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

Question:1 Evaluate $\small \left [ i^{18}+\left ( \frac{1}{i} \right )^2^5\right ]^3$ .

Answer:

The given problem is
$\small \left [ i^{18}+\left ( \frac{1}{i} \right )^2^5\right ]^3$
Now, we will reduce it into
$\small \left [ i^{18}+\left ( \frac{1}{i} \right )^2^5\right ]^3= \left [ (i^4)^4.i^2+\frac{1}{(i^4)^6.i} \right ]^3$
$=\left [ 1^4.(-1)+\frac{1}{1^6.i} \right ]^3$ $(\because i^4 = 1, i^2 = -1 )$
$= \left [ -1+\frac{1}{i} \right ]^3$
$= \left [ -1+\frac{1}{i} \times \frac{i}{i}\right ]^3$
$= \left [ -1+\frac{i}{i^2} \right ]^3$
$= \left [ -1+\frac{i}{-1} \right ]^3 = \left [ -1-i \right ]^3$
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 $\small z_1$ and $\small z_2$ , prove that $\small Re (z_1z_2)=Re\hspace {1mm}z_1\hspace {1mm}Re\hspace {1mm}z_2-Imz_1\hspace {1mm}Imz_2$

Answer:

Let two complex numbers are
$z_1=x_1+iy_1$
$z_2=x_2+iy_2$
Now,
$z_1.z_2=(x_1+iy_1).(x_2+iy_2)$
$=x_1x_2+ix_1y_2+iy_1x_2+i^2y_1y_2$
$=x_1x_2+ix_1y_2+iy_1x_2-y_1y_2$ $(\because i^2 = -1)$
$=x_1x_2-y_1y_2+i(x_1y_2+y_1x_2)$
$Re(z_1z_2)= x_1x_2-y_1y_2$
$=Re(z_1z_2)-Im(z_1z_2)$

Hence proved

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

Answer:

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

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

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

Answer:

the given problem is

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

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

Hence proved

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

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

Answer:

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

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
$r\cos\theta = -1 \ \ and \ \ r\sin \theta = 1$
On squaring both and then add
$r^2(\cos^2\theta+\sin^2\theta)= (-1)^2+1^2$
$r^2=2$
$r = \sqrt2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because r > 0)$
Now,
$\sqrt2 \cos \theta = -1 \ \ and \ \ \sqrt2\sin \theta = 1$
$\cos \theta = -\frac{1}{\sqrt2} \ \ and \ \ \sin \theta = \frac{1}{\sqrt2}$
Since the value of $\cos \theta$ is negative and $\sin \theta$ is positive this is the case in II quadrant
Therefore,
$\theta = \pi - \frac{\pi}{4}= \frac{3\pi}{4} \ \ \ \ \ \ \ \ \ \ \ \ (lies \ in \ II \ quadrant)$
$z = r\cos \theta + ir\sin \theta$
$=\sqrt2\cos \frac{3\pi}{4} + i\sqrt2\sin \frac{3\pi}{4}$
$=\sqrt2\left ( \cos \frac{3\pi}{4} + i\sin \frac{3\pi}{4} \right )$
Therefore, the required polar form is

$\sqrt2\left ( \cos \frac{3\pi}{4} + i\sin \frac{3\pi}{4} \right )$

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

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

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
$r\cos\theta = -1 \ \ and \ \ r\sin \theta = 1$
On squaring both and then add
$r^2(\cos^2\theta+\sin^2\theta)= (-1)^2+1^2$
$r^2=2$
$r = \sqrt2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because r > 0)$
Now,
$\sqrt2 \cos \theta = -1 \ \ and \ \ \sqrt2\sin \theta = 1$
$\cos \theta = -\frac{1}{\sqrt2} \ \ and \ \ \sin \theta = \frac{1}{\sqrt2}$
Since the value of $\cos \theta$ is negative and $\sin \theta$ is positive this is the case in II quadrant
Therefore,
$\theta = \pi - \frac{\pi}{4}= \frac{3\pi}{4} \ \ \ \ \ \ \ \ \ \ \ \ (lies \ in \ II \ quadrant)$
$z = r\cos \theta + ir\sin \theta$
$=\sqrt2\cos \frac{3\pi}{4} + i\sqrt2\sin \frac{3\pi}{4}$
$=\sqrt2\left ( \cos \frac{3\pi}{4} + i\sin \frac{3\pi}{4} \right )$
Therefore, the required polar form is

