NCERT Solutions Complex Numbers and Quadratic Equations Miscellaneous Exercise

NCERT Solutions for Complex Numbers and Quadratic Equations Miscellaneous Exercise provide detailed and step-by-step solutions to important problems based on complex numbers, quadratic equations, roots of equations, algebraic identities, and related concepts. These solutions help students understand problem-solving techniques, strengthen conceptual clarity, and improve analytical skills required for board examinations and competitive exams. The chapter is an important part of Class 11 Mathematics and plays a significant role in preparing for CBSE, JEE Main, JEE Advanced, NDA, and other competitive examinations.

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

Solution : Given

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

Rewrite powers as multiples of 4:

$i^{18}=i^{4\cdot 4+2}=(i^4{)}^4\cdot i^2$

$(1/i{)}^{25}=1/i^{25}=1/(i^{4\cdot 6+1})=1/((i^4{)}^6\cdot i)$

Since ($i^4=1$) and ($i^2=-1$):

$[i^{18}+(1/i{)}^{25}{]}^3=[-1+1/i{]}^3$

Rationalize (1/i):

$\frac{1}{i}\cdot \frac{i}{i}=\frac{i}{i^2}=-i$

Hence:

$[-1+1/i{]}^3=(-1-i{)}^3$

Expand the cube:

$(-1-i{)}^3=(-1{)}^3+3(-1{)}^2(-i)+3(-1)(-i{)}^2+(-i{)}^3$

Compute powers of (i) (($i^2=-1,i^3=-i$)):

$(-1-i{)}^3=-1-3i+3+i=2-2i$

Master related concepts such as SETS NCERT Solutions Exercise 1.4 for Class 11 Maths

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Q.2 : Prove that $Re(z_1{z}_2)=Re{z}_{1}Re{z}_2-Im{z}_{1}Im{z}_2$

Solution : Let

$z_1=x_1+i{y}_1,z_2=x_2+i{y}_2$

Multiply:

$z_1{z}_2=(x_1+i{y}_1)(x_2+i{y}_2)$

Step by step:

$z_1{z}_2=x_1{x}_2+i{x}_1{y}_2+i{x}_2{y}_1+i^2{y}_1{y}_2$

Since ($i^2=-1$):

$z_1{z}_2=(x_1{x}_2-y_1{y}_2)+i(x_1{y}_2+x_2{y}_1)$

Hence:

$Re(z_1{z}_2)=x_1{x}_2-y_1{y}_2$

Also:

$Re{z}_{1}Re{z}_2-Im{z}_{1}Im{z}_2=x_1{x}_2-y_1{y}_2$

Hence proved.

$Re(z_1{z}_2)=Re{z}_{1}Re{z}_2-Im{z}_{1}Im{z}_2$

Note : Similar property :

$Im(z_1{z}_2)=(x_1{y}_2+x_2{y}_1)$

$Re{z}_{1}Im{z}_2+Re{z}_{2}Im{z}_1=(x_1{y}_2+x_2{y}_1)$

$Im(z_1{z}_2)=Re{z}_{1}Im{z}_2+Re{z}_{2}Im{z}_1$

Students should also study SETS NCERT Solutions Miscellaneous Exercise for Class 11 Maths : Maths Anand Classes

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Q.3 : Reduce to standard form : $(\frac{1}{1-4i}-\frac{2}{i+1})\frac{3-4i}{5+i}$

Solution

Step 1: Compute the bracket:

$\frac{1}{1-4i}-\frac{2}{i+1}=\frac{-1+9i}{5-3i}$

Step 2: Multiply by $(\frac{3-4i}{5+i})$

$\frac{-1+9i}{5-3i}\cdot \frac{3-4i}{5+i}$

Step 3: Split numerator and denominator:

Numerator:
$(-1+9i)(3-4i)=$
$=(-1\cdot 3)+(-1\cdot -4i)+(9i\cdot 3)+(9i\cdot -4i)$
$=33+31i$
Denominator:
$(5-3i)(5+i)=$
$=(5\cdot 5)+(5\cdot i)+(-3i\cdot 5)+(-3i\cdot i)$
$=28-10i$

Step 4: Rationalize:

$\frac{33+31i}{28-10i}\cdot \frac{28+10i}{28+10i}=\frac{614+1198i}{884}$

Step 5: Simplify:

$\frac{614+1198i}{884}=\frac{307}{442}+i\frac{599}{442}$

Build strong concepts by studying Sets Exercise 1.2 NCERT Solutions Class 11 Maths : Maths Anand Classes

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Q.4 : If $x-iy=\sqrt{\frac{a-ib}{c-id}}$ prove that $(x^2+y^2{)}^2=\frac{a^2+b^2}{c^2+d^2}.$

