NCERT Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations

Question:1 Express each of the complex numbers in the form $a+ib$.

(1) $(5i)(-\frac{3}{5}i)$

(2) $i^9+i^{19}$

(3) $i^{-39}$

(4) $3(7+7i)+i(7+7i)$

(5) $(1-i)-(-1+6i)$

(6) $(\frac{1}{5}+i\frac{2}{5})-(4+i\frac{5}{2})$

(7) $[(\frac{1}{3}+i\frac{7}{3})+(4+i\frac{1}{3})]-(-\frac{4}{3}+i)$

(8) $(1-i{)}^4$

(9) ${(\frac{1}{3}+3i)}^3$

(10)${(-2-\frac{1}{3}i)}^3$

Answer:

(1) On solving

$(5i)(-\frac{3}{5}i)$
we will get

$(5i)(-\frac{3}{5}i)=5\times (-\frac{3}{5})\times i\times i$
$=-3\times i^2$ $(\because i^2=-1)$
$=-3\times -1$
$=3$

Now, in the form of $a+ib$, we can write it as
$=3+0i$

(2)

We know that $i^4=1$
Now, we will reduce $i^9+i^{19}$ into

$\begin{array}{rl}& =(1{)}^2\cdot i+(1{)}^3\cdot (-i){(\because i^4=1,i^3=-i)}^2\cdot i+{\left(i^4\right)}^3\cdot i^3\\ & =i-i=0\end{array}$

Now, in the form of $a+ib$ we can write it as

$o+io$

Therefore, the answer is $o+io$

(3)

We know that $i^4=1$
Now, we will reduce $i^{-39}$ into

$\begin{array}{rl}& i^{-39}={\left(i^4\right)}^{-9}\cdot i^{-3}\\ & =(1{)}^{-9}\cdot (-i{)}^{-1}(\because i^4=1,i^3=-i)\\ & =\frac{1}{-i}\\ & =\frac{1}{-i}\times \frac{i}{i}\\ & =\frac{i}{-i^2}(\because i^2=-1)\\ & =\frac{i}{-(-1)}\\ & =i\end{array}$

Now, in the form of $a+ib$ we can write it as

$o+i1$

Therefore, the answer is $o+i1$

(4)

Given problem is

$3(7+7i)+i(7+7i)$

Now, we will reduce it into

$\begin{array}{rl}& 3(7+7i)+i(7+7i)=21+21i+7i+7i^2\\ & =21+21i+7i+7(-1)(\because i^2=-1)\\ & =21+21i+7i-7\\ & =14+28i\end{array}$

Therefore, the answer is $14+i28$

(5)

$(1-i)-(-1+6i)$

Answer:
Given problem is

$(1-i)-(-1+6i)$

Now, we will reduce it into

$\begin{array}{rl}& (1-i)-(-1+6i)=1-i+1-6i\\ & =2-7i\end{array}$

Therefore, the answer is $2-7i$

(6)

Given problem is

$(\frac{1}{5}+i\frac{2}{5})-(4+i\frac{5}{2})$

Now, we will reduce it into

$\begin{array}{rl}& (\frac{1}{5}+i\frac{2}{5})-(4+i\frac{5}{2})=\frac{1}{5}+i\frac{2}{5}-4-i\frac{5}{2}\\ & =\frac{1-20}{5}+i\frac{(4-25)}{10}\\ & =-\frac{19}{5}-i\frac{21}{10}\end{array}$

Therefore, the answer is $-\frac{19}{5}-i\frac{21}{10}$

(7)

Given problem is

$[(\frac{1}{3}+i\frac{7}{3})+(4+i\frac{1}{3})]-(-\frac{4}{3}+i)$

Now, we will reduce it into

$\begin{array}{rl}& [(\frac{1}{3}+i\frac{7}{3})+(4+i\frac{1}{3})]-(-\frac{4}{3}+i)=\frac{1}{3}+i\frac{7}{3}+4+i\frac{1}{3}+\frac{4}{3}-i\\ & =\frac{1+4+12}{3}+i\frac{(7+1-3)}{3}\\ & =\frac{17}{3}+i\frac{5}{3}\end{array}$

Therefore, the answer is $\frac{17}{3}+i\frac{5}{3}$

(8)

The given problem is

$(1-i{)}^4$
Now, we will reduce it into

$\begin{array}{rl}& (1-i{)}^4={((1-i{)}^2)}^2\\ & ={(1^2+i^2-2.1.i)}^2(u\mathrm{sing}(a-b{)}^2=a^2+b^2-2ab)\\ & =(1-1-2i{)}^2(\because i^2=-1)\\ & =(-2i{)}^2\\ & =4i^2\\ & =-4\end{array}$

