Miscellaneous Exercise (Revised) - Chapter 4 Complex Numbers & Quadratic Equations class 11 ncert solutions Maths - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Chapter 4 - Complex Numbers & Quadratic Equations | NCERT Solutions Class 11 Maths

Miscellaneous Exercise Question 1.

Evaluate: $\left[i^{18}+\left(\frac{1}{i}\right)^{25}\right]^3$

Answer.

Given: $\left[i^{18}+\left(\frac{1}{i}\right)^{25}\right]^3$
$=\left[\left(i^2\right)^9+\frac{1}{\left(i^2\right)^{12} i}\right]^3$
$=\left[(-1)^9+\frac{1}{(-1)^{12} i}\right]^3$
$=\left[-1+\frac{1}{i}\right]^3$
$=(-1-i)^3$
$=-(1+i)^3$
$=-\left(1+i^3+3 i+3 i^2\right)$
$=-(1-i+3 i-3)$

$\begin{aligned}
& =-(-2+2 i) \\
& =2-2 i
\end{aligned}$

Miscellaneous Exercise Question 2.

For any two complex numbers $z_1$ and $z_2$ prove that $\operatorname{Re}\left(z_1 z_2\right)=\operatorname{Re} z_1 \operatorname{Re} z_2-\operatorname{Im} z_1 \operatorname{Im} z_2$

Answer.

Let $z_1=a_1+i b_1$ and $z_2=a_2+i b_2$
Then $\operatorname{Re}\left(z_1\right)=a_1, \operatorname{Re}\left(z_2\right)=a_2, \operatorname{Im}\left(z_1\right)=b_1$ and $\operatorname{Im}\left(z_2\right)=b_2$
Now, $z_1 z_2=\left(a_1+i b_1\right)\left(a_2+i b_2\right)$
$
\begin{aligned}
& =a_1 a_2+i a_1 b_2+i a_2 b_1+i^2 b_1 b_2 \\
& =\left(a_1 a_2-b_1 b_2\right)+i\left(a_1 b_2+a_2 b_1\right) \\
& \therefore \operatorname{Re}\left(z_1 z_2\right)=a_1 a_2-b_1 b_2 \\
& =\operatorname{Re} z_1 \operatorname{Re} z_2-\operatorname{Im} z_1 \operatorname{Im} z_2
\end{aligned}
$
Miscellaneous Exercise Question 3.

Reduce $\left[\frac{1}{1-4 i}-\frac{2}{1+i}\right]\left[\frac{3-4 i}{5+i}\right]$ to the standard form.

Answer.

Here: $\left[\frac{1}{1-4 i}-\frac{2}{1+i}\right]\left[\frac{3-4 i}{5+i}\right]$

$\begin{aligned}
& =\left[\frac{1+i-2+8 i}{(1-4 i)(1+i)}\right]\left[\frac{3-4 i}{5+i}\right] \\
& =\left[\frac{-1+9 i}{1+i-4 i-4 i^2}\right]\left[\frac{3-4 i}{5+i}\right] \\
& =\left[\frac{-1+9 i}{5-3 i}\right]\left[\frac{3-4 i}{5+i}\right] \\
& =\frac{-3+4 i+27 i-36 i^2}{25+5 i-15 i-3 i^2} \\
& =\frac{33+31 i}{28-10 i} \times \frac{28+10 i}{28+10 i}
\end{aligned}$

$\begin{aligned}
& =\frac{924+330 i+868 i+310 i^2}{(28)^2-(10 i)^2} \\
& =\frac{614+1198 i}{784+100} \\
& =\frac{2(307+599 i)}{884} \\
& =\frac{307+599 i}{442}
\end{aligned}$

Miscellaneous Exercise Question 4.

If $x-i y=\sqrt{\frac{a-i b}{c-i d}}$ prove that $\left(x^2+y^2\right)^2=\frac{a^2+b^2}{c^2+d^2}$.

