# NCERT Solutions for Class 11 Maths Chapter 4 - Complex Numbers and Quadratic Equations

Exercise 5.1

Express each of the complex number given in the Exercises 1 to 10 in the form a + ib:

$$\textbf{1. (5i)}\bigg(\frac{\textbf{-3}}{\textbf{5}}\textbf{i}\bigg).\\\textbf{Sol.}\space(5i)×\bigg(\frac{-3}{5}i\bigg)=\frac{5}{5}i×i×(-3)$$

= i2 × (– 3)

= – 3 × (– 1) [∵ i2 = – 1]

= 3

$$\text{Hence,}\space(5i)\bigg(\frac{-3}{5}i\bigg)=3+i0.$$

2. i9 + i19.

Sol. i9 + i19 = [(i2)4 × i] + [(i2)9 × i]

= (– 1)4 × i + (– 1)9 × i [∵ i2 = – 1]

= 1 × i + (– 1) × i

= i – i

= 0

Hence, i9 + i19 = 0 + i0.

3. i– 39.

$$\textbf{Sol.}\space i^{\normalsize-39}=\frac{1}{i^{39}}=\frac{1}{(i^{4})^{9}i^{3}}\\=\frac{1}{(1)^{9}×i^{3}}\space\begin{bmatrix}\because\space i^{2}=-1\\\therefore\space i^{4}=1\\\text{and}\space i^{3}=-i\end{bmatrix}\\=\frac{1}{-i}$$

Now, multiplying the numerator and denominator by i, we get

$$=\frac{i}{-i×1}=\frac{i}{-i^{2}}=\frac{i}{-(-1)}\\=\frac{i}{1}$$

= i

Hence, i– 39 = 0 + i

4. 3(7 + i7) + i(7 + i7).

Sol. 3(7 + i7) + i(7 + i7) = 21 + 21i + 7i + 7i2

= 21 + 28i – 7 [·.· i2 = – 1]

= 14 + 28i

5. (1 – i) – (– 1 + 6i).

Sol. (1 – i) – (– 1 + i6) = 1 – i + 1 – 6i

= 2 – i7.

$$\textbf{6.\space}\bigg(\frac{\textbf{1}}{\textbf{5}}+i\frac{\textbf{2}}{\textbf{5}}\bigg)-\bigg(\textbf{4}+i\frac{\textbf{5}}{\textbf{2}}\bigg)\\\textbf{Sol.}\space\bigg(\frac{1}{5}+\frac{2}{5}i\bigg)-\bigg(4+\frac{5}{2}i\bigg)\\=\frac{1}{5}+\frac{2}{5}i-4-\frac{5}{2}i\\=\bigg(\frac{1}{5}-4\bigg)+i\bigg(\frac{2}{5}+\frac{5}{2}\bigg)\\=\bigg(\frac{1-20}{5}\bigg)+i\bigg(\frac{4-25}{10}\bigg)\\=\bigg(\frac{-19}{5}\bigg)+i\bigg(\frac{-21}{10}\bigg)\\=\frac{-19}{5}=\frac{i21}{10}.$$

$$\textbf{7.}\bigg[\bigg(\frac{\textbf{1}}{\textbf{3}}+\textbf{i}\frac{\textbf{7}}{\textbf{3}}\bigg)+\bigg(\textbf{4}+\textbf{i}\frac{\textbf{1}}{\textbf{3}}\bigg)\bigg]\\-\bigg(\frac{\textbf{-4}}{\textbf{3}}+\textbf{i}\bigg).\\\textbf{Sol.}\bigg[\bigg(\frac{1}{3}+\frac{7}{3}i\bigg)+\bigg(4+\frac{1}{3}i\bigg)\bigg]\\-\bigg(\frac{-4}{3}+i\bigg).\\=\frac{1}{3}+\frac{7}{3}i+4+\frac{1}{3}i+\frac{4}{3}-i\\=\bigg(\frac{1}{3}+4+\frac{4}{3}\bigg)+i\bigg(\frac{7}{3}+\frac{1}{3}-1\bigg)\\=\bigg(\frac{1+12+4}{3}\bigg)+i\bigg(\frac{7+1-3}{3}\bigg)$$

$$=\frac{17}{3}+i\frac{5}{3}.$$

8. (1 – i)4.

Sol. (1 – i)4 = [(1 – i)2]2

= [1 + i2 – 2i]2

[∵ (a – b)2 = a2 + b2 – 2ab]

= [1 – 1 – 2i]2 [∵ i2 = – 1]

= [– 2i]2

= (– 2)2 (i)2

= 4(– 1)  [∵ i2 = – 1]

= – 4

Hence, (1 – i)4 = – 4 + i0.

$$\textbf{9.}\space\bigg(\frac{\textbf{1}}{\textbf{3}}+\textbf{3i}\bigg)^{\textbf{3}}.\\\textbf{Sol.}\bigg(\frac{1}{3}+3i\bigg)^{3}=\\\bigg(\frac{1}{3}\bigg)^{3}+(3i)^{3}+3\bigg(\frac{1}{3}\bigg)(3i)\bigg(\frac{1}{3}+3i\bigg)$$

[∵ (a + b)3 = a3 + b3 + 3ab(a + b)]

$$=\frac{1}{27}+27i^{3}+3i\bigg(\frac{1}{3}+3i\bigg)\\=\frac{1}{27}+27(-i)+i+9i^{2}\\\lbrack\because i^{3}=-i\rbrack\\=\frac{1}{27}-27i+i-9\\\lbrack\because\space i^{2}=-1\rbrack\\=\bigg(\frac{1}{27}-9\bigg)+i(-27+1)\\=\bigg(\frac{1-243}{27}\bigg)+i(-26)\\=\frac{-242}{27}-i26.$$

$$\textbf{10.}\space\bigg(\textbf{-2-}\frac{\textbf{1}}{\textbf{3}}i\bigg)^{\textbf{3}}.\\\textbf{Sol.}\space\bigg(-2-\frac{1}{3}i\bigg)^{3}=(-1)^{3}\bigg(2+\frac{1}{3}i\bigg)^{3}\\=(-1)^{3}\bigg[2^{3}+\bigg(\frac{i}{3}\bigg)^{3}+3(2)\bigg(\frac{i}{3}\bigg)\bigg(2+\frac{i}{3}\bigg)\bigg]$$

[∵ (a + b)3 = a3 + b3 + 3ab(a + b)]

$$=-1\bigg[8+\frac{i^{3}}{27}+2i\bigg(2+\frac{i}{3}\bigg)\bigg]\\=-\bigg[8-\frac{1}{27}+4i+\frac{2i^{2}}{3}\bigg]\\\lbrack\because i^{3}=-1\rbrack\\=-\bigg[8-\frac{i}{27}+4i-\frac{2}{3}\bigg]\\\lbrack\because\space i^{2}=-1\rbrack\\=-\bigg[\bigg(8-\frac{2}{3}\bigg)+i\bigg(\frac{-1}{27}+4\bigg)\bigg]\\=-\bigg[\bigg(\frac{24-2}{3}\bigg)+i\bigg(\frac{-1+108}{27}\bigg)\bigg]\\=-\bigg[\frac{22}{3}+\frac{107}{27}i\bigg]$$

$$=\frac{-22}{3}-i\frac{107}{27}.$$

Find the multiplicative inverse of each of the complex numbers given in the Exercises 11 to 13.

11. 4 – 3i.

