NCERT Solutions Class 11 Chapter 4 Complex Numbers And Quadratic Equations – Miscellaneous Exercise (original) (raw)

Last Updated : 19 Apr, 2024

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

**Solution:

\displaystyle\left[i^{18}+\left(\frac{1}{i}\right)^{25}\right]^3\\ =\left[i^{4\times4+2}+\frac{1}{i^{4\times6+1}}\right]^3\\ =\left[(i^4)^{4}.i^2+\frac{1}{(i^4)^6.i}\right]^3\\ =\left[i^2+\frac{1}{i}\right]^3\ \ \ \ \ \ [i^4=1]\\ =\left[-1+\frac{1}{i}\times\frac{i}{i}\right]^3\ \ \ \ \ \ [i^2=-1]\\ =\left[-1+\frac{i}{i^2}\right]^3

= [-1 - i]3

= (-1)3 [1 + i]3

= -[13 + i3 + 3 × 1 × i (1 + i)]

= -[1 + i3 + 3i + 3i2]

= -[1 - i + 3i - 3]

= -[2 + 2i]

= 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

**Solution:

Let's assume z1 = x1 + iy1 and z2 = x2 + iy2 as two complex numbers

Product of these complex numbers, z1z2

z1z2 = (x1 + iy1)(x2 + iy2)

= x1(x2 + iy2) + iy1(x2 + iy2)

= x1x2 + ix1y2 + iy1x2 + i2y1y2

= x1x2 + ix1y2 + iy1x2 - y1y2 [i2 = -1]

= (x1x2 - y1y2) + i(x1y2 + y1x2)

Now,

Re(z1z2) = x1x2 - y1y2

⇒ Re(z1z2) = Rez1Rez2 - Imz1Imz2

Hence, proved.

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

**Solution:

\displaystyle\left(\frac{1}{1-4i}-\frac{2}{1+i}\right)\left(\frac{3-4i}{5+i}\right)=\left[\frac{(1+i)-2(1-4i)}{(1-4i)(1+i)}\right]\left[\frac{3-4i}{5+i}\right]\\ = \left[\frac{1+i-2+8i}{1+i-4i-4i^2}\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]=\frac{33+31i}{28-10i}=\frac{33+31i}{2(14-5i)}\\ = \frac{(33+31i)}{2(14-5i)}\times\frac{14+5i}{14+5i}

On multiplying numerator and denominator by ****(14+5i)**

\\ =\frac{462+165i+434i+155i^2}{2[(14)^2-(5i)^2]}=\frac{307+599i}{2(196-25i^2)}\\ =\frac{307+599i}{2(221)}=\frac{307+599i}{442}=\frac{307}{442}+\frac{599i}{442}

Hence, this is the required standard form.

**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}

**Solution:

Given:

x - iy = \sqrt{\frac{a-ib}{c-id}}

=\sqrt{\frac{a-ib}{c-id}\times\frac{c+id}{c+id}}

On multiplying numerator and denominator by ****(c+id)**

\\ =\sqrt{\frac{(ac+bd)+i(ad-bc)}{c^2+d^2}}

So,

(x - iy)2 = =\frac{(ac+bd)+i(ad-bc)}{c^2+d^2}

x2 - y2 - 2ixy =\frac{(ac+bd)+i(ad-bc)}{c^2+d^2}

On comparing real and imaginary parts, we get

x2 - y2 = \frac{ac+bd}{c^2+d^2} , -2xy = \frac{ad-bc}{c^2+d^2} (1)

(x2 + y2)2 = (x2 - y2)2 + 4x2y2

= \left( \frac{ac+bd}{c^2+d^2}\right)^2+\left(\frac{ad-bc}{c^2+d^2} \right)^2\ \ \ \ \ [Using (1)\\ =\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{(c^2+d^2)(a^2+b^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|

**Solution:

Given, z1 = 2 – _i, z2 = 1 + _i

\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}{2-2i}\right|=\left|\frac{4}{2(1-i)}\right|\\ =\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|\ \ \ \ \ [i^2=-1]\\ =\left|\frac{2(1+i)}{2}\right|\\ =|1+i|=\sqrt{1^2+1^2}=\sqrt2

Hence, the value of \left|\frac{z_1+z_2+1}{z_1-z_2+1}\right| is √2

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

**Solution:

Given:

a + ib = \frac{(x+i)^2}{2x^2+1}\\ =\frac{x^2+i^2+2xi}{2x^2+1}\\ =\frac{x^2-1+i2x}{2x^2+1}\\ =\frac{x^2-1}{2x^2+1}+i\left(\frac{2x}{2x^2+1}\right)

On comparing the real and imaginary parts, we have

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

Therefore,

a2 + b2 = \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+1)^2}\\ =\frac{x^4+1+2x^2}{(2x^2+1)^2}\\ =\frac{(x^2+1)^2}{(2x^2+1)^2}

Hence, proved,

a2 + b2 = \frac{(x^2+1)^2}{(2x^2+1)^2}

Question 7. Let z 1 = 2 – **i**, z** 2 = -2 + **i****. Find**

****(i)** Re\left(\frac{z_1z_2}{\overline{z_1}}\right)

****(ii)** Im\left(\frac{1}{z_1\overline{z_2}}\right)

**Solution:

****(i)** Given:

z1 = 2 - i, z2 = -2 + i

(i) z1z2 = (2 - i)(-2 + i) = -4 + 2i + 2i - i2 = -4 + 4i - (-1) = -3 + 4i

\overline{z_1} = 2 + i

Therefore,

\frac{z_1z_2}{\overline{z_1}}=\frac{-3+4i}{2+i}

On multiplying numerator and denominator by (2 - i), we get

\frac{z_1z_2}{\overline{z_1}}=\frac{(-3+4i)(2-i)}{(2+i)(2-i)}=\frac{-6+3i+8i-4i^2}{2^2+1^2}=\frac{-6+11i-4(-1)}{2^2+1^2}\\ =\frac{-2+11i}{5}=\frac{-2}{5}+\frac{11}{5}i

