Complex Numbers and Quadratic Equations Class 11 – NCERT Solutions PDF

Class 11 Complex Numbers Miscellaneous Exercise Solution

Q1. Evaluate: ${\left[i^{18}+{\left(\frac{1}{i}\right)}^{23}\right]}^3$

A.1. ${\left[i^{18}+{\left(\frac{1}{i}\right)}^{23}\right]}^3$

=  ${\left[i^{\mathrm{4\prime }<br>4+2}+\frac{1}{i^{\mathrm{4\prime }<br>6+1}}\right]}^3$

=  ${\left[-1+\frac{1}{i}\right]}^3$ [as i^{4 ×k + 2} = –1 and i^{4 ×k + 1} = i]

=  ${\left[-1+\frac{i}{i^2}\right]}^3$

= [–1 – i]³                        [as i² = –1]

= (-1)³ (1 + i)³

= –1 [1³ + i³ + 3 × 1 ×i(1 + i)]                        [since, (a + b)³ = a³ + b³ + 3ab(a + b)]

= –1 [1 – i³ + 3i(1 + i)]

= –1 [1 – i³ + 3i + 3i²]

= –1 [1 – i + 3i – 3]                           [Since, i² = –1 and i³ = i².i = –i]

= –1 [–2 + 2i]

= 2 – 2i

A.2. To proof, Re (z1z2) = Re z1 Re z2 – Imz1 Imz2

Let z­­₁ = x₁ + iy₁ and z₂ = x₂ + iy₂ be two complex number.

Then, z₁.z₂ = (x₁ + iy₁)(x₂ + iy₂)

=x₁x₂ + ix₁y₂ + ix₂y₁ + i²y₁y₂

= x₁x₂ + ix₁y₂ + ix₂y₁ – y₁y₂            [since, i² = -1]

= (x₁x₂ – y₁y₂) + i(x₁y₂ + x₂y₁)

As, Re(z₁z₂) = (x₁)(x₂) – (y₁)(y₂)

Now, RHS = Re z₁ Re z₂ – Imz₁Imz₂ = x₁x₂ – y₁y₂

Therefore, Re(z₁z₂) = Rez₁Rez₂ – Imz₁Imz₂

Hence proved.

A.3. $\left(\frac{1}{1-4i}-\frac{2}{1+i}\right)$  $\left(\frac{3-4i}{5+i}\right)$

=  $\left[\frac{1+i-2\left(1-4i\right)}{\left(1-4i\right)\left(1+i\right)}\right]$  $\left[ \frac{3-4i}{5+i}\right]$

=  $\left(\frac{1+i-2+8i}{1+i-4i-4i^2}\right)$  $\left( \frac{3-4i}{5+i}\right)$

=  $\left(\frac{9i-1}{5-3i}\right)$  $\left( \frac{3-4i}{5+i}\right)$

=  $\frac{\left(9i-1\right)\left(3-4i\right)}{\left(5-3i\right)\left(5+i\right)}$

=  $\frac{27i-36i^2-3+4i}{25+5i-15i-3i^2}$

= .. [since, i² = –1]

=  $\frac{33+31i}{28-10i}$

=  $\frac{33+31i}{28-10i}$ ×  $\frac{28+10i}{28+10i}$ [multiplying denominator and numerator by 28 + 10i]

=  $\frac{\left(3\mathrm{3\prime } <br>28\right)+\left(3\mathrm{3\prime }<br>10i\right)+\left(31\mathrm{i\prime } <br>28\right)+\left(31\mathrm{i\prime }<br>10i\right)}{28^2-{\left(10i\right)}^2}$ [since (a – b)(a + b) = a² – b²]

=  $\frac{924+330i+868i+310i^2}{784-100i^2}$

=  $\frac{924+1198i-310}{784+100}$ [since, i² = –1]

=  $\frac{614+1198i}{884}$

=  $\frac{2\left(307+599i\right)}{884}$

=  $\frac{307+599i}{442}$

=  $\frac{307}{442}$ +  $i\frac{599}{442}$

Solve each of the equation in Exercises 4 to 7.

Q4.

A.8. z1 = 2 – i ,z2 = 1 + i

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

=  $\left|\frac{2-i+1+i+1}{2-i-1-i+1}\right|$

=  $\left|\frac{4}{2-2i}\right|$

=  $\left|\frac{4}{2\left(1-i\right)}\right|$

=  $\left|\frac{2}{1i}\right|$

=  $|\frac{2}{1-i}$ ×  $\frac{1+i}{1+i}|$ [multiply numerator and denominator by (1 + i)]

=  $\left|\frac{2+2i}{1^2-i^2}\right|$

=  $\left|\frac{2+2i}{1-\left(-1\right)}\right|$ [since, i2 = –1]

=  $\left|\frac{2+2i}{1+1}\right|$

=  $\left|\frac{2\left(1+i\right)}{2}\right|$

= |1 + i|

Q9. If a + ib $=\frac{{\left(x+i\right)}^2}{2x^2+1}$ , prove thata2 + b2  $=\frac{{\left(x^2+1\right)}^2}{{\left(2x^2+1\right)}^2}.$

A.9. Let, z = a + ib

=  $\frac{{\left(x+i\right)}^2}{2x^2+ 1}$

=  $\frac{x^2+ i^2+2xi}{2x^2+ 1}$ [since, (a + b)2 = a2 + b2 + 2ab]

=  $\frac{x^2- 1+2xi}{2x^2+ 1}$ [since, i2 = –1]

=  $\frac{x^2- 1}{2x^2+ 1}$ +  $\frac{i2x}{2x^2+ 1}$

So, |z|2 = a2 + b2

=  $\frac{{\left(x^2- 1\right)}^2}{{\left(2x^2+ 1\right)}^2}$ +  $\frac{{(2x)}^2}{{\left(2x^2+ 1\right)}^2}$

=  $\frac{{\left(x^2\right)}^2+ 1^2- 2.x^2.1+ 4x^2}{{\left(2x^2+ 1 \right)}^2}$ [since, (a + b)2 = a2 + b2 + 2ab]

=  $\frac{x^4+ 1-2x^2+ 4x^2}{{\left(2x^2+ 1\right)}^2}$

=  $\frac{x^4+ 2x^2+ 1}{{\left(2x^2+ 1\right)}^2}$

=  $\frac{{\left(x^2+ 1\right)}^2}{{\left(2x^2+ 1\right)}^2}$ [as, (a + b)2 = a2 + b2 + 2ab]

Hence proved.