Sol. Let, z = 4 – 3i

∴  | z |2 = 42 + (– 3)2

= 16 + 9 = 25

and z = 4 + 3i

So, multiplicative inverse of 4 – 3i is given by

$$z^{\normalsize-1}=\frac{z}{|z|^{2}}\\=\frac{4+3i}{25}\\=\frac{4}{25}+\frac{3}{25}i.$$

$$\textbf{12.}\space\sqrt{\textbf{5}}\textbf{+3i}.\\\textbf{Sol.}\space\text{Let}\space z=\sqrt{5}+3i\\\therefore\space|z|^{2}=(\sqrt{5})^{2}+3^{2}$$

$$\text{and}\space\bar z=\sqrt{5}-3i\\\text{So, multiplicative inverse of}\space\sqrt{3}+3i\\\text{is given by}\\z^{\normalsize-1}=\frac{\bar z}{|z|^{2}}\\=\frac{\sqrt{5}-3i}{14}\\=\frac{\sqrt{5}}{14}-\frac{3}{14}i.$$

13. – i.

Sol. Let, z = – i

∴ | z |2 = (0)2 + (– 1)2 = 1

$$\text{and}\space\bar{z} = i$$

So, multiplicative inverse of (– i) is given by

$$z^{\normalsize-1}=\frac{\bar{z}}{|z|^{2}}\\=\frac{i}{1}=i.$$

14. Express the following expression in the terms of a + ib:

$$\frac{(\textbf{3+i}\sqrt{\textbf{5}})(\textbf{3-i}\sqrt{\textbf{5}})}{(\sqrt{\textbf{3}}+\sqrt{\textbf{2}}\textbf{i})-(\sqrt{\textbf{3}}-\textbf{i}\sqrt{\textbf{2}})}.$$

$$\textbf{Sol.}\space\frac{(3+i\sqrt{5})(3-i\sqrt{5})}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})}\\=\frac{(3)^{2}-(i\sqrt{5})^{2}}{\sqrt{3}+\sqrt{2}i-\sqrt{3}+i\sqrt{2}}$$

[∵ (a + b) (a – b) = a2 – b2]

$$=\frac{9-5i^{2}}{2\sqrt{2}i}\\=\frac{9-5(-1)}{2\sqrt{2}i}\space[\because i^{2}=1]\\=\frac{14}{2\sqrt{2}i}$$

Now, multiplying numerator and denominator by i

$$=\frac{14i}{2\sqrt{2}i×i}\\=\frac{14i}{2\sqrt{2}i^{2}}\\=\frac{14i}{2\sqrt{2}(\normalsize-1)}\space[\because i^{2}=-1]\\=\frac{-14i}{2\sqrt{2}}=\frac{-7i}{\sqrt{2}}$$

Now, multiplying numerator and denominator by

$$\text{Now, multiplying numerator and}\\ \text{denominator by}\space\sqrt{2}\\=\frac{-7i}{\sqrt{2}}×\frac{\sqrt{2}}{\sqrt{2}}\\=\frac{-7\sqrt{2}i}{2}\\\text{Hence}\frac{(3+\sqrt{5}i)(3-\sqrt{5}i)}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-\sqrt{2}i)}\\=0+i\bigg(\frac{-7\sqrt{2}}{2}\bigg).$$

Exercise 5.2

Find the modulus and the arguments of each of the complex numbers in Exercises 1 to 2:

$$\textbf{1.}\space \textbf{z\space= -1-i}\sqrt{\textbf{3}}.\\\textbf{Sol.}\space\text{Given}\space z=-1-i\sqrt{3}$$

Let, r cos θ = – 1

$$\text{and}\space r\text{sin}\space\theta=-\sqrt{3}$$

On squaring and adding both, we get

(r cos θ)2 + (r sin θ)2 = (-1)2+($$-\sqrt{3})2$$

⇒ r2 (cos2 θ + sin2 θ) = 1 + 3

(∵ cos2 θ + sin2 θ = 1)

⇒ r2 = 4 (r > 0)

⇒ r = 2 (·.· r > 0)

Thus, modulus = 2

So, we have

$$\text{2 cos}\space\theta = – \text{1 and 2 sin}\space\theta =-\sqrt{3}\\\Rarr\space\text{cos}\space\theta=\frac{\normalsize-1}{2}\text{and sin}\space\theta =\frac{-\sqrt{3}}{2}$$

As the values of both sin  θ and cos  θ are negative,  θ lies in III quadrant.

$$\text{Argument = -}\bigg(\pi-\frac{\pi}{3}\bigg)\\=\frac{-2\pi}{3}$$

Thus, the modulus and argument of the complex number $$-1-\sqrt{3}i\space\text{are 2 and}\frac{-2\pi}{3}\\\text{respectively.}$$

$$\textbf{2. z =}-\sqrt{\textbf{3}}+\textbf{i}.\\\textbf{Sol.}\space \text{Given, z = -}\sqrt{3}+i\\\text{Let, r cos}\space\theta=-\sqrt{3}\space\text{and}\text{r sin}\space\theta=1$$

On squaring and adding both, we get

$$r^{2}(\text{cos}^{2}\theta)+r^{2}(\text{sin}^{2}\theta)\\=(-\sqrt{3})^{2}+(1)^{2}\\\Rarr\space r^{2}(\text{cos}^{2}\theta + sin^{2}\theta)\\=(-\sqrt{3})^{2}+(1)^{2}$$

⇒ r2 (1) = 3 + 1 (∵ cos2 θ + sin2 θ = 1)

⇒ r2 = 4

⇒ r = 2 (∵ r > 0)

Thus modulus = 2

$$\text{So, \space2 cos θ =}-\sqrt{3}\\\text{and 2 sin θ = 1}\\\Rarr\space\text{cos}\space\theta=\frac{-\sqrt{3}}{2}\text{and sin}\theta=\frac{1}{2}\\\therefore\space\theta=\pi=\frac{\pi}{6}=\frac{5\pi}{6}$$

[As θ lies in the II quadrant]

Thus, the modulus and argument of the complex number $$-\sqrt{3}+i\space\text{are 2 and}\space\frac{5\pi}{6}\text{respectively.}$$

Convert each of the complex numbers given in Exercises 3 to 8 in the polar form.

3. 1 – i.

Sol. Let, z = 1 – i

So, r cos θ = 1 and r sin θ = – 1

On squaring and adding both, we get

r2 cos2 θ + r2 sin2 θ = 12 + (– 1)2

⇒ r2 (cos2 θ + sin2 θ) = 1 + 1

(∵ cos2 θ + sin2 θ = 1)

⇒ r2 = 2

$$\Rarr\space r=\sqrt{2}\space(\because r>0)\\\text{So,\space}\sqrt{2}\space\text{cos}\space\theta=1\text{and}\\\sqrt{2}\space\text{sin}\space\theta=-1\\\Rarr\space\text{cos}\space\theta=\frac{1}{\sqrt{2}}\\\text{and sin}\space\theta=\frac{\normalsize-1}{\sqrt{2}}\\\therefore\space\theta=\frac{-\pi}{4}$$

[As  θ lies in the IV quadrant]

So, in polar form

1 – i = r cos  θ + ir sin  θ

$$=\sqrt{2}\text{cos}\bigg(\frac{-\pi}{4}\bigg)+i\sqrt{2}\text{sin}\bigg(\frac{-\pi}{4}\bigg)\\=\sqrt{2}\bigg[\text{cos}\bigg(\frac{-\pi}{4}\bigg)+i\space\text{sin}\bigg(\frac{-\pi}{4}\bigg)\bigg]$$

4. – 1 + i.