Therefore, the answer is $-4+i0$

Given problem is

${(\frac{1}{3}+3i)}^3$

Now, we will reduce it into

$\begin{array}{rl}& {(\frac{1}{3}+3i)}^3={\left(\frac{1}{3}\right)}^3+(3i{)}^3+3\cdot {\left(\frac{1}{3}\right)}^2\cdot 3i+3\cdot \frac{1}{3}\cdot (3i{)}^2(u\mathrm{sing}(a+b{)}^3=a^3+b^3+3a^{2}b+3a{b}^2)\\ & =\frac{1}{27}+27i^3+i+9i^2\\ & =\frac{1}{27}+27(-i)+i+9(-1)(\because i^3=-i\text{ and }{i}^2=-1)\\ & =\frac{1}{27}-27i+i-9\\ & =\frac{1-243}{27}-26i\\ & =-\frac{242}{27}-26i\end{array}$

Therefore, the answer is

$-\frac{242}{27}-26i$
(10)

Given problem is

${(-2-\frac{1}{3}i)}^3$

Now, we will reduce it into

$\begin{array}{rl}& {(-2-\frac{1}{3}i)}^3=-((2{)}^3+{\left(\frac{1}{3}i\right)}^3+3\cdot (2{)}^2\frac{1}{3}i+3\cdot {\left(\frac{1}{3}i\right)}^2\cdot 2)(u\mathrm{sing}(a+b{)}^3=a^3+b^3+3a^{2}b+3a{b}^2)\\ & =-(8+\frac{1}{27}{i}^3+3\cdot 4\cdot \frac{1}{3}i+3\cdot \frac{1}{9}{i}^2\cdot 2)\\ & =-(8+\frac{1}{27}(-i)+4i+\frac{2}{3}(-1))(\because i^3=-i\text{ and }{i}^2=-1)\\ & =-(8-\frac{1}{27}i+4i-\frac{2}{3})\\ & =-(\frac{(-1+108)}{27}i+\frac{24-2}{3})\\ & =-\frac{22}{3}-i\frac{107}{27}\end{array}$

Therefore, the answer is $-\frac{22}{3}-i\frac{107}{27}$

Question:11 Find the multiplicative inverse of each of the complex numbers.

$4-3i$

Answer:

Let $z=4-3i$
Then,
$\overline{z}=4+3i$
And
$|z{|}^2=4^2+(-3{)}^2=16+9=25$
Now, the multiplicative inverse is given by
$z^{-1}=\frac{\overline{z}}{|z{|}^2}=\frac{4+3i}{25}=\frac{4}{25}+i\frac{3}{25}$

Therefore, the multiplicative inverse is

$\frac{4}{25}+i\frac{3}{25}$

Question:12 Find the multiplicative inverse of each of the complex numbers.

$\sqrt{5}+3i$

Answer:

Let $z=\sqrt{5}+3i$
Then,
$\overline{z}=\sqrt{5}-3i$
And
$|z{|}^2=(\sqrt{5}{)}^2+(3{)}^2=5+9=14$
Now, the multiplicative inverse is given by
$z^{-1}=\frac{\overline{z}}{|z{|}^2}=\frac{\sqrt{5}-3i}{14}=\frac{\sqrt{5}}{14}-i\frac{3}{14}$

Therefore, the multiplicative inverse is $\frac{\sqrt{5}}{14}-i\frac{3}{14}$

Question 13 Find the multiplicative inverse of each of the complex numbers.

$-i$

Answer:

Let $z=-i$
Then,
$\overline{z}=i$
And
$|z{|}^2=(0{)}^2+(1{)}^2=0+1=1$
Now, the multiplicative inverse is given by
$z^{-1}=\frac{\overline{z}}{|z{|}^2}=\frac{i}{1}=0+i$

Therefore, the multiplicative inverse is $0+i1$

Question:14 Express the following expression in the form of $a+ib:$

$\frac{(3+i\sqrt{5})(3-i\sqrt{5})}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})}$

Answer:

Given problem is
$\frac{(3+i\sqrt{5})(3-i\sqrt{5})}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})}$
Now, we will reduce it into

$\frac{(3+i\sqrt{5})(3-i\sqrt{5})}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})}=\frac{3^2-(\sqrt{5}i{)}^2}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})}$ $(using(a-b)(a+b)=a^2-b^2)$
$=\frac{9-5i^2}{\sqrt{3}+\sqrt{2}i-\sqrt{3}+\sqrt{2}i}$
$=\frac{9-5(-1)}{2\sqrt{2}i}$ $(\because i^2=-1)$
$=\frac{14}{2\sqrt{2}i}\times \frac{\sqrt{2}i}{\sqrt{2}i}$
$=\frac{7\sqrt{2}i}{2i^2}$
$=-\frac{7\sqrt{2}i}{2}$
Therefore, answer is $0-i\frac{7\sqrt{2}}{2}$