Q10. Let z1 = 2 – i, z2 = –2 + i. Find

$RE\left(\frac{z_1{z}_2}{z_1}\right)$ (ii)  $IM\left(\frac{1}{z_1{z}_1}\right)$

A.10.  Z1 = 2 – i, z2 = –2 + i

Z1z2 = (2 – i)(–2 + i)

= –4 + 2i + 2i – i2

= –4 + 4i + 1[since, i2 = –1]

= –3 + 4i

$\overline{z_1}$ = 2 + i

i.  $\frac{z_1{z}_2}{\overline{z_1}}$ =  $\frac{- 3+4i}{2+i}$

=  $\frac{-3+4i}{2+i}$ ×  $\frac{2-i}{2-i}$ [multiply denominator and numerator by (2 – i)]

=  $\frac{-6+3i+8i-4i^2}{2^2- i^2}$

=  $\frac{-6+11i+4}{4-\left(-1\right)}$ [since, i2 = –1]

=  $\frac{-6+4+11i}{4+1 }$

=  $\frac{-2+11i}{5}$

=  $\frac{-2}{5}$ +  $\frac{11}{5}i$

So, Re(  $\frac{z_1{z}_2}{\overline{z_1}}$ ) =  $\frac{-2}{5}$

ii.  $\frac{1}{z_1\overline{z_1}}$ =  $\frac{1}{\left(2-i\right)\left(2+i\right)}$

=  $\frac{1}{2^2- i^2}$

=  $\frac{1}{4+1}$ [since, i2 = –1]

=  $\frac{1}{5}$ + 0i

Therefore, Im  $\left(\frac{1}{z_1\overline{z_1}}\right)$ = 0

Q11. Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of –6 – 24i.

A.11. Let z = (x – iy)(3 + 5i)

= 3x + 5xi – 3yi – 5yi²

= (3x + 5y) + (5x – 3y)i

Given,  $\bar{z}$ = –6 – 24i

=> (3x + 5y) – (5x – 3y)i = –6 – 24i

Equating real and imaginary part,

3x + 5y = –6 ----------- (1)

5x – 3y = 24 ----------- (2)

Multiplying (1) by 3 and (2) by 5 and adding them, we get

9x + 15y + 25x – 15y = –18 + 120

=> 34x = 102

=>x = 102/34 = 3

Putting x = 3 in (1) we get,

3 × 3 + 5y = –6

=> 9 + 5y = –6

=> 5y = –6 – 9

=> 5y = –15

=>y = –15/5 = –3

Hence, the values of x and y are 3 and –3 respectively.

A.12. $\frac{1+i}{1-i}$ –  $\frac{1-i}{1+i}$

=  $\frac{{\left(1+i\right)}^2- {\left(1-i\right)}^2}{\left(1-i\right)\left(1+i\right)}$

=  $\frac{1^2+i^2+2.1.i –\left(1^2+i^2- 2.1.i\right)}{1^2- i^2}$ [Since, (a + b)² = a² + b² + 2ab;

(a – b)² = a² + b² – 2ab;

a² – b² = (a + b)(a – b)]

=  $\frac{1-1+2i-1+1+2i}{1+1}$ [Since, i² = –1]

=  $\frac{4i}{2}$

= 2i

A.13. (x + iy)3 = u + iv

=>x³ + (iy)³ + 3.x.iy(x + iy) = u + iv   [since, (a + b)³ = a³ + b³ + 3ab(a + b)]

=>x³ – iy³ + 3x²yi + 3xy²i² = u + iv

=>x³ – iy³ + 3x²yi – 3xy² = u + iv                [since, i² = -1]

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

Equating real and imaginary part we get,

u = x³ – 3xy² and v = 3x²y – y³

Now,  $\frac{u}{x}$ +  $\frac{v}{y}$

=  $\frac{x^3-3x{y}^2}{x}$ +  $\frac{3x^{2}y- y^3}{y}$

=  $\frac{x\left(x^2- 3y^2\right)}{x}$ +  $\frac{y\left(3x^2- y^2\right)}{y}$

= x² – 3y² + 3x² – y²

= 4x² – 4y²

= 4(x² – y²)

Hence proved.

So, the only solution of the given equation is 0.

Hence, there is no non – zero integral solution of the given equation.

A15. Given,

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

We know that,

Hence proved.

A.16. We have,

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

=>  ${\left(\frac{1+i}{1-i} \times \frac{1+}{1+}\right)}^m$ = 1 [multiply denominator and numerator of LHS by (1 + i)]

=>  ${\left(\frac{1+i+i+i^2}{1^2- i^2}\right)}^m$ = 1[since, (a – b)(a + b) = a² – b²]

=>  ${\left(\frac{1+2i-1}{1+1}\right)}^m$ = 1[since, i² = –1]

=>  ${\left(\frac{2i}{2}\right)}^m$ = 1

=>i^m = 1

=>i^m = i^4k              [since, i^4k = 1]

So, m = 4k where k = integer

Therefore, least positive integral value of m is,

m = 4 × 1

m = 4