$\sqrt2\left ( \cos \frac{3\pi}{4} + i\sin \frac{3\pi}{4} \right )$

Question:6 Solve each of the equation: $\small 3x^2-4x+\frac{20}{3}=0$

Answer:

Given equation is
$\small 3x^2-4x+\frac{20}{3}=0$
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

$a=3,b=-4 \ and \ c= \frac{20}{3}$
Therefore,
$\frac{-(-4)\pm \sqrt{(-4)^2-4.3.\frac{20}{3}}}{2.3}= \frac{4\pm\sqrt{16-80}}{6} = \frac{4\pm\sqrt{-64}}{6}$ $=\frac{4\pm8i}{6}= \frac{2}{3}\pm i\frac{4}{3}$
Therefore, the solutions of requires equation are

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

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

Answer:

Given equation is
$\small x^2-2x+\frac{3}{2}=0$
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 $a=1,b=-2 \ and \ c= \frac{3}{2}$
Therefore,
$\frac{-(-2)\pm \sqrt{(-2)^2-4.1.\frac{3}{2}}}{2.1}= \frac{2\pm\sqrt{4-6}}{2} = \frac{2\pm\sqrt{-2}}{2}$ $=\frac{2\pm i\sqrt2}{2}=1\pm i\frac{\sqrt2}{2}$
Therefore, the solutions of requires equation are

$1\pm i\frac{\sqrt2}{2}$

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

Answer:

Given equation is
$\small 27x^2-10x+1=0$
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 $a=27,b=-10 \ and \ c= 1$
Therefore,
$\frac{-(-10)\pm \sqrt{(-10)^2-4.27.1}}{2.27}= \frac{10\pm\sqrt{100-108}}{54} = \frac{10\pm\sqrt{-8}}{54}$ $=\frac{10\pm i2\sqrt2}{54}=\frac{5}{27}\pm i\frac{\sqrt2}{27}$
Therefore, the solutions of requires equation are $\frac{5}{27}\pm i\frac{\sqrt2}{27}$

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

Answer:

Given equation is
$\small 21x^2-28x+10=0$
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 $a=21,b=-28 \ and \ c= 10$
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 $\small z_1=2-i, z_2=1+i$ , find $\small \left |\frac{z_1+z_2+1}{z_1-z_2+1} \right |$ .

Answer:

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

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

Question:11 If $\small a+ib=\frac{(x+i)^2}{2x^2+1}$ , prove that $\small a^2+b^2=\frac{(x^2+1)^2}{(2x^2+1)^2}$ .

Answer:

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

$\small a+ib=\frac{(x+i)^2}{2x^2+1} = \frac{x^2+i^2+2xi}{2x^2+1}=\frac{x^2-1+2xi}{2x^2+1}=\frac{x^2-1}{2x^2+1}+i\frac{2x}{2x^2+1}$
On comparing real and imaginary part. we will get
$a=\frac{x^2-1}{2x^2+1}\ and \ b=\frac{2x}{2x^2+1}$
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 $\small z_1=2-i,z_2=-2+i.$ Find

$\small Re\left ( \frac{z_1z_2}{\bar{z_1}} \right )$

Answer:

It is given that
$\small z_1=2-i \ and \ z_2=-2+i$
Now,
$z_1z_2= (2-i)(-2+i)= -4+2i+2i-i^2=-4+4i+1= -3+4i$
And
$\bar z_1 = 2+i$
Now,
$\frac{z_1z_2}{\bar z_1}= \frac{-3+4i}{2+i}= \frac{-3+4i}{2+i}\times \frac{2-i}{2-i}= \frac{-6+3i+8i-4i^2}{2^2-i^2}= \frac{-6+11i+4}{4+1}$ $= \frac{-2+11i}{5}= -\frac{2}{5}+i\frac{11}{5}$
Now,
$Re\left ( \frac{z_1z_2}{z_1} \right )= -\frac{2}{5}$
Therefore, the answer is

$-\frac{2}{5}$

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

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

Answer:

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

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

Answer:

