right|^2=\left|a z_1\right|^2+\left|b z_2\right|^2-2 \operatorname{Re}\left(a z_1 b \bar{z}_2\right)+\left|b z_1\right|^2+\left|a z_2\right|^2+2 \operatorname{Re}\left(b z_1 a \bar{z}_2\right) \\
& =a^2\left|z_1\right|^2+b^2\left|z_2\right|^2+b^2\left|z_1\right|^2+a^2\left|z_2\right|^2 \\
& =\left(a^2+b^2\right)\left|z_1\right|^2+\left(b^2+a^2\right)\left|z_2\right|^2 \\
& =\left(a^2+b^2\right)\left(\left|z_1\right|^2+\left|z_2\right|^2\right) \\
&
\end{aligned}
$

Question 5.
Multiplicative inverse of $1+i$ is .....
Solution:
Multiplicative inverse of $1+i=\frac{1}{1+i} \times \frac{1-i}{1-i}=\frac{1-i}{1-i^2}=\frac{1-i}{1+1}=\frac{1-i}{2}=\frac{1}{2}-\frac{i}{2}$
Question 6.
$\arg (z)+\arg (\bar{z})(\bar{z} \neq 0)$ is
Solution:
If $|z|=4$ and $\arg (z)=\frac{5 \pi}{6}$, then $z=$
Question 7.
If $|z|=4$ and $\arg (z)=\frac{5 \pi}{6}$, then $z=$
Solution:
Let
$
\begin{aligned}
z & =r(\cos \theta+i \sin \theta), \text { then } \\
r & =|z|=4 ; \arg z=\frac{5 \pi}{6} \Rightarrow \theta=\frac{5 \pi}{6} \\
z & =4\left(\cos \frac{5 \pi}{6}+i \sin \frac{5 \pi}{6}\right)=4\left[\cos \left(\pi-\frac{\pi}{6}\right)+i \sin \left(\pi-\frac{\pi}{6}\right)\right] \\
& =4\left(-\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)=4\left(-\frac{\sqrt{3}}{2}+\frac{i}{2}\right)=-2 \sqrt{3}+2 i
\end{aligned}
$

Question 8.
State true or false for the following:
(i) The order relation is defined on the set of complex numbers.
Solution:
The given statement is false because the order relation "greater than" and "less than" are not defined on the set of complex numbers.
(ii) For any complex number 2, the minimum value of $|z|+|z-1|$ is 1 .
Solution:
Let $\mathrm{z}=\mathrm{x}+$ iy be any complex number.
Let $z=r(\cos \theta+i \sin \theta)$. Where $\theta$ is the argument or amplitude of $z$.
$
\begin{aligned}
\bar{z} & =r(\cos \theta-i \sin \theta) \\
& =r[\cos (-\theta)+i \sin (-\theta)](-\theta) \text { is the argument of } \bar{z} . \\
\therefore \quad \arg z+\arg (\bar{z}) & =\theta+(-\theta)=0
\end{aligned}
$
When $x=0, y=0$, the value of $|z|+|z-1|$ is minimum and the minimum value.
$
=0+\sqrt{(0-1)^2+0}=0+\sqrt{1}=1
$
Hence, the given statement is true.
(iii) 2 is not a complex number.
Solution:
It is a false statement because $2=2+\mathrm{i}(0)$, which is of the form $\alpha+\mathrm{ib}$, and the number of the form $\mathrm{a}+\mathrm{ib}$, where $\mathrm{a}$ and $\mathrm{b}$ are real numbers is called a complex number.
(iv) The locus represented by $|z-1|=|z-i|$ is a line perpendicular to the join of $(1,0)$ and $(0,1)$.
Solution:

Equation of a straight line passing through $(1,0)$ and $(0,1)$ is
$
\begin{aligned}
& y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right) \Rightarrow y-0=\frac{1-0}{0-1}(x-1) \\
& \Rightarrow \quad y=-(x-1) \quad \Rightarrow x+y=1 \quad \Rightarrow y=-x+1 \\
& \text { Again } \quad|z-1|=|z-i| \\
& \text { Let } \quad z=x+i y \quad \therefore|x+i y-1|=|x+i y-i| \\
& \Rightarrow|(x-1)+i y|=|x+i(y-1)| \quad \Rightarrow \sqrt{(x-1)^2+y^2}=\sqrt{x^2+(y-1)^2} \\
& \Rightarrow \quad(x-1)^2+y^2=x^2+(y-1)^2 \\
& \Rightarrow x^2-2 x+1+y^2=x^2+y^2-2 y+1 \Rightarrow-2 x=-2 y \\
& \Rightarrow \quad y=x \\
&
\end{aligned}
$
Now, the lines $y=-x+1$ and $y=x$ are perpendicular to each other, so the given statement is true.