Sol. Let, z = – 1 + i

So, r cos θ = – 1 and r sin θ = 1

On squaring and adding both, we get

r2 cos2 θ + r2 sin2 θ = (– 1)2 + 1

⇒ r2 (cos2 θ + sin2 θ) = 1 + 1

⇒ r2 (1) = 2 (·.· cos2 θ + sin2 θ = 1)

⇒ r2 = 2

$$\Rarr\space \text{r}=\sqrt{2}\space(\because r\gt0)\\\text{So,}\space\sqrt{2}\space\text{cos}\space\theta=-1\space\text{and}\sqrt{2}\text{sin}\space\theta=1\\\text{cos}\space\theta=\frac{\normalsize-1}{\sqrt{2}}\space\text{and sin}\space\theta=\frac{1}{\sqrt{2}}\\\theta=\pi-\frac{\pi}{4}=\frac{3\pi}{4}$$

[As θ lies in the II quadrant]

So, in polar form

– 1 + i = r cos θ + ir sin θ

$$=\sqrt{2}\space\text{cos}\bigg(\frac{3\pi}{4}\bigg)+i\sqrt{2}\space\text{sin}\bigg(\frac{3\pi}{4}\bigg)\\=\sqrt{2}\bigg[\text{cos}\bigg(\frac{3\pi}{4}\bigg)+ i\space\text{sin}\bigg(\frac{3\pi}{4}\bigg)\bigg]$$

5. – 1 – i.

Sol. Let z = – 1 – i

So, r cos θ = – 1 and r sin θ = – 1

On squaring and adding both, we get

r2 cos2 θ + r2 sin2 θ = (– 1)2 + (– 1)2

⇒ r2 (cos2 θ + sin2 θ) = 1 + 1

⇒ r2 = 2 (∵ cos2 θ + sin2 θ = 1)

(∵ r > 0)

$$r=\sqrt{2}\\\text{So,}\space\sqrt{2}\space\text{cos}\space\theta=-1\space\text{and}\\\sqrt{2}\space\text{sin}\space\theta=-1\\\Rarr\space\text{cos}\space\theta=\frac{\normalsize-1}{\sqrt{2}}\text{and sin}\space\theta=\frac{\normalsize-1}{\sqrt{2}}\\\text{Therefore,}\space\theta=-\bigg(\pi-\frac{\pi}{4}\bigg)=\frac{-3\pi}{4}$$

(As θ lies in the III quadrant)

Hence, in polar form

– 1 – i = r cos θ + ir sin θ

$$=\sqrt{2}\space\text{cos}\bigg(\frac{-3\pi}{4}\bigg)+i\sqrt{2}\space\text{sin}\bigg(\frac{-3\pi}{4}\bigg)\\=\sqrt{2}\bigg[\text{cos}\bigg(\frac{3\pi}{4}\bigg)+ i\space\text{sin}\bigg(\frac{3\pi}{4}\bigg)\bigg]$$

6. – 3.

Sol. Let z = – 3 = – 3 + 0i

So, r cos θ = – 3 and r sin θ = 0

On squaring and adding both, we get

r2 cos2 θ + r2 sin2 θ = (– 3)2 + 02

⇒ r2 (cos2 θ + sin2 θ) = 9

⇒ r2 = 9    (∵ cos2 θ + sin2 θ = 1)

⇒ r = 3     (∵ r > 0)

So,  3 cos θ = – 3 and 3 sin θ = 0

cos θ = – 1 and sin θ = 0

Therefore,  θ = π

So, In polar form

– 3 = r cos θ + ir sin θ

= 3 cos π + i3 sin π

= 3(cos π + i sin π)

$$\textbf{7.\space}\sqrt{\textbf{3}}\textbf{+i.}\\\textbf{Sol.}\space\text{Let} z=\sqrt{3}+i\\\text{So,\space r cos}\space\theta=\sqrt{3}\space\text{and r sin q = 1}$$

On squaring and adding both, we get

$$\text{r}^{2}\text{cos}^{2}\theta\space+\space r^{2}\text{sin}^{2}\theta=\sqrt{3 ^{2}+1^{2}}$$

⇒ r2 (cos2 θ + sin2 θ) = 3 + 1

(∵ cos2 θ + sin2 θ = 1)

⇒ r2 = 4  (∵ r > 0)

⇒ r = 2

$$\text{So,}\space\text{2 cos}\space\theta=\sqrt{3}\space\text{and 2 sin}\space\theta = 1\\\Rarr\space\text{cos}\space\theta=\frac{\sqrt{3}}{2}\text{and sin}\space\theta =\frac{1}{2}\\\text{Therefore,}\space\theta=\frac{\pi}{6}\space\text{(As q lies in I quadrant)}$$

Hence, in polar form

$$\sqrt{3}+i=\text{r cos}\space\theta+i\space\text{sin}\space\theta\\=\text{2 cos}\bigg(\frac{\pi}{6}\bigg)+2i\space\text{sin}\bigg(\frac{\pi}{6}\bigg)\\=2\bigg[\text{cos}\bigg(\frac{\pi}{6}\bigg)+i\space \text{sin}\bigg(\frac{\pi}{6}\bigg)\bigg]$$

8. i.

Sol. Let, z = i = 0 + i

So, r cos  θ = 0 and r sin  θ = 1

On squaring and adding both, we get

r2 cos2  θ + r2 sin2  θ = (0)2 + (1)2

⇒ r2 (cos2  θ + sin2  θ) = 1

(∵ cos2  θ + sin2  θ = 1)

⇒ r2 = 1

$$\Rarr\space r=\sqrt{1}=1$$

(∵ r > 0)

So, 1 cos θ = 0 and 1 sin θ = 1

⇒ cos θ = 0 and sin θ = 1

$$\text{Therefore,}\space\theta=\frac{\pi}{2}$$

Hence, in ploar form

i = r cos θ + ri sin θ

$$=\text{cos}\frac{\pi}{2}+i\space\text{sin}\frac{\pi}{2}.$$

Exercise 5.3

Solve each of the following equations:

1. x2 + 3 = 0.

Sol. x2 + 3 = 0
On comparing it with ax2 + bx + c = 0, we get

a = 1, b = 0 and c = 3

Therefore, the discriminant of the given equation is

D = b2 – 4ac

⇒ D = 02 – 4 × 1 × 3 = – 12

Hence, the required solution are:

$$x=\frac{-b\pm\sqrt{D}}{2a}\\=\pm\frac{\sqrt{-12}}{2×1}\\=\pm\frac{\sqrt{12}}{2}i\space[\because\space\sqrt{-1}=i]\\\therefore\space x=\pm\frac{2\sqrt{3}}{2}i\\\Rarr\space x=\pm\sqrt{3i}$$

2. 2x2 + x + 1 = 0.

Sol. 2x2 + x + 1 = 0

On comparing it with ax2 + bx + c = 0, we get

a = 2, b = 1 and c = 1

Therefore, the discriminant of the given equation is

D = b2 – 4ac

⇒ D = 12 – 4 × 2 × 1 = 1 – 8 = – 7

Hence, the required solutions are:

$$x=\frac{-b\pm\sqrt{D}}{2a}\\=\frac{-1\pm\sqrt{-7}}{2×2}=\frac{-1+\sqrt{7}i}{4}\\(\because\sqrt{-1}=i)\\x=\frac{\normalsize-1}{4}\pm\frac{\sqrt{7}}{4}i.$$

3. x2 + 3x + 9 = 0.