Let
$z = \small \frac{1+2i}{1-3i}$
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,
$r\cos \theta= -\frac{1}{2} \ \ and \ \ r\sin \theta =\frac{1}{2}$
Square and add both the sides
$r^2\cos^2\theta +r^2\sin^2\theta= \left ( -\frac{1}{2} \right )^2+\left ( \frac{1}{2} \right )^2$
$r^2(\cos^2\theta +\sin^2\theta)= \left ( \frac{1}{4} \right )+\left ( \frac{1}{4} \right )$
$r^2= \frac{1}{2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because \sin^2\theta +\cos^2\theta = 1)$
$r = \frac{1}{\sqrt2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because r > 0)$
Therefore, the modulus is $\frac{1}{\sqrt2}$
Now,
$\frac{1}{\sqrt2} \cos\theta = -\frac{1}{2} \ \ and \ \ \frac{1}{\sqrt2} \sin\theta = \frac{1}{2}$
$\cos\theta = -\frac{1}{\sqrt2} \ \ and \ \ \sin\theta = \frac{1}{\sqrt2}$
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 $=\left ( \pi-\frac{\pi}{4} \right )= \frac{3\pi}{4}$
Therefore, Argument and modulus are $\frac{3\pi}{4} \ \ and \ \ \frac{1}{\sqrt2}$ respectively

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

Answer:

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

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

Question:15 Find the modulus of $\small \frac{1+i}{1-i}-\frac{1-i}{1+i}$ .

Answer:

Let
$z =\small \frac{1+i}{1-i}-\frac{1-i}{1+i}$
Now, we will reduce it into
$z =\small \frac{1+i}{1-i}-\frac{1-i}{1+i} = \frac{(1+i)^2-(1-i)^2}{(1+i)(1-i)}= \frac{1^2+i^2+2i-1^2-i^2+2i}{1^2-i^2}$ $= \frac{4i}{1+1}= \frac{4i}{2}=2i$
Now,
$r\cos\theta = 0 \ \ and \ \ r\sin \theta = 2$
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

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

Question:16 If $\small (x+iy)^3=u+iv$ , then show that $\small \frac{u}{x}+\frac{v}{y}=4 (x^2-y^2).$

Answer:

it is given that
$\small (x+iy)^3=u+iv$
Now, expand the Left-hand side
$x^3+(iy)^3+3.(x)^2.iy+3.x.(iy)^2= u + iv$
$x^3+i^3y^3+3x^2iy+3xi^2y^2= u + iv$
$x^3-iy^3+3x^2iy-3xy^2= u + iv$ $(\because i^3 = -i \ \ and \ \ i^2 = -1)$
$x^3-3xy^2+i(3x^2y-y^3)= u + iv$
On comparing real and imaginary part. we will get,
$u = x^3-3xy^2 \ \ \ and \ \ \ v = 3x^2y-y^3$
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 $\small \alpha$ and $\small \beta$ are different complex numbers with $\small |\beta|=1$ , then find $\small \left | \frac{\beta -\alpha }{1-\bar{\alpha }\beta } \right |$ .

Answer:

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

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

Question:18 Find the number of non-zero integral solutions of the equation $\small |1-i|^x=2^x$ .

Answer:

Given problem is
$\small |1-i|^x=2^x$
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 $\small |1-i|^x=2^x$

Question:19 If $\small (a+ib)(c+id)(e+if)(g+ih)=A+iB,$ then show that $\small (a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2)=A^2+B^2$

Answer:

It is given that
$\small (a+ib)(c+id)(e+if)(g+ih)=A+iB,$
Now, take mod on both sides
$\left | (a+ib)(c+id)(e+if)(g+ih) \right |= \left | A+iB \right |$
$|(a+ib)||(c+id)||(e+if)||(g+ih)|= \left | A+iB \right |$ $(\because |z_1z_2|=|z_1||z_2|)$
$(\sqrt{a^2+b^2})(\sqrt{c^2+d^2})(\sqrt{e^2+f^2})(\sqrt{g^2+h^2})= (\sqrt{A^2+B^2})$
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:20 If $\small \left ( \frac{1+i}{1-i} \right )^m=1,$ then find the least positive integral value of $\small m$ .

Answer:

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