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

Answer:

The given problem is
${[i^{18}+{\left(\frac{1}{i}\right)}^{25}]}^3$
Now, we will reduce it into
${[i^{18}+{\left(\frac{1}{i}\right)}^{25}]}^3={[(i^4{)}^4.i^2+\frac{1}{(i^4{)}^6.i}]}^3$
$={[1^4.(-1)+\frac{1}{1^6.i}]}^3$ $(\because i^4=1,i^2=-1)$
$={[-1+\frac{1}{i}]}^3$
$={[-1+\frac{1}{i}\times \frac{i}{i}]}^3$
$={[-1+\frac{i}{i^2}]}^3$
$={[-1+\frac{i}{-1}]}^3={[-1-i]}^3$
Now,
$-(1+i{)}^3=-(1^3+i^3+{3.1}^2.i+3.1.i^2)$ $(using(a+b{)}^3=a^3+b^3+3.a^2.b+3.a.b^2)$
$=-(1-i+3i+3(-1))$ $(\because i^3=-i,i^2=-1)$
$=-(1-i+3i-3)=-(-2+2i)$
$=2-2i$
Therefore, the answer is $2-2i$

Question:2 For any two complex numbers $z_1$ and $z_2$ , prove that $Re(z_1{z}_2)=Re{z}_{1}Re{z}_2-Im{z}_{1}Im{z}_2$

Answer:

Let two complex numbers are
$z_1=x_1+i{y}_1$
$z_2=x_2+i{y}_2$
Now,
$z_1.z_2=(x_1+i{y}_1).(x_2+i{y}_2)$
$=x_1{x}_2+i{x}_1{y}_2+i{y}_1{x}_2+i^2{y}_1{y}_2$
$=x_1{x}_2+i{x}_1{y}_2+i{y}_1{x}_2-y_1{y}_2$ $(\because i^2=-1)$
$=x_1{x}_2-y_1{y}_2+i(x_1{y}_2+y_1{x}_2)$
$Re(z_1{z}_2)=x_1{x}_2-y_1{y}_2$
$=Re(z_1{z}_2)-Im(z_1{z}_2)$

Hence proved

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

Answer:

Given problem is
$(\frac{1}{1-4i}-\frac{2}{1+i})\left(\frac{3-4i}{5+i}\right)$
Now, we will reduce it into

$(\frac{1}{1-4i}-\frac{2}{1+i})\left(\frac{3-4i}{5+i}\right)=\left(\frac{(1+i)-2(1-4i)}{(1+i)(1-4i)}\right)\left(\frac{3-4i}{5+i}\right)$
$=\left(\frac{1+i-2+8i}{1-4i+i-4i^2}\right)\left(\frac{3-4i}{5+i}\right)$
$=\left(\frac{-1+9i}{1-3i-4(-1)}\right)\left(\frac{3-4i}{5+i}\right)$
$=\left(\frac{-1+9i}{5-3i}\right)\left(\frac{3-4i}{5+i}\right)$
$=\left(\frac{-3+4i+27i-36i^2}{25+5i-15i-3i^2}\right)=\left(\frac{-3+31i+36}{25-10i+3}\right)=\frac{33+31i}{28-10i}=\frac{33+31i}{2(14-5i)}$

Now, multiply the numerator and denominator by $(14+5i)$
$\Rightarrow \frac{33+31i}{2(14-5i)}\times \frac{14+5i}{14+5i}$
$\Rightarrow \frac{462+165i+434i+155i^2}{2({14}^2-(5i{)}^2)}$ $(using(a-b)(a+b)=a^2-b^2)$
$\Rightarrow \frac{462+599i-155}{2(196-25i^2)}$
$\Rightarrow \frac{307+599i}{2(196+25)}=\frac{307+599i}{2\times 221}=\frac{307+599i}{442}=\frac{307}{442}+i\frac{599}{442}$