Sol. x2 + 3x + 9 = 0

On comparing it with ax2 + bx + c = 0, we get

a = 1, b = 3 and c = 9

Therefore, the discriminant of the given equation is

D = b2 – 4ac

D = 32 – 4 × 1 × 9

= 9 – 36 = – 27

Hence, the required solutions are:

$$x=\frac{-b\pm\sqrt{D}}{2a}\\=\frac{-3\pm\sqrt{-27}}{2×1}\\=\frac{-3\pm\sqrt{-3×3×3}}{2×1}\\=\frac{-3\pm3\sqrt[]{-3}}{2}\\=\frac{-3\pm3\sqrt{3}i}{2}\\\lbrack\because\space\sqrt{-1}=i\rbrack\\x=\frac{-3}{2}\pm\frac{3\sqrt{3i}}{2}.$$

4. – x2 + x – 2 = 0.

Sol. – x2 + x – 2 = 0

On comparing it with ax2 + bx + c = 0, we get

a = – 1, b = 1 and c = – 2

Therefore, the discriminant of the given equation is

D = b2 – 4ac

D = (1)2 – 4 × (– 1) × (– 2)

= 1 – 8 = – 7

Hence, the required solutions are:

$$x=\frac{-b\pm\sqrt{D}}{2a}=\frac{-1\pm\sqrt{-7}}{2×(\normalsize-1)}\\=\frac{-1\pm\sqrt{7}i}{-2}\bigg[\because\space\sqrt{-1}=i\bigg]\\\Rarr\space x=\frac{1}{2}\pm\frac{\sqrt{7}}{-2}i\\\Rarr\space x=\frac{1}{2}\pm\frac{\sqrt{7}}{-2}i.$$

5. x2 + 3x + 5 = 0.

Sol.  x2 + 3x + 5 = 0

On comparing it with ax2 + bx + c = 0, we get

a = 1, b = 3 and c = 5

Therefore, the discriminant of the given equation is

D = b2 – 4ac

⇒ D = 32 – 4 × 1 × 5

= 9 – 20 = – 11

Hence, the required solutions are

$$x=\frac{-b\pm\sqrt{D}}{2a}=\frac{-3\pm\sqrt{-11}}{2×1}\\=\frac{-3\pm\sqrt{-11i}}{2}[\because\space\sqrt{-1}=i]\\x=\frac{-3}{2}\pm\frac{\sqrt{11}}{2}i.$$

6. x2 – x + 2 = 0.

Sol. x2 – x + 2 = 0

On comparing it with ax2 + bx + c = 0, we have

a = 1, b = – 1 and c = 2

Therefore, the discriminant of the given equation is

D = b2 – 4ac

⇒ D = (– 1)2 – 4 × 1 × 2 = 1 – 8 = – 7

Hence, the required solutions are

$$x=\frac{-b\pm\sqrt{D}}{2a}=\frac{-(-1)\pm\sqrt{-7}}{2}\\=\frac{1\pm\sqrt{7i}}{2}[\because\space\sqrt{-1}=i]\\x=\frac{1}{2}\pm\frac{\sqrt{7}}{2}i.$$

$$\textbf{7.}\sqrt{\textbf{2x} ^{2}}+\textbf{x}+\sqrt{\textbf{2}}\textbf{= 0.}\\\textbf{Sol.}\space\sqrt{2}x^{2}+x+\sqrt{2}=0$$

On comparing it with ax2 + bx + c = 0, we get

$$a=\sqrt{2}, b=1\text{and c}=\sqrt{2}$$

Therefore, the discriminant of the given equation is

D = b2 – 4ac

$$\Rarr\space\text{D = (1)}^{2}-4×\sqrt{2}×\sqrt{2}\\=1-8=-7$$

Hence, the required solutions are

$$x=\frac{-b\pm\sqrt{D}}{2a}=\frac{-1\pm\sqrt{-7}}{2\sqrt{2}}\\=\frac{-1\pm\sqrt{7}i}{2\sqrt{2}}\space[\because\sqrt{-1}=i]\\\Rarr\space x=\frac{-1}{2\sqrt{2}}\pm\frac{\sqrt{7}}{2\sqrt{2}}i.$$

$$\textbf{8.\space}\sqrt{\textbf{3}}\textbf{x}^{\textbf{2}}-\sqrt{\textbf{2}}\textbf{x+3}\sqrt{\textbf{3}}\textbf{=0.}\\\textbf{Sol.}\space\sqrt{3}x^{2}-\sqrt{2x}+3\sqrt{3}=0$$

On comparing it with ax2 + bx + c = 0,

$$\text{we get}\space a=\sqrt{3}, b=\sqrt{2},\text{and c}=3\sqrt{3}$$

Therefore, the discriminant of the given equation is

D = b2 – 4ac

$$\Rarr\space\text{D}=(-\sqrt{2})^{2}-4×\sqrt{3}×3\sqrt{3}$$

= 2 – 36 = – 34

Hence, the required solutions are

$$x=\frac{-b\pm\sqrt{D}}{2a}=\frac{-(-2)\pm\sqrt{-34}}{2×\sqrt{3}}\\=\frac{\sqrt{2}\pm\sqrt{34}i}{2\sqrt{3}}\space[\because\space\sqrt{-1}=i]\\\Rarr\space x=\frac{\sqrt{2}}{2\sqrt{3}}\pm\frac{\sqrt{34}}{2\sqrt{3}}i.$$

$$\textbf{9. x}^{\textbf{2}}\textbf{+x+}\frac{\textbf{1}}{\sqrt{\textbf{2}}}\textbf{=0.}\\\textbf{Sol.}\space x^{2}+x+\frac{1}{\sqrt{2}}=0\\\Rarr\space\sqrt{2}x^{2}+\sqrt{2}x+1=0$$

On comparing it with ax2 + bx + c, we have

$$a=\sqrt{2},b=\sqrt{2}\text{and c = 1}$$

Therefore, the discriminant of the given equation is

D = b2 – 4ac

$$\Rarr\space D=(\sqrt{2})^{2}-4×\sqrt{2}×1\\=2-4\sqrt{2}=2(1-2\sqrt{2})$$

Hence, the required solutions are

$$x=\frac{-b\pm\sqrt{D}}{2a}=\frac{\sqrt{2}\pm\sqrt{2(1-2\sqrt{2})}}{2\sqrt{2}}\\=\frac{-\sqrt{2}\pm\sqrt{2}(\sqrt{(2\sqrt{2}-1)})i}{2\sqrt{2}}\\=x=\frac{-1\pm(\sqrt{2\sqrt{2}-1})i}{2}.$$

$$\textbf{10. x}^\textbf{2}+\frac{\textbf{x}}{\sqrt{\textbf{2}}}\textbf{+1=0}\\\textbf{Sol.}\space x^{2}+\frac{x}{\sqrt{2}}+1=0\\\Rarr\space\sqrt{2}x^{2}+x+\sqrt{2}=0$$

On comparing it with ax2 + bx + c = 0, we get,

$$a=\sqrt{2},b=1\text{and c}=\sqrt{2}$$

Therefore, the discriminant of the given equation is

D = b2 – 4ac

$$\Rarr\space\text{D}=(1)^{2}-4×\sqrt{2}×\sqrt{2}$$

= 1 – 8 = – 7

Hence, the required solutions are

$$x=\frac{-b\pm\sqrt{D}}{2a}=\frac{-1\pm\sqrt{-7}}{2\sqrt{2}}\\\Rarr\space x=\frac{-1}{2\sqrt{2}}\pm\frac{\sqrt{7}}{2\sqrt{2}}i\space[\because\space\sqrt{-1}=i]$$