On comparing the real parts, we have

Re\left(\frac{z_1z_2}{\overline{z_1}}\right)=\frac{-2}{5}

****(ii)** \frac{1}{z_1\overline{z_2}}=\frac{1}{(2-1)(2+i)}=\frac{1}{(2)^2+(1)^2}=\frac{1}{5}\\

On comparing the imaginary part, we get

Im\left(\frac{1}{z_1\overline{z_2}}\right) = 0

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

**Solution:

Let us assume z = (_x – iy) (3 + 5__i)

z = 3x + xi - 3yi - 5yi2 = 3x + 5xi - 3yi + 5y = (3x + 5y) + i(5x - 3y)

Therefore,

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

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

And,

(3x + 5y) – i(5x – 3y) = -6 -24__i

After equating real and imaginary parts, we get

3x + 5y = -6 …… (i)

5x – 3y = 24 …… (ii)

After doing (i) x 3 + (ii) x 5, we have

(9x + 15y) + (25x – 15y) = -18 + 120

34x = 102

x = 102/34 = 3

Putting the value of _x in equation (i), we get

3(3) + 5y = -6

5y = -6 – 9 = -15

y = -3

Therefore, the values 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}

**Solution:

\frac{1+i}{1-i}-\frac{1-i}{1+i}=\frac{(1+i)^2-(1-i)^2}{(1-i)(1+i)}\\ =\frac{1+i^2+2i-1-i^2+2i}{1^2+1^2}\\ =\frac{4i}{2}=2i\\ \therefore\left|\frac{1+i}{1-i}-\frac{1-i}{1+i}\right|=|2i|=\sqrt{2^2}=2

Question 10. If ( x + **iy**)** 3 = **u + **iv****, then show that** \frac{u}{y}+\frac{v}{y} = 4(x2 - y2)

**Solution:

(x + iy)3 = u + iv

x3 + (iy)3 + 3 × x × iy(x + iy) = u + iv

x3 + i3y3 + 3x2yi + 3xy2 = u + iv

x3 - iy3 + 3x2yi - 3xy2 = u + iv

(x3 - 3xy2) + i(3x2y - y3) = u + iv

On equating real and imaginary parts, we get

u = x3 - 3xy2, v = 3x2y - y3

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

= x2 - 3y2 + 3x2 - y2

= 4x2 - 4y2

= 4(x2 - y2)

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

Hence proved

**Question 11. If α and β are different complex numbers with |β| = 1, then find \left|\frac{\beta-\alpha}{1-\overline{\alpha}\beta}\right|

**Solution:

Assume α = a + ib and β = x + iy

Given: |β| = 1

So, \sqrt{x^2+y^2}=1

= x2 + y2 = 1 ....(1)

\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+aiy-ibx+by)}\right|\\ =\left|\frac{(x-a)+i(y-b)}{(1-ax-by)+i(bx-ay)}\right|\\ =\left|\frac{(x-a)+i(y-b)}{(1-ax-by)+i(bx-ay)}\right|\ \ \ \ \ \left[\left|\frac{z_1}{z_2}\right|=\left|\frac{z_1}{z_2}\right|\right]\\ =\frac{\sqrt{(x-a)^2+(y-b)^2}}{\sqrt{(1-ax-by)^2+(bx-ay)^2}}\\ =\frac{\sqrt{x^2+a^2-2ax+y^2+b^2-2by}}{\sqrt{1+a^2x^2+b^2y^2-2ax+2abxy-2by+b^2x^2+a^2y^2-2abxy}}\\ =\frac{\sqrt{(x^2+y^2)+a^2+b^2-2ax-2by}}{\sqrt{1+a^2(x^2+y^2)+b^2(y^2+x^2)-2ax-2by}}\\ =\frac{\sqrt{1+a^2+b^2-2ax-2by}}{\sqrt{1+a^2+b^2-2ax-2by}}\ \ \ \ \ \ \ \ [Using\ (1)]

= 1

\therefore\left|\frac{\beta-\alpha}{1-\overline{\alpha}\beta}\right|=1

**Question 12. Find the number of non-zero integral solutions of the equation |1 – i| x = 2 x

**Solution:

|1 - i|x = 2t

(\sqrt{1^2+(-1)^2})^x=2^x\\ (\sqrt2)^x=2^x\\ 2^{\frac{x}{2}}=2^x\\ \frac{x}{2}=x

x = 2x

2x - x = 0

Thus, '0' is the only integral solution of the given equation.

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

**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 .

**Solution:

Given:

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

Therefore,

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

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

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

On squaring both sides, we get

(__a_2 + __b_2) (__c_2 + __d_2) (__e_2 + __f_2) (__g_2 + __h_2) = A2 + B2

Hence, proved.

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

**Solution:

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

\left(\frac{1+i}{1-i}\times\frac{1+i}{1+i}\right)^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

im = 1

Hence, m = 4k, where k is some integer.

Hence, the least positive integer is 1.

Thus, the least positive integral value of _m is 4 (= 4 × 1).