Answer.

Given: $x-i y=\sqrt{\frac{a-i b}{c-i d}}$
Squaring both sides, we get
$
\begin{aligned}
& \Rightarrow(x-i y)^2=\frac{a-i b}{c-i d} \\
& \Rightarrow\left|(x-i y)^2\right|=\left|\frac{a-i b}{c-i d}\right| \\
& \left.\Rightarrow|(x-i y)|(x-\dot{y})|=| \frac{a-i b}{c-i d} \right\rvert\, \\
& \Rightarrow\left(\sqrt{x^2+y^2}\right)\left(\sqrt{x^2+y^2}\right)=\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}} \\
& \Rightarrow\left(x^2+y^2\right)=\sqrt{\frac{a^2+b^2}{c^2+d^2}}
\end{aligned}
$

Squaring both sides, we get

$
\Rightarrow\left(x^2+y^2\right)^2=\frac{a^2+b^2}{c^2+d^2}
$
Miscellaneous Exercise Question 5.

If $z_1=2-i, z_2=1+i$, find $\left|\frac{z_1+z_2+1}{z_1-z_2+i}\right|$

Answer.

Here $z_1=2-i$ and $z_2=1+i$
$
=\left|\frac{z_1+z_2+1}{z_1-z_2+i}\right|
$

$
\begin{aligned}
& =\left|\frac{2-i+1+i+1}{2-i-1-i+i}\right| \\
& =\left|\frac{4}{1-i}\right| \\
& =\frac{|4|}{|1-i|}=\frac{4}{\sqrt{1^2+1^2}} \\
& =\frac{4}{\sqrt{2}}=2 \sqrt{2}
\end{aligned}
$
Miscellaneous Exercise Question 6.

If $a+i b=\frac{(x+i)^2}{2 x^2+1}$, prove that $a^2+b^2=\frac{\left(x^2+1\right)^2}{\left(2 x^2+1\right)^2}$

Answer.

Here $a+i b=\frac{(x+i)^2}{2 x^2+1}$
$
=\frac{x^2+i^2+2 i x}{2 x^2+1}=\frac{x^2-1}{2 x^2+1}+i \frac{2 x}{2 x^2+1}
$

Comparing both sides, we have
$
\begin{aligned}
& a=\frac{x^2-1}{2 x^2+1} \text { and } b=\frac{2 x}{2 x^2+1} \\
& \therefore a^2+b^2=\left(\frac{x^2-1}{2 x^2+1}\right)^2+\left(\frac{2 x}{2 x^2+1}\right)^2
\end{aligned}
$

$\begin{aligned}
& =\frac{\left(x^2-1\right)^2+(2 x)^2}{\left(2 x^2+1\right)^2} \\
& =\frac{x^4+1-2 x^2+4 x^2}{\left(2 x^2+1\right)^2} \\
& \Rightarrow a^2+b^2=\frac{x^4+1+2 x^2}{\left(2 x^2+1\right)^2}
\end{aligned}$

$=\frac{\left(x^2+1\right)^2}{\left(2 x^2+1\right)^2}$ Proved
Miscellaneous Exercise Question 7.

Let $z_1=2-i, z_2=-2+i$ find:
(i) $\operatorname{Re}\left(\frac{z_1 z_2}{\overline{z_1}}\right)$
(ii) I $m\left(\frac{1}{z_1 \overline{z_1}}\right)$

Answer.