Question 9.
If $p+i q=\frac{a+i b}{a-i b}$ then $p^2+q^2=$
Solution:
1
Question 10.
If $-\bar{z}$ lies in the third quadrant, then $z$ lies in the
(a) first quadrant
(b) second quadrant
(c) third quadrant
(d) fourth quadrant
Solution:
(d) fourth quadrant
Hint:
$-\bar{z}$ lies in III quadrant. i.e., $-\bar{z}=-x-i y ; \bar{z}=x+i y$
So, $z=x-i y$ which is in IV quadrant
Question 11.
If $a=\cos \alpha-i \sin \alpha, b=\cos \beta-i \sin \beta, c=\cos \gamma-i \sin \gamma$, then $\frac{\left(a^2 c^2-b^2\right)}{a b c}$ is
(a) $\cos 2(\alpha-\beta+\gamma)+i \sin 2(\alpha-\beta+\gamma)$
(b) $-2 \cos (\alpha-\beta+\gamma)$
(c) $-2$ i $\sin (\alpha-\beta+\gamma)$
(d) $2 \cos (\alpha-\beta+\gamma)$
Solution:
(c) $-2$ i $\sin (\alpha-\beta+\gamma)$

Hint: $\alpha \beta \gamma$
$
\begin{aligned}
\frac{a^2 c^2-b^2}{a b c} & =\frac{a c}{b}-\frac{b}{a c} \cdot \text { Here } a=e^{-i a} ; b=e^{i \beta} ; c=e^{-i \gamma} \\
\frac{a c}{} & =\frac{e^{-i \alpha} \cdot e^{-i \gamma}}{e^{-i \beta}}=e^{-i(\alpha-\beta+\gamma)}=\cos (\alpha-\beta+\gamma)-i \sin (\alpha-\beta+\gamma) \\
\frac{b}{a c} & =\frac{1}{a c}=\cos (\alpha-\beta+\gamma)+i \sin (\alpha-\beta+\gamma) \\
\therefore \quad \therefore \quad \frac{a c}{b}-\frac{b}{a c} & =\cos (\alpha-\beta+\gamma)-i \sin (\alpha-\beta+\gamma)-[\cos (\alpha-\beta+\gamma)+i \sin (\alpha-\beta+\gamma)] \\
& =-2 i \sin (\alpha-\beta+\gamma)
\end{aligned}
$

Question 12.
$z_1=4+5 i, z_2=-3+2 i$, then $\frac{z_1}{z_2}$ is
(a) $\frac{2}{13}-\frac{22}{13} i$
(b) $\frac{-2}{13}+\frac{22}{13} i$
(c) $\frac{-2}{13}-\frac{23}{13} i$
(d) $\frac{2}{13}+\frac{22}{13} i$
Solution:
(c) $\frac{-2}{13}-\frac{23}{13} i$
Hint:
$
\begin{aligned}
z_1 & =4+5 i ; z_2=-3+2 i \\
\frac{z_1}{z_2} & =\frac{4+5 i}{-3+2 i}=\frac{4+5 i}{-3+2 i} \times \frac{-3-2 i}{-3-2 i}=\frac{-12-8 i-15 i-10 i^2}{9+4} \\
& =\frac{-12-23 i+10}{13}=\frac{-2-23 i}{13}=\frac{-2}{13}-\frac{23}{13} i
\end{aligned}
$
Question 13.
The conjugate of $i^{13}+i^{14}+i^{15}+i^{16}$ is
(a) 1
(b) $-1$
(c) 0
(d)- $\mathrm{i}$
Solution:
(c) 0
Hint:
$i^{13}+i^{14}+i^{15}+i^{16}=0$
Conjugate of 0 is $0-i 0=0$