Solution

Let $u=a-ib$ and $v=c-id$. Then

$x-iy=\sqrt{\frac{u}{v}}$

$x-iy=\frac{\sqrt{u}}{\sqrt{v}}.$

Take modulus of both sides. Using $|\sqrt{Z}|=\sqrt{|Z|}$ and $|Z_1/Z_2|=|Z_1|/|Z_2|$ we get

$|x-iy|=\frac{|\sqrt{u}|}{|\sqrt{v}|}=\frac{\sqrt{|u|}}{\sqrt{|v|}}=\sqrt{\frac{|u|}{|v|}}.$

Square both sides:

$|x-iy{|}^2=\frac{|u|}{|v|}.$

But $|x-iy{|}^2=x^2+y^2$, and

$|u|=|a-ib|=\sqrt{a^2+b^2},|v|=|c-id|=\sqrt{c^2+d^2}.$

Therefore

$x^2+y^2=\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}=\sqrt{\frac{a^2+b^2}{c^2+d^2}}.$

Finally square both sides to obtain the required identity:

$(x^2+y^2{)}^2=\frac{a^2+b^2}{c^2+d^2}.$

Important exam-related topics include NCERT Solutions for SETS Exercise 1.3 Class 11 Maths : Maths Anand Classes

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Q.5 : If $z_1=2-i$, $z_2=1+i$, find $\left|{\displaystyle \frac{z_1+z_2+1}{z_1-z_2+1}}\right|$

Solution:

Given, $z_1=2-i$, $z_2=1+i$

$\left|{\displaystyle \frac{z_1+z_2+1}{z_1-z_2+1}}\right|=\left|{\displaystyle \frac{(2-i)+(1+i)+1}{(2-i)-(1+i)+1}}\right|$

$=\left|{\displaystyle \frac{4}{2-2i}}\right|=\left|{\displaystyle \frac{4}{2(1-i)}}\right|=|{\displaystyle \frac{2}{1-i}}\times {\displaystyle \frac{1+i}{1+i}}|$

$=\left|{\displaystyle \frac{2(1+i)}{1^2-i^2}}\right|=\left|{\displaystyle \frac{2(1+i)}{1+1}}\right|[i^2=-1]$

$=\left|{\displaystyle \frac{2(1+i)}{2}}\right|=|1+i|=\sqrt{1^2+1^2}=\sqrt{2}$

Hence, the value of $\left|{\displaystyle \frac{z_1+z_2+1}{z_1-z_2+1}}\right|$ is $\sqrt{2}$.

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Q.6 : If $a+ib=\frac{(x+i{)}^2}{2x^2+1}$ prove that $a^2+b^2=\frac{(x^2+1{)}^2}{(2x^2+1{)}^2}.$

Solution : Given

$a+ib=\frac{(x+i{)}^2}{2x^2+1}$

$a+ib=\frac{x^2+i^2+2xi}{2x^2+1}$

$a+ib=\frac{x^2-1+i2x}{2x^2+1}$

$a+ib=\frac{x^2-1}{2x^2+1}+i\frac{2x}{2x^2+1}$

On comparing real and imaginary parts:

$a=\frac{x^2-1}{2x^2+1},b=\frac{2x}{2x^2+1}.$

Now compute $a^2+b^2$:

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

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

Simplify numerator:

$(x^2-1{)}^2+(2x{)}^2=x^4-2x^2+1+4x^2$
$(x^2-1{)}^2+(2x{)}^2=x^4+2x^2+1$
$(x^2-1{)}^2+(2x{)}^2=(x^2+1{)}^2$

Hence,

$a^2+b^2=\frac{(x^2+1{)}^2}{(2x^2+1{)}^2}.$

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Q.7 : Let $z_1=2-i,z_2=-2+i.$ Find : (i) $\mathrm{Re}\left({\displaystyle \frac{z_1{z}_2}{\overline{z_1}}}\right)$ (ii) $\mathrm{Im}\left({\displaystyle \frac{1}{z_1\overline{z_2}}}\right)$

Solution : Given:

$z_1=2-i,z_2=-2+i$

(i) Compute $z_1{z}_2$:

$z_1{z}_2=(2-i)(-2+i)$
$z_1{z}_2=-4+2i+2i-i^2$
$z_1{z}_2=-4+4i-(-1)$
$z_1{z}_2=-3+4i$

Now, $\overline{z_1}=2+i.$

Therefore,

${\displaystyle \frac{z_1{z}_2}{\overline{z_1}}}={\displaystyle \frac{-3+4i}{2+i}}.$

Multiply numerator and denominator by $(2-i)$:

${\displaystyle \frac{z_1{z}_2}{\overline{z_1}}}={\displaystyle \frac{(-3+4i)(2-i)}{(2+i)(2-i)}}$

${\displaystyle \frac{z_1{z}_2}{\overline{z_1}}}={\displaystyle \frac{-6+3i+8i-4i^2}{2^2+1^2}}$

