NCERT Solutions for Miscellaneous Exercise Chapter 5 Class 11 - Complex Numbers and Quadratic Equations

Question:1 Evaluate $\text{Double exponent: use braces to clarify}$ .

Answer:

The given problem is
$\text{Double exponent: use braces to clarify}$
Now, we will reduce it into
$\text{Double exponent: use braces to clarify}$
$={[1^4.(-1)+\frac{1}{1^6.i}]}^3$ $(\because i^4=1,i^2=-1)$
$={[-1+\frac{1}{i}]}^3$
$={[-1+\frac{1}{i}\times \frac{i}{i}]}^3$
$={[-1+\frac{i}{i^2}]}^3$
$={[-1+\frac{i}{-1}]}^3={[-1-i]}^3$
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, 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 numerator an 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 part, 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(i) Convert the following in the polar form:

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

Answer:

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

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

Question:5(ii) Convert the following in the polar form:

$\frac{1+3i}{1-2i}$

Answer:

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 Solve each of the equation: $3x^2-4x+\frac{20}{3}=0$

Answer:

Given equation is
$3x^2-4x+\frac{20}{3}=0$
Now, we know that the roots of the quadratic equation are given by the formula
$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
In this case the value of

$a=3,b=-4andc=\frac{20}{3}$
Therefore,
$\frac{-(-4)\pm \sqrt{(-4{)}^2-4.3.\frac{20}{3}}}{2.3}=\frac{4\pm \sqrt{16-80}}{6}=\frac{4\pm \sqrt{-64}}{6}$ $=\frac{4\pm 8i}{6}=\frac{2}{3}\pm i\frac{4}{3}$
Therefore, the solutions of requires equation are

$\frac{2}{3}\pm i\frac{4}{3}$

Question:7 Solve each of the equation: $x^2-2x+\frac{3}{2}=0$

Answer:

Given equation is
$x^2-2x+\frac{3}{2}=0$
Now, we know that the roots of the quadratic equation are given by the formula
$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
In this case the value of $a=1,b=-2andc=\frac{3}{2}$
Therefore,
$\frac{-(-2)\pm \sqrt{(-2{)}^2-4.1.\frac{3}{2}}}{2.1}=\frac{2\pm \sqrt{4-6}}{2}=\frac{2\pm \sqrt{-2}}{2}$ $=\frac{2\pm i\sqrt{2}}{2}=1\pm i\frac{\sqrt{2}}{2}$
Therefore, the solutions of requires equation are

$1\pm i\frac{\sqrt{2}}{2}$

Question:8 Solve each of the equation: $27x^2-10x+1=0$ .

Answer:

Given equation is
$27x^2-10x+1=0$
Now, we know that the roots of the quadratic equation are given by the formula
$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
In this case the value of $a=27,b=-10andc=1$
Therefore,
$\frac{-(-10)\pm \sqrt{(-10{)}^2-4.27.1}}{2.27}=\frac{10\pm \sqrt{100-108}}{54}=\frac{10\pm \sqrt{-8}}{54}$ $=\frac{10\pm i2\sqrt{2}}{54}=\frac{5}{27}\pm i\frac{\sqrt{2}}{27}$
Therefore, the solutions of requires equation are $\frac{5}{27}\pm i\frac{\sqrt{2}}{27}$

Question:9 Solve each of the equation: $21x^2-28x+10=0$

Answer:

Given equation is
$21x^2-28x+10=0$
Now, we know that the roots of the quadratic equation are given by the formula
$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
In this case the value of $a=21,b=-28andc=10$
Therefore,
$\frac{-(-28)\pm \sqrt{(-28{)}^2-4.21.10}}{2.21}=\frac{28\pm \sqrt{784-840}}{42}=\frac{28\pm \sqrt{-56}}{42}$ $=\frac{28\pm i2\sqrt{14}}{42}=\frac{2}{3}\pm i\frac{\sqrt{14}}{21}$
Therefore, the solutions of requires equation are

$\frac{2}{3}\pm i\frac{\sqrt{14}}{21}$

Question:10 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:11 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 part. 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:12(i) Let $z_1=2-i,z_2=-2+i.$ Find

$Re\left(\frac{z_1{z}_2}{\overline{z_1}}\right)$

Answer:

It is given that
$z_1=2-iand{z}_2=-2+i$
Now,
$z_1{z}_2=(2-i)(-2+i)=-4+2i+2i-i^2=-4+4i+1=-3+4i$
And
${\overline{z}}_1=2+i$
Now,
$\frac{z_1{z}_2}{{\overline{z}}_1}=\frac{-3+4i}{2+i}=\frac{-3+4i}{2+i}\times \frac{2-i}{2-i}=\frac{-6+3i+8i-4i^2}{2^2-i^2}=\frac{-6+11i+4}{4+1}$ $=\frac{-2+11i}{5}=-\frac{2}{5}+i\frac{11}{5}$
Now,
$Re\left(\frac{z_1{z}_2}{z_1}\right)=-\frac{2}{5}$
Therefore, the answer is

$-\frac{2}{5}$

Question:12(ii) Let $z_1=2-i,z_2=-2+i.$ Find

$Im\left(\frac{1}{z_1\overline{z_1}}\right)$

Answer:

It is given that
$z_1=2-i$
Therefore,
${\overline{z}}_1=2+i$
NOw,
$z_1{\overline{z}}_1=(2-i)(2+i)=2^2-i^2=4+1=5$ $(using(a-b)(a+b)=a^2-b^2)$
Now,
$\frac{1}{z_1{\overline{z}}_1}=\frac{1}{5}$
Therefore,
$Im\left(\frac{1}{z_1{\overline{z}}_1}\right)=0$
Therefore, the answer is 0

Question:13 Find the modulus and argument of the complex number $\frac{1+2i}{1-3i}$ .