Miscellaneous Exercise

$$\textbf{1. Evaluate}\bigg[\textbf{i}^{\textbf{18}}+\bigg(\frac{\textbf{1}}{\textbf{i}}\bigg)^{\textbf{25}}\bigg]^{\textbf{3}}.$$

$$\textbf{Sol.}\bigg[i^{18}+\bigg(\frac{1}{i}\bigg)^{25}\bigg]^{3}\\=\bigg[i^{4×4+2}+\frac{1}{i^{4×6+1}}\bigg]^{3}\\=\bigg[(i^{4})^{4}×i^{2}+\frac{1}{(i^{4})^{6}×i}\bigg]^{3}\\=\bigg[i^{2}+\frac{1}{i}\bigg]^{3}\space\begin{bmatrix}i^{2}=-1\\i^{4}=1\end{bmatrix}\\=\bigg[-1+\frac{1}{i}\bigg]^{3}\\=\bigg[-1+\frac{i}{i×i}\bigg]^{3}$$

[Multiply numerator and denominator by i]

$$=\bigg[-1+\frac{i}{i^{2}}\bigg]^{3}\\=\bigg[-1+\frac{i}{-1}\bigg]^{3}\space\lbrace\because\space i^{2}=-1\rbrace$$

= [– 1 – i]3

= (– 1)3 [1 + i]3

= – [13 + i3 + 3 × 1 × i(1 + i)]

[(a + b)3 = a3 + b3 + 3ab(a + b)]

= – [1 – i + 3i(1 + i)] [i3 = – i]

= – [1 – i + 3i + 3i2] [i2 = – 1]

= – [1 – i + 3i – 3]

= – [– 2 + 2i]

= 2 – 2i.

2. For any two complex numbers z1 and z2, prove that Re (z1z2) = Re z1 Re z2 – Imz1 Imz2.

Sol. Let 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) = Rez1 Rez2 – Imz1 Imz2.

$$\textbf{3. Reduce}\bigg(\frac{\textbf{1}}{\textbf{1-4i}}-\frac{\textbf{2}}{\textbf{1+i}}\bigg)\bigg(\frac{\textbf{3-4i}}{\textbf{5+i}}\bigg)\\\textbf{to the standard form.}$$

$$\textbf{Sol.}\space\bigg(\frac{1}{1-4i}-\frac{2}{1+i}\bigg)\bigg(\frac{3-4i}{5+i}\bigg)$$

Taking L.C.M.

$$=\bigg[\frac{(1+i)-2(1-4i)}{(1-4i)(1+i)}\bigg]\bigg[\frac{3-4i}{5+i}\bigg]\\=\bigg[\frac{1+i-2+8i}{1+i-4i-4i^{2}}\bigg]\bigg[\frac{3-4i}{5+i}\bigg]\\=\bigg[\frac{-1+9i}{5-3i}\bigg]\bigg[\frac{3-4i}{5+i}\bigg]\\=\bigg[\frac{-3+4i+27i-36i^{2}}{25+5i-15i-3i^{2}}\bigg](t^{2}=-1)\\=\frac{33+31i}{28-10i}=\frac{33+31}{2(14-5i)}\\=\frac{(33+31i)}{2(14-5i)}×\frac{(14+5i)}{(14+5i)}$$

[on multiplying numerator and denominator by (14 + 5i)]

$$=\frac{462+165i+434i+155i}{2[(14)^{2}-(5i)^{2}]}\\=\frac{307+599i}{2(196-25i^{2})}\space\lbrace\because\space i^{2}=-1\rbrace\\=\frac{307+599i}{2(221)}=\frac{307+599i}{442}\\=\frac{307}{442}+\frac{599 i }{442}$$

Hence, this is the required standard form.

$$\textbf{4. If x-iy}=\sqrt{\frac{\textbf{a-ib}}{\textbf{c-id}}},\\\textbf{prove that}(\textbf{x}^{\textbf{2}}+\textbf{y}^{\textbf{2}})^{\textbf{2}}=\frac{\textbf{a}^{\textbf{2}}\textbf{+b}^{\textbf{2}}}{\textbf{c}^{\textbf{2}}\textbf{+d}^{\textbf{2}}}.$$

$$\textbf{Sol.}\space x-iy=-\sqrt{\frac{\text{a-ib}}{\text{c-id}}}\\=\sqrt{\frac{(a-ib)}{(c-id)}×\frac{(c+id)}{(c+id)}}$$

[on multiplying numerator and denominator by (c + id)]

$$=\sqrt{\frac{a(c+id)-ib(c+id)}{c^{2}+(id)^{2}}}$$

[a2 – b2 = (a + b) (a – b)]

$$=\sqrt{\frac{(ac+bd)+i(ad+bc)}{c^{2}+d^{2}}}\space\lbrace\because i^{2}=-1\rbrace$$

So, on squaring both sides we get

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

On comparing real and imaginary parts, we get

$$x^{2}-y^{2}=\frac{ac+bd}{c^{2}+d^{2}}\\\text{and}-2xy=\frac{ad-ac}{c^{2}+d^{2}}\space\text{...(i)}$$

L.H.S. = (x2 + y2)2 = (x2 – y2)2 + 4x2y2

$$=\bigg(\frac{ac+bd}{c^{2}+d^{2}}\bigg)^{2}+\bigg(\frac{ad-bc}{c^{2}+d^{2}}\bigg)^{2}\\\lbrack\text{from equation }(1)\rbrack\\=\frac{a^{2}c^{2}+b^{2}d^{2}+2acbd+a^{2}d^{2}+b^{2}c^{2}-2adbc}{(c^{2}+d^{2})^{2}}\\=\frac{a^{2}c^{2}+b^{2}d^{2}+a^{2}d^{2}+b^{2}c^{2}}{(c^{2}+d^{2})^{2}}\\=\frac{a^{2}(c^{2}+d^{2})+b^{2}(c^{2}+d^{2})}{(c^{2}+d^{2})^{2}}\\=\frac{(c^{2}+d^{2})(a^{2}+b^{2})}{(c^{2}+d^{2})^{2}}\\=\frac{a^{2}+b^{2}}{c^{2}+d^{2}}$$

= R.H.S.

5. Convert the following in the polar form:

$$\textbf{(i)}\space\frac{\textbf{1+7i}}{\textbf{(2-i)}^{\textbf{2}}}\\\textbf{(ii)\space}\frac{\textbf{1+3i}}{\textbf{1-2i}}$$

$$\textbf{Sol.}\space(i)\space z=\frac{1+7i}{(2-i)^{2}}\\=\frac{1+7i}{4+i^{2}-4i}=\frac{1+7i}{4-1-4i}\\\lbrack i^{2}=1\rbrack\\=\frac{1+7i}{3-4i}\\=\frac{(1+7i)}{(3-4i)}×\frac{(3+4i)}{(3+4i)}$$

[on multiplying numerator and donominator by 3 + 4i]

$$=\frac{1(3+4i)+7i(3+4i)}{(3)^{2}-(4i)^{2}}\\\lbrack(a+b)(a-b)=a^{2}-b^{2}\rbrack\\=\frac{3+4i+21i-28i^{2}}{3^{2}-4^{2}(\normalsize-1)}\space[i^{2}=-1]\\=\frac{3+4i+21i-28}{3^{2}+4^{2}}\space\lbrace\because\space i^{2}=-1\rbrace\\=\frac{-25+25i}{25}$$

