NCERT Solutions for Miscellaneous Exercise Chapter 4 Class 11 - Complex Numbers and Quadratic Equations

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
${[i^{18}+{\left(\frac{1}{i}\right)}^{25}]}^3={[{\left(i^4\right)}^4\cdot i^2+\frac{1}{{\left(i^4\right)}^6\cdot i}]}^3$
$={[1^4.(-1)+\frac{1}{1^6.i}]}^3$ $(\because i^4=1,i^2=-1)$
$={[-1+\frac{1}{i}]}^3$
$={[-1+\frac{1}{i}\times \frac{i}{i}]}^3$
$={[-1+\frac{i}{i^2}]}^3$
$={[-1+\frac{i}{-1}]}^3={[-1-i]}^3$
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, 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: 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 6: If $a+ib=\frac{(x+i{)}^2}{2x^2+1}$ , prove that $a^2+b^2=\frac{(x^2+1{)}^2}{(2x^2+1{)}^2}$ .

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 7: (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 7:(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 8: 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 9: 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 10: 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 11: 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

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

Answer:

Given problem is
$|1-i{|}^x=2^x$
Now,
$(\sqrt{1^2+(-1{)}^2}{)}^x=2^x$
$(\sqrt{1+1}{)}^x=2^x$
${\left(\sqrt{2}\right)}^x=2^x$
$2^{\frac{x}{2}}=2^x$
$\frac{x}{2}=x$
$\frac{x}{2}=0$
x = 0 is the only possible solution to the given problem

Therefore, there are 0 number of non-zero integral solutions of the equation $|1-i{|}^x=2^x$

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

Answer:

It is given that
$(a+ib)(c+id)(e+if)(g+ih)=A+iB,$
Now, take mod on both sides
$|(a+ib)(c+id)(e+if)(g+ih)|=|A+iB|$
$|(a+ib)||(c+id)||(e+if)||(g+ih)|=|A+iB|$ $(\because |z_1{z}_2|=|z_1||z_2|)$
$(\sqrt{a^2+b^2})(\sqrt{c^2+d^2})(\sqrt{e^2+f^2})(\sqrt{g^2+h^2})=(\sqrt{A^2+B^2})$
Square both the sides. we will get

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

Hence proved

Question 14: If ${\left(\frac{1+i}{1-i}\right)}^m=1,$ then find the least positive integral value of $m$ .

Answer:

Let
$z={\left(\frac{1+i}{1-i}\right)}^m$
Now, multiply both numerator and denominator by $(1+i)$
We will get,
$z={(\frac{1+i}{1-i}\times \frac{1+i}{1+i})}^m$
$={\left(\frac{(1+i{)}^2}{1^2-i^2}\right)}^m$
$={\left(\frac{1^2+i^2+2i}{1+1}\right)}^m$
$={\left(\frac{1-1+2i}{2}\right)}^m$ $(\because i^2=-1)$
$={\left(\frac{2i}{2}\right)}^m$
$=i^m$
We know that $i^4=1$
Therefore, the least positive integral value of $m$ is 4