NCERT Solutions Class 11 – Chapter 4 Complex Numbers And Quadratic Equations – Miscellaneous Exercise

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

Solution:

[Tex]\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[/Tex]

= [-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 z₁ and z₂, prove that, Re (z₁z₂) = Re z₁ Re z₂ – Im z₁ Im z₂

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 [Tex]\displaystyle\left(\frac{1}{1-4i}-\frac{2}{1+i}\right)\left(\frac{3-4i}{5+i}\right)[/Tex]

Solution:

[Tex]\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}[/Tex]

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

[Tex]\\ =\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}[/Tex]

Hence, this is the required standard form.

Question 4. If x – iy = [Tex]\sqrt{\frac{a-ib}{c-id}} [/Tex] prove that (x² + y²)²[Tex]=\frac{a^2+b^2}{c^2+d^2}[/Tex]

Solution:

Given:

x – iy = [Tex]\sqrt{\frac{a-ib}{c-id}}[/Tex]

[Tex]=\sqrt{\frac{a-ib}{c-id}\times\frac{c+id}{c+id}}[/Tex]

On multiplying numerator and denominator by (c+id)

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

So,

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

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

On comparing real and imaginary parts, we get

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

(x² + y²)² = (x² – y²)² + 4x²y²

= [Tex]\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}[/Tex]

Hence proved

Question 5. If z₁ = 2 – i, z₂ = 1 + i, find [Tex]\left|\frac{z_1+z_2+1}{z_1-z_2+1}\right|[/Tex]

Solution:

Given, z₁ = 2 – i, z₂ = 1 + i

[Tex]\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[/Tex]

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

Question 6. If a + ib =  [Tex]\frac{(x+i)^2}{2x^2+1}[/Tex], prove that a² + b² = [Tex]\frac{(x^2+i)^2}{(2x^2+1)^2}[/Tex]

Solution:

Given:

a + ib = [Tex]\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)[/Tex]

On comparing the real and imaginary parts, we have

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

Therefore,

a² + b² = [Tex]\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}[/Tex]

Hence, proved,

a² + b² = [Tex]\frac{(x^2+1)^2}{(2x^2+1)^2}[/Tex]

Question 7. Let z₁ = 2 – i, z₂ = -2 + i. Find

(i) [Tex]Re\left(\frac{z_1z_2}{\overline{z_1}}\right)[/Tex]

(ii) [Tex]Im\left(\frac{1}{z_1\overline{z_2}}\right)[/Tex]

Solution:

(i) Given:

z₁ = 2 – i, z₂ = -2 + i

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

[Tex]\overline{z_1}[/Tex] = 2 + i

Therefore,

[Tex]\frac{z_1z_2}{\overline{z_1}}=\frac{-3+4i}{2+i}[/Tex]

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

[Tex]\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[/Tex]

On comparing the real parts, we have

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

(ii) [Tex]\frac{1}{z_1\overline{z_2}}=\frac{1}{(2-1)(2+i)}=\frac{1}{(2)^2+(1)^2}=\frac{1}{5}\\[/Tex]

On comparing the imaginary part, we get

[Tex]Im\left(\frac{1}{z_1\overline{z_2}}\right)[/Tex] = 0

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

Solution:

Let us assume z = (x – iy) (3 + 5i)

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

Therefore,

[Tex]\overline{z}[/Tex] =(3x + 5y) – i(5x – 3y)

Also given, [Tex]\overline{z}  [/Tex] = -6 – 24i 

And,

(3x + 5y) – i(5x – 3y) = -6 -24i

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 [Tex]\frac{1+i}{1-i}-\frac{1-i}{1+i}[/Tex]

Solution:

[Tex]\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[/Tex]

Question 10. If (x + iy)³ = u + iv, then show that [Tex]\frac{u}{y}+\frac{v}{y}[/Tex] = 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

[Tex]\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}[/Tex]

= x² – 3y² + 3x² – y²

= 4x² – 4y²

= 4(x² – y²)

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

Hence proved

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

Solution:

Assume α = a + ib and β = x + iy

Given: |β| = 1

So, [Tex]\sqrt{x^2+y^2}=1[/Tex]

= x2 + y2 = 1            ….(1)

[Tex]\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)] [/Tex]

= 1

[Tex]\therefore\left|\frac{\beta-\alpha}{1-\overline{\alpha}\beta}\right|=1[/Tex]

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

Solution:

|1 – i|^x = 2^t

[Tex](\sqrt{1^2+(-1)^2})^x=2^x\\ (\sqrt2)^x=2^x\\ 2^{\frac{x}{2}}=2^x\\ \frac{x}{2}=x[/Tex]

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 (a² + b²) (c² + d²) (e² + f²) (g² + h²) = A² + B².

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|

[Tex]\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}[/Tex]

On squaring both sides, we get

(a² + b²) (c² + d²) (e² + f²) (g² + h²) = A² + B²

Hence, proved.

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

Solution:

[Tex]\left(\frac{1+i}{1-i}\right)^m=1[/Tex]

[Tex]\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[/Tex]

i^m = 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).