${\displaystyle \frac{z_1{z}_2}{\overline{z_1}}}={\displaystyle \frac{-6+11i-4(-1)}{5}}$

${\displaystyle \frac{z_1{z}_2}{\overline{z_1}}}={\displaystyle \frac{-2+11i}{5}}$

Hence,

$\mathrm{Re}\left({\displaystyle \frac{z_1{z}_2}{\overline{z_1}}}\right)={\displaystyle \frac{-2}{5}}$

(ii) Compute:

${\displaystyle \frac{1}{z_1\overline{z_2}}}={\displaystyle \frac{1}{(2-i)(-2-i)}}.$

Now,

$(2-i)(-2-i)=-4-2i+2i+i^2$

$(2-i)(-2-i)=-4-1=-5$

Therefore,

${\displaystyle \frac{1}{z_1\overline{z_2}}}={\displaystyle \frac{1}{-5}}=-{\displaystyle \frac{1}{5}}$

This is a purely real number.

Hence,

$\mathrm{Im}\left({\displaystyle \frac{1}{z_1\overline{z_2}}}\right)=0.$

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Question 8. Find the real numbers x and y if $(x-iy)(3+5i)$ is the conjugate of $-6-24i$.

Solution: Let

$z=(x-iy)(3+5i)$

Then,

$z=3x+5xi-3yi-5y{i}^2$

Since $i^2=-1$,

$z=3x+5xi-3yi+5y$

$z=(3x+5y)+i(5x-3y)$

Therefore,

$\overline{z}=(3x+5y)-i(5x-3y)$

Also given that $\overline{z}=-6-24i$

So,

$(3x+5y)-i(5x-3y)=-6-24i$

By comparing real and imaginary parts:

$3x+5y=-6\text{…(i)}$

$5x-3y=24\text{…(ii)}$

Multiply (i) by 3 and (ii) by 5:

$9x+15y=-18$

$25x-15y=120$

Add both equations:

$(9x+25x)+(15y-15y)=-18+120$

$34x=102$

$x=\frac{102}{34}=3$

Substitute $x=3$ in equation (i):

$3(3)+5y=-6$

$9+5y=-6$

$5y=-15$

$y=-3$

Hence, the values of $x$ and $y$ are:

$x=3,y=-3$

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Q.9 : Find the modulus of ${\displaystyle \frac{1+i}{1-i}}-{\displaystyle \frac{1-i}{1+i}}$

Solution:

${\displaystyle \frac{1+i}{1-i}}-{\displaystyle \frac{1-i}{1+i}}={\displaystyle \frac{(1+i{)}^2-(1-i{)}^2}{(1-i)(1+i)}}$

Now,

$(1+i{)}^2=1+i^2+2i=2i$

and

$(1-i{)}^2=1+(-i{)}^2+2(1)(-i)=-2i$

So,

${\displaystyle \frac{(1+i{)}^2-(1-i{)}^2}{(1-i)(1+i)}}={\displaystyle \frac{2i-(-2i)}{1^2+1^2}}={\displaystyle \frac{4i}{2}}=2i$

Therefore,

$|{\displaystyle \frac{1+i}{1-i}}-{\displaystyle \frac{1-i}{1+i}}|=|2i|=\sqrt{(2{)}^2}=2$

Hence, the modulus is 2.

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Q.10 : If $(x+iy{)}^3=u+iv$, then show that ${\displaystyle \frac{u}{x}}+{\displaystyle \frac{v}{y}}=4(x^2-y^2)$

Solution:

Given,

$(x+iy{)}^3=u+iv$

Expanding the left-hand side:

$x^3+(iy{)}^3+3x(iy)(x+iy)=u+iv$

Simplifying:

$x^3+i^3{y}^3+3x^{2}yi+3x{y}^2{i}^2=u+iv$

Since $i^2=-1$ and $i^3=-i$, we get

$x^3-i{y}^3+3x^{2}yi-3x{y}^2=u+iv$

Separating real and imaginary parts:

$(x^3-3x{y}^2)+i(3x^{2}y-y^3)=u+iv$

On comparing,
$$u=x^3-3x{y}^2,v=3x^{2}y-y^3$$

Now,

${\displaystyle \frac{u}{x}}+{\displaystyle \frac{v}{y}}={\displaystyle \frac{x^3-3x{y}^2}{x}}+{\displaystyle \frac{3x^{2}y-y^3}{y}}$

Simplify each term:

${\displaystyle \frac{u}{x}}+{\displaystyle \frac{v}{y}}=(x^2-3y^2)+(3x^2-y^2)$

${\displaystyle \frac{u}{x}}+{\displaystyle \frac{v}{y}}=4x^2-4y^2$

${\displaystyle \frac{u}{x}}+{\displaystyle \frac{v}{y}}=4(x^2-y^2)$

Therefore,

${\displaystyle \frac{u}{x}}+{\displaystyle \frac{v}{y}}=4(x^2-y^2)$

Hence proved.