Question 14.
If $-i+2$ is one root of the equation $\mathrm{ax}^2-b x+c=0$, then the other root is
(a) $-i-2$
(b) $i-2$
(c) $2+\mathrm{i}$
(d) $2 i+i$
Solution:
(c) $2+i$
Hint:
Complex roots occur in pairs when $-i+2$ is one root, i.e., when $2-i$ is a root, the other root is $2+i$
Question 15.
The equation having $4-3 i$ and $4+3 i$ as roots is ......
(a) $x^2+8 x+25=0$
(b) $x^2+8 x-25=0$
(c) $x^2-8 x+25=0$
(d) $x^2-8 x-25=0$
Solution:
(c) $x^2-8 x+25=0$
Hint.
$
\begin{aligned}
\alpha & =4-3 i, \beta=4+3 i ; \alpha+\beta=8 \\
\alpha \beta & =4^2+3^2=25
\end{aligned}
$
Equation is $x^2-(8) x+25=0 ; \quad$ i.e., $x^2-8 x+25=0$
Question 16.
If $\frac{1-i}{1+i}$ is a root of the equation $a x^2+b x+1=0$. where $a, b$ are real, then $(a, b)$ is
(a) $(1,1)$
(b) $(1,-1)$
(c) $(0,1)$
(d) $(1,0)$
Solution:
(d) $(1,0)$

Hint:
$\frac{1-i}{1+i}=\frac{1-i}{1+i} \times \frac{1-i}{1+i}=\frac{1+i^2-2 i}{2}=-i$. When $-i$ is a root the other root is $+i$.
Here, the given quadratic equation is $a x^2+b x+1=0$ and the roots are $-i$ and $+i$.
Sum of the roots $=\frac{-b}{a}=0 \quad \Rightarrow b=0$
Product of the roots $=\frac{1}{a}=1 \quad \Rightarrow a=1$
$
\therefore \quad(a, b)=(1,0)
$

Question 17.
If $-i+3$ is a root of $x^2-6 x+k=0$, then the value of $k$ is
(a) 5
(b) $\sqrt{5}$
(c) $\sqrt{10}$
(d) 10
Solution:
(d) 10
Hint:
$\alpha=-i+3=3-7 \Rightarrow$ The other root is $\beta=3+i$
Product of the roots $=\alpha \beta=(3-i)(3+i)=10$
Product of the roots of $\mathrm{x}^2-6 \mathrm{x}+\mathrm{k}=0$ is $\mathrm{k} \Rightarrow \mathrm{k}=10$
Question 18.
If $\omega$ is a cube root of unity, then the value of $\left(1-\omega+\omega^2\right)^4+\left(1+\omega-\omega^2\right)^4$ is $\ldots \ldots$
(a) 0
(b) 32
(c) $-16$
(d) $-32$
Solution:
(c) $-16$
Hint.
$
\begin{aligned}
\left(1-\omega+\omega^2\right)^4+(1+\omega & \left.-\omega^2\right)^4 \\
& =\left[\left(1+\omega^2\right)-\omega\right]^4+\left[(1+\omega)-\omega^2\right]^4=(-2 \omega)^4+\left(-2 \omega^2\right)^4 \\
& =16 \omega^4+16 \omega^8 \\
& =16 \omega+16 \omega^2=16\left(\omega+\omega^2\right)=16(-1)=-16
\end{aligned}
$
Question 19.
If $\omega$ is a cube root of unity, then the value of $(1-\omega)\left(1-\omega^2\right)\left(1-\omega^4\right)\left(1-\omega^8\right)$ is $\ldots \ldots$
(a) 9
(b) $-9$
(c) 16
(d) 32
Solution:
(a) 9
Hint.
$
\begin{aligned}
(1-\omega)\left(1-\omega^2\right)\left(1-\omega^4\right) & \left(1-\omega^8\right) \\
= & (1-\omega)\left(1-\omega^2\right)(1-\omega)\left(1-\omega^2\right) \\
= & {\left[(1-\omega)\left(1-\omega^2\right)\right]^2=\left(1-\omega^2-\omega+\omega^3\right)^2 } \\
= & {\left[2-\left(\omega+\omega^2\right)\right]^2=[2-(-1)]^2=(2+1)^2=3^2=9 }
\end{aligned}
$

Question 20.
If $\frac{z-1}{z+1}$ is purely imaginary, then
(a) $|z|=1$
(b) $|z|>1$
(c) $|z|<1$
(d) None of these.
Solution:
(a) $|z|=1$