= – 1 + i

Let r cos θ = – 1 and r sin θ = 1

On squaring and adding both, we get

r2 cos2 θ + r2 sin2 θ = (– 1)2 + 12

r2 (cos2 θ + sin2 θ) = 1 + 1

(cos2 θ + sin2 θ = 1)

r2 = 2

$$r=\sqrt{2}\space(r>0)$$

(conventionally, r > 0)

$$\text{So,}\space\sqrt{2}\space\text{cos}\space\theta=-1\space\text{and}\\\sqrt{2}\text{sin}\space\theta=1\\\Rarr\space\text{cos}\space\theta=\frac{\normalsize-1}{\sqrt{2}}\text{and}\space\text{sin}\space\theta=\frac{1}{\sqrt{2}}\\\theta=\pi-\frac{\pi}{4}=\frac{3\pi}{4}$$

[As θ lies in II quadrant]

Expressing as,

z = r cos θ + ir sin θ

$$z=\sqrt{2}\text{cos}\frac{3\pi}{4}+\sqrt{2i}\space\text{sin}\frac{3\pi}{4}\\z=\sqrt{2}\bigg(\text{cos}\frac{3\pi}{4}+i\space\text{sin}\frac{3\pi}{4}\bigg)$$

Therefore, this is the required polar form.

$$\text{(ii)}\space z=\frac{1+3i}{1-2i}\\z=\frac{(1+3i)}{(1-2i)}×\frac{(1+2i)}{(1+2i)}$$

[on multiplying numerator and denominator by 1 + 2i]

$$z=\frac{1+2i+3i-6}{(1)^{2}-(2i)^{2}}\\\lbrack(a+b)(a-b)=a^{2}-b^{2}\rbrack\\z=\frac{-5+5i}{1-4i}\space[i^{2}=-1]\\z=\frac{-5+5i}{1-4(-1)}=\frac{-5+5i}{1+4}\\z=\frac{-5+5i}{5}=-1+i$$

Now, let r cos θ = – 1 and r sin θ = – 1

On squaring and adding both we get

r2 cos2 θ + r2 sin2 θ = (– 1)2 + (1)2

r2 (cos2 θ + sin2 θ) = 1 + 1

r2 = 2 [cos2 θ + sin2 θ = 1]

$$r=\sqrt{2}\space\text{(conventionally r > 0)}$$

$$\text{So,\space}\sqrt{2}\space\text{cos}\space\theta=-1\space\\\text{and}\space\sqrt{2}\space\text{sin}\space\theta=1\\\text{cos}\space\theta=\frac{-1}{\sqrt{2}}\text{and sin}\space\theta=\frac{1}{\sqrt{2}}\\\text{Therefore,}\space\theta=\pi-\frac{\pi}{4}=\frac{3\pi}{4}$$

[As  θ lies in II quadrant]

Therefore, expressing as,

z = r cos  θ + ir sin  θ

$$z=\sqrt{2}\space\text{cos}\frac{3\pi}{4}+\sqrt{2}\space\text{sin}\bigg(\frac{3\pi}{4}\bigg)\\=z=\sqrt{2}\bigg(\text{cos}\frac{3\pi}{4}+i\space\text{sin}\space\frac{3\pi}{4}\bigg)$$

Therefore, this is the required polar form:

Solve each of the equation in Exercises 6 to 9:

$$\textbf{6. 3x}^{\textbf{2}}-\textbf{4x}+\frac{\textbf{20}}{\textbf{3}}\space\textbf{= 0.}\\\textbf{Sol.}\space\text{Let}\space z=3x^{2}-4x+\frac{20}{3}=0$$

Multiply by 3 on both side

9x2 – 12x + 20 = 0

On comparing it with ax2 + bx + c = 0, we get

a = 9, b = – 12 and c = 20

So, the discriminant of the given equation is

D = b2 – 4ac

D = (– 12)2 – 4 × 9 × 20

= 144 – 720

= – 576

Hence, the required solutions are

$$x=\frac{-b\pm\sqrt{D}}{2a}\\x=\frac{-(-12)\pm\sqrt{-576}}{2×9}\\x=\frac{12\pm\sqrt{576}i}{18}\space[\sqrt{-1}=i]\\x=\frac{12\pm24i}{18}=\frac{6(2\pm4i)}{18}\\x=\frac{2\pm4i}{3}=\frac{2}{3}\pm\frac{4i}{3}$$

Hence, this is the required answer.

$$\textbf{7.}\space\textbf{x}^{\textbf{2}}-\textbf{2x}+\frac{\textbf{3}}{\textbf{2}}\textbf{= 0.}\\\textbf{Sol.}\space\text{Let}\space z=x^{2}-2x+\frac{3}{2}=0$$

Multiply by 2 on both side

2x2 – 4x + 3 = 0

On comparing it with ax2 + bx + c = 0, we get

a = 2, b = – 4 and c = 3

So, the discriminant of the given equation is

D = b2 – 4ac

D = (– 4)2 – 4 × 2 × 3

= 16 – 24

= – 8

Hence, the required solutions are

$$x=\frac{-b\pm\sqrt{D}}{2a}\\x=\frac{-(-4)\pm\sqrt{-8}}{2×2}\\x=\frac{4\pm2\sqrt{2}i}{4}\space[\sqrt{-1}=i]\\x=1\pm\frac{\sqrt{2}}{2}i$$

Hence, this is the required answer.

8. 27x2 – 10x + 1 = 0.

Sol. Let z = 27x2 – 10x + 1

On comparing it with ax2 + bx + c = 0, we get

a = 27, b = – 10 and c = 1

So, the discriminant of the given equation is

D = b2 – 4ac

D = (– 10)2 – 4 × 27 × 1

= 100 – 108 = – 8

Hence, the required solutions are

$$x=\frac{-b\pm\sqrt{D}}{2a}\\ x=\frac{-(-10)\pm\sqrt{-8}}{2×27}\\=\frac{10\pm2\sqrt{2i}}{54}\space[\because\sqrt{\normalsize-1}=i]\\x=\frac{5\pm \sqrt{2i}}{27}\\x=\frac{5}{27}\pm\frac{\sqrt{2}}{\sqrt{27}}i$$

Hence, this is the required answer.

9. 21x2 – 28x + 10 = 0.

Sol. 21x2 – 28x + 10 = 0

On comparing it with ax2 + bx + c = 0, we have

a = 21, b = – 28 and c = 10

So, the discriminant of the given equation is

D = b2 – 4ac

D = (– 28)2 – 4 × 21 × 10

D = 784 – 840

D = – 56

Hence, the required solutions are

$$x=\frac{-b\pm\sqrt{D}}{2a}\\x=\frac{-(-28)\pm\sqrt{-56}}{2×21}\\x=\frac{28\pm\sqrt{56}i}{42}\space[\sqrt{-1}=i]\\x=\frac{28\pm2\sqrt{14}i}{42}\\x=\frac{28}{42}\pm\frac{2\sqrt{14}}{42}i\\x=\frac{2}{3}\pm\frac{\sqrt{14}}{21}i$$

Hence, this is the required answer.

$$\textbf{10.}\space \textbf{If z}_\textbf{1} \textbf{= 2 – i}, \textbf{z}_\textbf{2} \textbf{= 1 + i, find}\\\begin{vmatrix}\frac{\textbf{z}_\textbf{1}\textbf{+}\textbf{z}_\textbf{2}\textbf{+1}}{\textbf{z}_\textbf{1}\textbf{-}\textbf{z}_\textbf{2}\textbf{+1}}\end{vmatrix}.$$