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Q.11 : If $\alpha$ and $\beta$ are different complex numbers with $|\beta |=1$, then find $$\left|\frac{\beta -\alpha }{1-\overline{\alpha }\beta }\right|$$

Solution:

Assume $\alpha =a+ib$ and $\beta =x+iy$

Given: $|\beta |=1$

So,

$\sqrt{x^2+y^2}=1$

which gives

$x^2+y^2=1\text{…(1)}$

Now,

$\left|\frac{\beta -\alpha }{1-\overline{\alpha }\beta }\right|=\left|\frac{(x+iy)-(a+ib)}{1-(a-ib)(x+iy)}\right|$

Simplify the numerator and denominator:

$\left|\frac{\beta -\alpha }{1-\overline{\alpha }\beta }\right|=\left|\frac{(x-a)+i(y-b)}{(1-ax-by)+i(bx-ay)}\right|$

We know that

$\left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}$

Therefore,

$\left|\frac{\beta -\alpha }{1-\overline{\alpha }\beta }\right|=\frac{\sqrt{(x-a{)}^2+(y-b{)}^2}}{\sqrt{(1-ax-by{)}^2+(bx-ay{)}^2}}$

Expanding both numerator and denominator:

$\left|\frac{\beta -\alpha }{1-\overline{\alpha }\beta }\right|=\frac{\sqrt{x^2+a^2-2ax+y^2+b^2-2by}}{\sqrt{1+a^2{x}^2+b^2{y}^2-2ax+2abxy-2by+b^2{x}^2+a^2{y}^2-2abxy}}$

Simplify:

$\left|\frac{\beta -\alpha }{1-\overline{\alpha }\beta }\right|=\frac{\sqrt{(x^2+y^2)+a^2+b^2-2ax-2by}}{\sqrt{1+a^2(x^2+y^2)+b^2(x^2+y^2)-2ax-2by}}$

Using equation (1), $x^2+y^2=1$,

$\left|\frac{\beta -\alpha }{1-\overline{\alpha }\beta }\right|=\frac{\sqrt{1+a^2+b^2-2ax-2by}}{\sqrt{1+a^2+b^2-2ax-2by}}=1$

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Q.12 : Find the number of non-zero integral solutions of the equation $|1-i{|}^x=2^x$.

Solution: We have, $$|1-i{|}^x=2^x$$

Now,

$|1-i|=\sqrt{1^2+(-1{)}^2}=\sqrt{2}$

So the equation becomes:

$(\sqrt{2}{)}^x=2^x$

Rewrite $\sqrt{2}$ as $2^{1/2}$:

$(2^{1/2}{)}^x=2^x$

Which gives:

$2^{x/2}=2^x$

Comparing the exponents:

$\frac{x}{2}=x$

Solving for $x$:

$2x-x=0\Rightarrow x=0$

Hence, $0$ is the only integral solution.

Therefore, the number of non-zero integral solutions is: 0.

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Q.13 : If $(a+ib)(c+id)(e+if)(g+ih)=A+iB$, then show that $(a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2)=A^2+B^2$

Solution:

Given:

$(a+ib)(c+id)(e+if)(g+ih)=A+iB$

Taking the modulus of both sides:

$|(a+ib)(c+id)(e+if)(g+ih)|=|A+iB|$

Using the property of modulus:

$|(a+ib)|\cdot |(c+id)|\cdot |(e+if)|\cdot |(g+ih)|=|A+iB|$

Now,

$\sqrt{a^2+b^2}\cdot \sqrt{c^2+d^2}\cdot \sqrt{e^2+f^2}\cdot \sqrt{g^2+h^2}=\sqrt{A^2+B^2}$

Squaring both sides, we get:

$(a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2)=A^2+B^2$

Hence, proved.

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Q.14 : Find the least positive integral value of $m$ if ${\left(\frac{1+i}{1-i}\right)}^m=1$

Solution:

We have:

${\left(\frac{1+i}{1-i}\right)}^m=1$

Multiply numerator and denominator by the conjugate of the denominator:

${(\frac{1+i}{1-i}\cdot \frac{1+i}{1+i})}^m=1$

${\left(\frac{(1+i{)}^2}{1^2+1^2}\right)}^m=1$

${\left(\frac{1-1+2i}{2}\right)}^m=1$

${\left(\frac{2i}{2}\right)}^m=1$

$(i{)}^m=1$

Since $i^4=1$, we have:

m = 4k, k is an integer

The least positive integer is $k=1$, so:

$m=4\times 1=4$

Hence, the least positive integral value of $m$ is: 4.

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