Here $z_1=2-i$ and $z_2=-2+i$
$
\therefore \overline{z_1}=2+i
$
$
\begin{aligned}
& \text { (i) } z_1 z_2=(2-i)(-2+i)=-4+2 i+2 i-i^2=-3+4 i \\
& \therefore \frac{z_1 z_2}{z_1}=\frac{-3+4 i}{2+i} \times \frac{2-i}{2-i}=\frac{-6+3 i+8 i-4 i^2}{4-i^2}=\frac{-2+11 i}{5}=\frac{-2}{5}+\frac{11}{5} i \\
& \therefore \operatorname{Re}\left(\frac{z_1 z_2}{z_1}\right)=\frac{-2}{5}
\end{aligned}
$

$
\begin{aligned}
& \text { (ii) } \frac{1}{z_1 \overline{z_1}}=\frac{1}{(2-i)(2+i)}=\frac{1}{4-i^1}=\frac{1}{5} \\
& \therefore \operatorname{Im}\left(\frac{1}{z_1 \overline{z_1}}\right)=0
\end{aligned}
$
Miscellaneous Exercise Question 8.

Find the real numbers $x$ and $y$ if $(x-i y)(3+5 i)$ is the conjugate of $-6-24 i$.

Answer.

Here $\overline{-6-24 i}=-6+24 i$

Now $(x-i y)(3+5 i)=-6+24 i$
$
\begin{aligned}
& \Rightarrow 3 x+5 x i-3 y i-5 y i^2=-6+24 i \\
& \Rightarrow(3 x+5 y)+(5 x-3 y) i=-6+24 i
\end{aligned}
$

Comparing both sides, we have $3 x+5 y=-6$ and $5 x-3 y=24$
Solving both equations, we have $x=3$ and $y=-3$
Miscellaneous Exercise Question 9.

Find the modulus of $\frac{1+i}{1-i}-\frac{1-i}{1+i}$.

Answer.

Here $\left|\frac{1+i}{1-i}-\frac{1-i}{1+i}\right|=\left|\frac{(1+i)^2-(1-i)^2}{(1-i)(1+i)}\right|$
$
\begin{aligned}
& =\left|\frac{1+i^2+2 i-1-i^2+2 i}{1-i^2}\right| \\
& =\left|\frac{4 i}{2}\right|=|2 i|=\sqrt{4}=2
\end{aligned}
$
Miscellaneous Exercise Question 10.

If $(x+i y)^3=u+i v$, then show that $\frac{u}{x}+\frac{v}{y}=4\left(x^2-y^2\right)$.

Answer.

Given: $(x+i y)^3=u+i v$
$
\begin{aligned}
& \Rightarrow x^3+i^3 y^3+3 x^2 y i+3 x y^2 i^2=u+i v \\
& \Rightarrow\left(x^3-3 x y^2\right)+\left(3 x^2 y-y^3\right) i=u+i v
\end{aligned}
$

Comparing both sides, we have
$
u=x\left(x^2-3 y^2\right) \text { and } v=y\left(3 x^2-y^2\right)
$

Now, $\frac{u}{x}+\frac{v}{y}=\frac{x\left(x^2-3 y^2\right)}{x}+\frac{y\left(3 x^2-y^2\right)}{y}$
$
=x^2-3 y^2+3 x^2-y^2=4 x^2-4 y^2=4\left(x^2-y^2\right)
$
Miscellaneous Exercise Question 11.

If $\alpha$ and $\beta$ are different complex numbers with $|\beta|=1$ then find $\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|$.

Answer.

Let $\mathrm{a}=\mathrm{a}+\mathrm{ib}$ and $\beta=\mathrm{x}+\mathrm{iy}$
It is given that, $|\beta|=1$
$
\begin{aligned}
& \therefore \sqrt{x^2+y^2}=1 \\
& \Rightarrow \mathrm{x}^2+\mathrm{y}^2=1 \text {..(i) }
\end{aligned}
$
$
\begin{aligned}
& \left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|=\left|\frac{(x+i y)-(a+i b)}{1-(a-i b)(x+i y)}\right| \\
& =\left|\frac{(x-a)+i(y-b)}{1-(a x+a i y-i b x+b y)}\right| \\
& =\left|\frac{(x-a)+i(y-b)}{(1-a x-b y)+i(b x-a y)}\right|
\end{aligned}
$