Sol. Given, z1 = 2 – i, z2 = 1 + i

$$\text{So,}\space \begin{vmatrix}\frac{z_1+z_2+1}{z_1-z_2+1}\end{vmatrix}=\begin{vmatrix}\frac{(2-i)+(1+i)+1}{(2-i)-(1+i)+1}\end{vmatrix}\\=\begin{vmatrix}\frac{4}{2-2i}\end{vmatrix}=\begin{vmatrix}\frac{4}{2(1-i)}\end{vmatrix}\\=\begin{vmatrix}\frac{2}{1-i}\end{vmatrix}\\=\begin{vmatrix}\frac{2}{(1-i)}×\frac{(1+i)}{(1+i)}\end{vmatrix}$$

[on multiplying numerator and denominator by (1 + i)]

$$\begin{vmatrix}\frac{2(1+i)}{(1)^{2}+(i)^{2}}\end{vmatrix}\\\lbrack(a+b)(a-b)=a^{2}-b^{2}\rbrack\\=\begin{vmatrix}\frac{2(1+i)}{1-(-1)}\end{vmatrix}\space[i^{2}=-1]\\=\begin{vmatrix}\frac{2(1+i)}{1+1}\end{vmatrix}\\=\begin{vmatrix}\frac{2(1+i)}{2}\end{vmatrix}\\=|1+i|=\sqrt{1^{2}+1^{2}}\\=\sqrt{2}$$

Hence, this is the required answer.

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

$$\textbf{Sol.}\space\text{Given,}\space\text{a+ib}=\frac{(x+i)^{2}}{2x^{2}+1}\\=\frac{x^{2}+i^{2}+2xi}{2x^{2}+1}\\\lbrace\because\space (a+b)^{2}=a^{2}+b^{2}=2ab\rbrace\\=\frac{x^{2}-1+2xi}{2x^{2}+1}\\=\frac{x^{2}-1}{2x^{2}+1}+i\bigg(\frac{2x}{2x^{2}+1}\bigg)$$

Comparing the real and imaginary parts, we have

$$a=\frac{x^{2}-1}{2x^{2}+1}\space\text{and}\\b=\frac{2x}{2x^{2}+1}$$

Therefore,

$$\text{L.H.S = a}^{2}+b^{2}=\\\bigg(\frac{x^{2}-1}{2x^{2}+1}\bigg)^{2}+\bigg(\frac{2x}{2x^{2}+1}\bigg)^{2}\\=\frac{x^{4}+1-2x^{2}+4x^{2}}{(2x^{2}+1)^{2}}\\=\frac{x^{4}+1+2x^{2}}{(2x^{2}+1)^{2}}=\frac{(x^{2}+1)^{2}}{(2x^{2}+1)^{2}}$$

= R.H.S.      Hence Proved.

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

$$\textbf{(i)\space Re}\bigg(\frac{\textbf{z}_\textbf{1}\textbf{z}_\textbf{2}}{\textbf{z}_\textbf{1}}\bigg),\textbf{(ii) Im}\bigg(\frac{\textbf{1}}{\textbf{z}_\textbf{1}\bar{\textbf{z}}_\textbf{1}}\bigg).$$

Sol. (i) Given, z1 = 2 – i, z2 = – 2 + i

z1z2 = (2 – i) (– 2 + i)

= – 4 + 2i + 2i – i2

= – 4 + 4i – (– 1) [i2 = – 1]

= – 4 + 4i + 1

= – 3 + 4i

$$\bar{z}_1=2+i\\\therefore\space\frac{z_1z_2}{\bar z_1}=\frac{-3+4i}{2+i}$$

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

$$\frac{z_1z_2}{\bar{z}_1}=\frac{(-3+4i)(2-i)}{(2+i)(2-i)}\\=\frac{-3(2-i)+4i(2-i)}{2^{2}-(i^{2})}\\\lbrack(a+b)(a-b)=a^{2}-b^{2}\rbrack\\=\frac{-6+3i+8i-4i^{2}}{4-(-1)}\space\lbrace i^{2}=-1\rbrace\\=\frac{-6+11i-4(-1)}{4+1}\space[i^{2}=-1]\\=\frac{-6+11i-4(-1)}{4+1}=\frac{-2+11i}{5}.\\=\frac{-2}{5}+\frac{11}{5}i$$

Comparing the real parts, we have

$$\text{Re}\bigg(\frac{z_1z_2}{\bar{z_1}}\bigg)=\frac{-2}{5}$$

Hence, this is the required answer.

(ii) Given, z1 = 2 – i, z2 = – 2 + i

$$\bar{z}_1=2+i$$

$$\frac{1}{z_1\bar z_1}=\frac{1}{(2-i)(2+i)}=\frac{1}{2^{2}-(-i^{2})}$$

[(a + b) (a – b) = a2 – b2]

$$=\frac{1}{4-(-1)}\space\lbrace i^{2}=1\rbrace\\=\frac{1}{4+1}\\=\frac{1}{5}\\\text{On comparing the imaginary part,}\\\text{Im}\bigg(\frac{1}{z_1 \bar z_1}\bigg)=0$$

Hence, this is the required answer.

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

$$\textbf{Sol.}\space\text{Let}\space z=\frac{1+2i}{1-3i},\text{then}\\z=\frac{(1+2i)(1+3i)}{(1-3i)(1+3i)},\text{then}\\z=\frac{1(1+3i)+2i(1+3i)}{(1)^{2}-(3i)^{2}}$$

[(a + b) (a – b) = a2 – b2]

$$z=\frac{1+3i+2i+6i^{2}}{1-(9)(i^{2})}\\z=\frac{1+5i+6(-1)}{1-9(-1)}\space[i^{2}=-1]\\=\frac{1+5i-6}{1+9}\\=\frac{-5+5i}{10}\\=\frac{-5}{10}+\frac{5}{10}i\\=\frac{-1}{2}+\frac{1}{2}i$$

Let z = r cos θ + ir sin θ

$$\text{So, r\space cos}\space\theta=\frac{\normalsize-1}{2}\\\text{and r sin}\theta=\frac{1}{2}$$

On squaring and adding both, we get

$$r^{2}\text{cos}^{2}\theta + r^{2}\space sin^{2}\theta\\=\bigg(\frac{\normalsize-1}{2}\bigg)^{2}+\bigg(\frac{1}{2}\bigg)^{2}\\\text{r}^{2}(\text{cos}^{2}\theta + \text{sin}^{2}\theta)=\\\frac{1}{4}+\frac{1}{4}$$

(cos2 θ + sin2 θ = 1)

$$r^{2}=\frac{2}{4}\\r^{2}=\frac{1}{2}\space\lbrack\text{conventionally} >0 \rbrack\\\text{r}=\frac{1}{\sqrt{2}}\\\text{Now,}\space\frac{1}{\sqrt{2}}\text{cos}\space\theta=\frac{-1}{2}\space\text{and}\\\frac{1}{\sqrt{2}}\space\text{sin}\space\theta=\frac{1}{2}\\\text{cos}\space\theta=\frac{-1}{\sqrt{2}}\space\text{and sin}\space\theta=\frac{1}{\sqrt{2}}\\\theta=\pi-\frac{\pi}{4}=\frac{3\pi}{4}$$

[As q lies in the II quadrant]

Hence, this is the required answer.