Therefore, answer is $\frac{307}{442}+i\frac{599}{442}$

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

Answer:

the given problem is

$x-iy=\sqrt{\frac{a-ib}{c-id}}$
Now, multiply the numerator and denominator by

$\sqrt{c+id}$
$x-iy=\sqrt{\frac{a-ib}{c-id}\times \frac{c+id}{c+id}}$
$=\sqrt{\frac{(ac+bd)+i(ad-bc)}{c^2-i^2{d}^2}}=\sqrt{\frac{(ac+bd)+i(ad-bc)}{c^2+d^2}}$
Now, square both the sides
$(x-iy{)}^2={\left(\sqrt{\frac{(ac+bd)+i(ad-bc)}{c^2+d^2}}\right)}^2$
$=\frac{(ac+bd)+i(ad-bc)}{c^2+d^2}$
$x^2-y^2-2ixy=\frac{(ac+bd)+i(ad-bc)}{c^2+d^2}$
On comparing the real and imaginary parts, we obtain

$x^2-y^2=\frac{ac+bd}{c^2+d^2}and-2xy=\frac{ad-bc}{c^2+d^2}-(i)$

Now,
$(x^2+y^2{)}^2=(x^2-y^2{)}^2+4x^2{y}^2$
$={\left(\frac{ac+bd}{c^2+d^2}\right)}^2+{\left(\frac{ad-bc}{c^2+d^2}\right)}^2(using(i))$
$=\frac{a^2{c}^2+b^2{d}^2+2acbd+a^2{d}^2+b^2{c}^2-2adbc}{(c^2+d^2{)}^2}$
$=\frac{a^2{c}^2+b^2{d}^2+a^2{d}^2+b^2{c}^2}{(c^2+d^2{)}^2}$
$=\frac{a^2(c^2+d^2)+b^2(c^2+d^2)}{(c^2+d^2{)}^2}$
$=\frac{(a^2+b^2)(c^2+d^2)}{(c^2+d^2{)}^2}$
$=\frac{(a^2+b^2)}{(c^2+d^2)}$

Hence proved

Question:5 Convert the following in the polar form:

(i) $\frac{1+7i}{(2-i{)}^2}$

(ii)$\frac{1+3i}{1-2i}$

Answer(i):

Let
$z=\frac{1+7i}{(2-i{)}^2}=\frac{1+7i}{4+i^2-4i}=\frac{1+7i}{4-1-4i}=\frac{1+7i}{3-4i}$

Now, multiply the numerator and denominator by $3+4i$
$\Rightarrow z=\frac{1+7i}{3-4i}\times \frac{3+4i}{3+4i}=\frac{3+4i+21i+28i^2}{3^2+4^2}=\frac{-25+25i}{25}=-1+i$
Now,
let
$r\cos \theta =-1andr\sin \theta =1$
On squaring both and then add
$r^2({\cos }^2\theta +{\sin }^2\theta )=(-1{)}^2+1^2$
$r^2=2$
$r=\sqrt{2}(\because r>0)$
Now,
$\sqrt{2}\cos \theta =-1and\sqrt{2}\sin \theta =1$
$\cos \theta =-\frac{1}{\sqrt{2}}and\sin \theta =\frac{1}{\sqrt{2}}$
Since the value of $\cos \theta$ is negative and $\sin \theta$ is positive this is the case in II quadrant
Therefore,
$\theta =\pi -\frac{\pi }{4}=\frac{3\pi }{4}(liesinIIquadrant)$
$z=r\cos \theta +ir\sin \theta$
$=\sqrt{2}\cos \frac{3\pi }{4}+i\sqrt{2}\sin \frac{3\pi }{4}$
$=\sqrt{2}(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4})$
Therefore, the required polar form is

$\sqrt{2}(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4})$

Answer(ii):

Let
$z=\frac{1+3i}{1-2i}$

Now, multiply the numerator and denominator by $1+2i$
$\Rightarrow z=\frac{1+3i}{1-2i}\times \frac{1+2i}{i+2i}=\frac{1+2i+3i-6}{1+4}=\frac{-5+5i}{5}=-1+i$
Now,
let
$r\cos \theta =-1andr\sin \theta =1$
On squaring both and then add
$r^2({\cos }^2\theta +{\sin }^2\theta )=(-1{)}^2+1^2$
$r^2=2$
$r=\sqrt{2}(\because r>0)$
Now,
$\sqrt{2}\cos \theta =-1and\sqrt{2}\sin \theta =1$
$\cos \theta =-\frac{1}{\sqrt{2}}and\sin \theta =\frac{1}{\sqrt{2}}$
Since the value of $\cos \theta$ is negative and $\sin \theta$ is positive this is the case in II quadrant
Therefore,
$\theta =\pi -\frac{\pi }{4}=\frac{3\pi }{4}(liesinIIquadrant)$
$z=r\cos \theta +ir\sin \theta$
$=\sqrt{2}\cos \frac{3\pi }{4}+i\sqrt{2}\sin \frac{3\pi }{4}$
$=\sqrt{2}(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4})$
Therefore, the required polar form is

$\sqrt{2}(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4})$

Question:6 If $z_1=2-i,z_2=1+i$ , find $\left|\frac{z_1+z_2+1}{z_1-z_2+1}\right|$ .