$\begin{aligned}
& =\frac{\sqrt{(x-a)^2+(y-b)^2}}{\sqrt{(1-a x-b y)^2+(b x-a y)^2}} \\
& =\frac{\sqrt{x^2+a^2-2 a x+y^2+b^2-2 b y}}{\sqrt{1+a^2 x^2+b^2 y^2-2 a x+2 a b x y-2 b y+b^2 x^2+a^2 y^2-2 a b x y} y} \\
& =\frac{\sqrt{\left(x^2+y^2\right)+a^2+b^2-2 a x-2 b y}}{\sqrt{1+a^2\left(x^2+y^2\right)+b^2\left(y^2+x^2\right)-2 a x-2 b y}} \\
& =\frac{\sqrt{1+a^2+b^2-2 a x-2 b y}}{\sqrt{1+a^2+b^2-2 a x-2 b y}}(\text { Using (i) ) }
\end{aligned}$

$
\begin{aligned}
& =1 \\
& \therefore\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|=1
\end{aligned}
$
Miscellaneous Exercise Question 12.

Find the number of non-zero integral solutions of the equation $|1-i|^x=2^x$.

Answer.

Here $|1-i|^x=2^x$
$
\begin{aligned}
& \Rightarrow\left|\sqrt{1^2+(-1)^2}\right|^x=2^x \\
& \Rightarrow[\sqrt{2}]^x=2^x \\
& \Rightarrow 2^{\frac{x}{2}}=2^x \\
& \Rightarrow \frac{x}{2}=x \\
& \Rightarrow \frac{x}{2}-x=0 \\
& \Rightarrow \frac{-x}{2}=0 \\
& \Rightarrow x=0
\end{aligned}
$

Miscellaneous Exercise Question 13.

If $(a+i b)(c+i d)(e+i f)(g+i h)=A+i B$ then show that:
$
\left(a^2+b^2\right)\left(c^2+d^2\right)\left(e^2+f^2\right)\left(g^2+h^2\right)=\mathrm{A}^2+\mathrm{B}^2 \text {. }
$

Answer.

Given: $(a+i b)(c+i d)(e+i f)(g+i h)=\mathrm{A}+i \mathrm{~B}$
Taking modulus on both sides,
$
|(a+i b)(c+i d)(e+i f)(g+i h)|=|A+i B|
$

$
\begin{aligned}
& \Rightarrow|(a+i b)|(c+i d)|(e+i f)|(g+i h)|=| \mathrm{A}+i \mathrm{~B} \mid \\
& \Rightarrow\left(\sqrt{a^2+b^2}\right)\left(\sqrt{c^2+d^2}\right)\left(\sqrt{e^2+f^2}\right)\left(\sqrt{g^2+h^2}\right)=\left(\sqrt{\mathrm{A}^2+\mathrm{B}^2}\right)
\end{aligned}
$

Squaring both sides, we get
$
\left(a^2+b^2\right)\left(c^2+d^2\right)\left(e^2+f^2\right)\left(g^2+h^2\right)=\mathrm{A}^2+\mathrm{B}^2
$
Miscellaneous Exercise Question 14.

If $\left[\frac{1+i}{1-i}\right]^m=1$, then find the least positive integral value of $m$

Answer.

Given: $\left[\frac{1+i}{1-i}\right]^m=1$
$
\begin{aligned}
& \Rightarrow\left[\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right]^m=1 \\
& \Rightarrow\left[\frac{(1+i)^2}{1-i^2}\right]^m=1 \\
& \Rightarrow\left[\frac{1+i^2+2 i}{1+1}\right]^m=1
\end{aligned}
$

$\begin{aligned}
& \Rightarrow\left[\frac{1-1+2 i}{2}\right]^m=1 \\
& \Rightarrow\left[\frac{2 i}{2}\right]^m=1 \\
& \Rightarrow[i]^m=1 \\
& \Rightarrow[i]^m=1^4
\end{aligned}$

$\Rightarrow m=4$