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

Sol. Given that

z = (x – iy) (3 + 5i)

= x(3 + 5i) – iy(3 + 5i)

= 3x + 5xi – 3yi – 5yi2

= 3x + 5xi – 3yi – 5y(– 1) (i2 = – 1)

= 3x + 5xi – 3yi + 5y

= (3x + 5y) + i(5x –3y)

$$\bar{z}=(3x-5y)-i(5x-3y)$$

Also given,

$$\bar{z}= -6-24i$$

And (3x + 5y) – i(5x – 3y) = – 6 – 24i

On equating real and imaginary parts,

3x + 5y = – 6 …(i)

5x – 3y = 24 …(ii)

Performing equation (i) × 3 + equation (ii) × 5, we get

(9x + 15y) + (25x – 15y) = – 18 + 120

34x = 102

$$x=\frac{102}{34}$$

x = 3

Putting the values 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.

$$\textbf{15. Find the modulus of}\space\frac{\textbf{1+i}}{\textbf{1-i}}-\frac{\textbf{1-i}}{\textbf{1+i}}\textbf{.}\\\textbf{Sol.}\space\frac{1+i}{1-i}-\frac{1-i}{1+i}\\=\frac{(1+i)^{2}-(1-i)^{2}}{(1-i)(1+i)}\\=\frac{1+i^{2}+2i-1-i^{2}+2i}{(1)^{2}-(i)^{2}}$$

[(a + b) (a – b) = a2 – b2]

$$=\frac{4i}{1-(-1)}\space(i^{2}=-1)\\=\frac{4i}{1+1}=\frac{4i}{2}=2i\\\begin{vmatrix}\frac{1+i}{1-i}-\frac{1-i}{1+i}\end{vmatrix}=|2i|=\sqrt{2^{2}}$$

= 2

Hence, the modulus is 2.

16. If (x + iy)3 = u + iv, then show that

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

Sol. Given (x + iy)3 = u + iv

x3 + (iy)3 + 3·x·iy(x + iy) = u + iv

[(a + b)3 = a3 + b3 + 3ab(a + b)]

x3 + iy3 + 3x2yi + 3xy2i2 = u + iv

x3 – iy3 + 3x2yi – 3xy2 = u + iv [i3 = – i]

(x3 – 3xy2) + i(3x2y – y3) = u + iv

On comparing real and imaginary part, we get

u = x3 – 3xy2, v = 3x2y – y3

$$\text{L.H.S}=\frac{u}{x}+\frac{v}{y}=\\\frac{x^{3}-3xy^{2}}{x}+\frac{3x^{2}y-y^{3}}{y}\\=\frac{x(x^{2}-3y^{2})}{x}+\frac{y(3x^{2}-y^{2})}{y}$$

= x2 – 3y2+ 3x2 – y2

= 4x2 – 4y2

= 4(x2 – y2)

=  R.H.S.        Hence Proved.

17. If α and β are different complex numbers with

$$|\beta|=1,\textbf{then find}\begin{vmatrix}\frac{\beta-\alpha}{1-\bar\alpha\beta}\end{vmatrix}.$$

Sol. Let a = a + ib

and b = x + iy

Given, | β | = 1

$$\text{So,}\space\sqrt{x^{2}+y^{2}}=1\\x^{2}+y^{2}\space\text{...(i)}\\\begin{vmatrix}\frac{\beta-\alpha}{1-\bar\alpha\beta}\end{vmatrix}=\begin{vmatrix}\frac{(x+iy)-(a+ib)}{1-(a-ib)(x+iy)}\end{vmatrix}\\=\begin{vmatrix}\frac{(x-a)+i(y-b)}{1-(ax+aiy-ibx+by)}\end{vmatrix}\\=\begin{vmatrix}\frac{(x-a)+i(y-b)}{(1-ax-by)+i(bx-ay)}\end{vmatrix}\\\begin{bmatrix}\frac{|Z_1|}{|Z_2|}=\frac{|z_1|}{|z_2|}\end{bmatrix}\\=\frac{|(x-a)+i(y-b)|}{|(1-ax-by)+i(bx-ay)|}\\=\frac{\sqrt{(x-a)^{2}+(y-b)^{2}}}{\sqrt{(1-ax-by)^{2}+(bx-ay)^{2}}}$$

$$=\space\frac{\sqrt{(x-a)^{2}+(y-b)^{2}}}{\sqrt{1+a^{2}x^{2}+b^{2}y^{2}-2ax+2abxy-2by+b^{2}x^{2}+a^{2}y-2abxy}}$$

$$=\space\frac{\sqrt{(x^{2}+y^{2})+a^{2}+b^{2}-2ax-2by}}{\sqrt{1+a^{2}(x^{2}+y^{2})+b^{2}(y^{2}+x^{2})-2ax-2by}}$$

$$=\frac{\sqrt{(x^{2}+y^{2})+a^{2}+b^{2}-2ax-2by}}{\sqrt{1+a^{2}(x^{2}+y^{2})+b^{2}(y^{2}+x^{2})-2ax-2by}}\\=\frac{\sqrt{1+a^{2}+b^{2}-2ax-2by}}{\sqrt{1+a^{2}(x^{2}+y^{2})+b^{2}(y^{2}+x^{2})-2ax-2by}}\\=\frac{\sqrt{1+a^{2}+b^{2}-2ax-2by}}{\sqrt{1+a^{2}+b^{2}-2ax-2by}}\\\lbrack\text{using equation (i)}\rbrack\\=1\\\text{Therefore,}\begin{vmatrix}\frac{\beta-\alpha}{1-\bar\alpha\beta}\end{vmatrix}=1.$$

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

Sol. | 1 – i |x = 2x

$$(\sqrt{1^{2}+(-1)^{2}})^{x}=2^{x}\\(\sqrt{2})^{x}=2^{x}\\2^{x/2}=2^{X}\\\lbrack\because\space (\sqrt{a})^{x}=(a^{1/2})^{x}=a^{x/2}\rbrack\\\frac{x}{2}=x$$

x = 2x

2x – x = 0

x = 0

Therefore, 0 is the only integral solution of the given equation.

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

19. If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that (a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2+ B2.

Sol. Given, (a + ib) (c + id) (e + if) (g + ih) = A + iB

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

⇒ | a + ib | × | c + id | × | e + if | × | g + ih |

= | A + iB |

[| z1z2 | = | z1 | | z2 |]

$$\Rarr\space\sqrt{a^{2}+b^{2}}×\sqrt{c^{2}+d^{2}}×\sqrt{e^{2}+f^{2}}×\sqrt{g^{2}+h^{2}}\\=\sqrt{A^{2}+B^{2}}$$

On squaring both sides, we get

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

Hence Proved.

$$\textbf{20.\space If}\space\bigg(\frac{\textbf{1+i}}{\textbf{1-i}}\bigg)^{\textbf{m}}=1,\\\textbf{then find the least positive integral}\\\textbf{value of m.}$$

$$\textbf{Sol.}\space\bigg(\frac{1+i}{1-i}\bigg)^{m}=1\\\bigg[\frac{(1+i)}{(1-i)}×\frac{(1+i)}{(1+i)}\bigg]^{m}=1\\\bigg[\frac{(1+i)^{2}}{1^{2}-(i)^{2}}\bigg]^{m}=1$$

[(a + b) (a – b) = a2 – b2]

$$\bigg[\frac{1^{2}+i^{2}+2i}{1+1}\bigg]^{m}=1\space[i^{2}=-1]\\\bigg(\frac{1-1+2i}{2}\bigg)^{m}=1\\\bigg(\frac{2i}{2}\bigg)^{m}=1$$

im = 1

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

Thus, the least positive integer is 1. Therefore, the least positive integral value of m is 4.