Answer:

It is given that
$z_1=2-i,z_2=1+i$
Then,
$\left|\frac{z_1+z_2+1}{z_1-z_2+1}\right|=\left|\frac{2-i+1+i+1}{2-i-1-i+1}\right|=\left|\frac{4}{2(1-i)}\right|=\left|\frac{2}{(1-i)}\right|$
Now, multiply the numerator and denominator by $1+i$
$\Rightarrow |\frac{2}{(1-i)}\times \frac{1+i}{1+i}|=\left|\frac{2(1+i)}{1^2-i^2}\right|=\left|\frac{2(1+i)}{1+1}\right|=|1+i|$
Now,
$|1+i|=\sqrt{1^2+1^2}=\sqrt{1+1}=\sqrt{2}$
Therefore, the value of

$\left|\frac{z_1+z_2+1}{z_1-z_2+1}\right|$ is $\sqrt{2}$

Question:7 If $a+ib=\frac{(x+i{)}^2}{2x^2+1}$ , prove that $a^2+b^2=\frac{(x^2+1{)}^2}{(2x^2+1{)}^2}$ .

Answer:

It is given that
$a+ib=\frac{(x+i{)}^2}{2x^2+1}$
Now, we will reduce it into

$a+ib=\frac{(x+i{)}^2}{2x^2+1}=\frac{x^2+i^2+2xi}{2x^2+1}=\frac{x^2-1+2xi}{2x^2+1}=\frac{x^2-1}{2x^2+1}+i\frac{2x}{2x^2+1}$
On comparing real and imaginary parts. we will get
$a=\frac{x^2-1}{2x^2+1}andb=\frac{2x}{2x^2+1}$
Now,
$a^2+b^2={\left(\frac{x^2-1}{2x^2+1}\right)}^2+{\left(\frac{2x}{2x^2+1}\right)}^2$
$=\frac{x^4+1-2x^2+4x^2}{(2x^2+1{)}^2}$
$=\frac{x^4+1+2x^2}{(2x^2+1{)}^2}$
$=\frac{(x^2+1{)}^2}{(2x^2+1{)}^2}$
Hence proved

Question:8 Find the modulus of $\frac{1+i}{1-i}-\frac{1-i}{1+i}$ .

Answer:

Let
$z=\frac{1+i}{1-i}-\frac{1-i}{1+i}$
Now, we will reduce it into
$z=\frac{1+i}{1-i}-\frac{1-i}{1+i}=\frac{(1+i{)}^2-(1-i{)}^2}{(1+i)(1-i)}=\frac{1^2+i^2+2i-1^2-i^2+2i}{1^2-i^2}$ $=\frac{4i}{1+1}=\frac{4i}{2}=2i$
Now,
$r\cos \theta =0andr\sin \theta =2$
square and add both sides. we will get,
$r^2{\cos }^2\theta +r^2{\sin }^2\theta =0^2+2^2$
$r^2({\cos }^2\theta +{\sin }^2\theta )=4$
$r^2=4(\because {\cos }^2\theta +{\sin }^2\theta =1)$
$r=2(\because r>0)$

Therefore, the modulus of

$\frac{1+i}{1-i}-\frac{1-i}{1+i}$ is 2

Question:9 If $(x+iy{)}^3=u+iv$ , then show that $\frac{u}{x}+\frac{v}{y}=4(x^2-y^2).$

Answer:

it is given that
$(x+iy{)}^3=u+iv$
Now, expand the Left-hand side
$x^3+(iy{)}^3+3.(x{)}^2.iy+3.x.(iy{)}^2=u+iv$
$x^3+i^3{y}^3+3x^{2}iy+3x{i}^2{y}^2=u+iv$
$x^3-i{y}^3+3x^{2}iy-3x{y}^2=u+iv$ $(\because i^3=-iand{i}^2=-1)$
$x^3-3x{y}^2+i(3x^{2}y-y^3)=u+iv$
On comparing real and imaginary parts. we will get,
$u=x^3-3x{y}^{2}andv=3x^{2}y-y^3$
Now,
$\frac{u}{x}+\frac{v}{y}=\frac{x(x^2-3y^2)}{x}+\frac{y(3x^2-y^2)}{y}$
$=x^2-3y^2+3x^2-y^2$
$=4x^2-4y^2$
$=4(x^2-y^2)$
Hence proved