Answer:

Let
$z=\frac{1+2i}{1-3i}$
Now, multiply the numerator and denominator by $(1+3i)$

$\Rightarrow z=\frac{1+2i}{1-3i}\times \frac{1+3i}{1+3i}=\frac{1+3i+2i+6i^2}{1^2-(3i{)}^2}=\frac{1+5i-6}{1-9i^2}=\frac{-5+5i}{10}$ $=-\frac{1}{2}+i\frac{1}{2}$
Therefore,
$r\cos \theta =-\frac{1}{2}andr\sin \theta =\frac{1}{2}$
Square and add both the sides
$r^2{\cos }^2\theta +r^2{\sin }^2\theta ={(-\frac{1}{2})}^2+{\left(\frac{1}{2}\right)}^2$
$r^2({\cos }^2\theta +{\sin }^2\theta )=\left(\frac{1}{4}\right)+\left(\frac{1}{4}\right)$
$r^2=\frac{1}{2}(\because {\sin }^2\theta +{\cos }^2\theta =1)$
$r=\frac{1}{\sqrt{2}}(\because r>0)$
Therefore, the modulus is $\frac{1}{\sqrt{2}}$
Now,
$\frac{1}{\sqrt{2}}\cos \theta =-\frac{1}{2}and\frac{1}{\sqrt{2}}\sin \theta =\frac{1}{2}$
$\cos \theta =-\frac{1}{\sqrt{2}}and\sin \theta =\frac{1}{\sqrt{2}}$
Since the value of $\cos \theta$ is negative and the value of $\sin \theta$ is positive and we know that it is the case in II quadrant
Therefore,
Argument $=(\pi -\frac{\pi }{4})=\frac{3\pi }{4}$
Therefore, Argument and modulus are $\frac{3\pi }{4}and\frac{1}{\sqrt{2}}$ respectively

Question:14 Find the real numbers x andy if $(x-iy)(3+5i)$ is the conjugate of $-6-24i$ .

Answer:

Let
$z=(x-iy)(3+5i)=3x+5xi-3yi-5y{i}^2=3x+5y+i(5x-3y)$
Therefore,
$\overline{z}=(3x+5y)-i(5x-3y)-(i)$
Now, it is given that
$\overline{z}=-6-24i-(ii)$
Compare (i) and (ii) we will get
$(3x+5y)-i(5x-3y)=-6-24i$
On comparing real and imaginary part. we will get
$3x+5y=-6and5x-3y=24$
On solving these we will get
$x=3andy=-3$

Therefore, the value of x and y are 3 and -3 respectively

Question:15 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 the 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, modulus of

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

Question:16 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 part. 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

Question:17 If $\alpha$ and $\beta$ are different complex numbers with $|\beta |=1$ , then find $\left|\frac{\beta -\alpha }{1-\overline{\alpha }\beta }\right|$ .

Answer:

Let
$\alpha =a+ib$ and $\beta =x+iy$
It is given that
$|\beta |=1\Rightarrow \sqrt{x^2+y^2}=1\Rightarrow x^2+y^2=1$
and
$\overline{\alpha }=a-ib$
Now,
$\left|\frac{\beta -\alpha }{1-\overline{\alpha }\beta }\right|=\left|\frac{(x+iy)-(a+ib)}{1-(a-ib)(x+iy)}\right|=\left|\frac{(x-a)+i(y-b)}{1-(ax+iay-ibx-i^{2}yb)}\right|$
$=\left|\frac{(x-a)+i(y-b)}{(1-ax-yb)-i(bx-ay)}\right|$
$=\frac{\sqrt{(x-a{)}^2+(y-b{)}^2}}{\sqrt{(1-ax-yb{)}^2+(bx-ay{)}^2}}$
$=\frac{\sqrt{x^2+a^2-2xa+y^2+b^2-yb}}{\sqrt{1+a^2{x}^2+b^2{y}^2-2ax+2abxy-by+b^2{x}^2+a^2{y}^2-2abxy}}$
$=\frac{\sqrt{(x^2+y^2)+a^2-2xa+b^2-yb}}{\sqrt{1+a^2(x^2+y^2)+b^2(x^2+y^2)-2ax+2abxy-by-2abxy}}$
$=\frac{\sqrt{1+a^2-2xa+b^2-yb}}{\sqrt{1+a^2+b^2-2ax-by}}$ $(\because x^2+y^2=1given)$
$=1$

Therefore, value of $\left|\frac{\beta -\alpha }{1-\overline{\alpha }\beta }\right|$ is 1