Students can track their progress and improvement through regular use of OP Malhotra Maths Class 11 Solutions Chapter 9 Complex Numbers Ex 9(d).

## S Chand Class 11 ICSE Maths Solutions Chapter 9 Complex Numbers Ex 9(d)

Find the modulus and amplitude of the following complex numbers and hence express them into polar form.

Question 1.
$$\sqrt{3}$$ + i
Solution:
Let z = $$\sqrt{3}$$ + i = x + iy
Here x = $$\sqrt{3}$$ > 0 and y = 1 > 0
∴ point (x, y) i.e. (-$$\sqrt{3}$$, 1)

Question 2.
– $$\sqrt{3}$$ + i
Solution:
Let z = – $$\sqrt{3}$$ + i …(1)
put – $$\sqrt{3}$$ = r cos θ …(2)
1 = r sin θ …(3)
On squaring and adding (1) and (2);
3 + 1 = r² (cos² θ + sin² θ)
⇒ r² = 4
⇒ r = 2 (r > 0)
On dividing (3) by (2); we have tan θ
tan θ = – $$\frac{1}{\sqrt{3}}$$ = – tan $$\frac { π }{ 6 }$$ = tan (π – $$\frac { π }{ 6 }$$)
[∵ cos θ < 0 and sin θ > 0
∴ θ lies in 2nd quad]
⇒ θ = $$\frac { 5π }{ 6 }$$
⇒ arg (z) = $$\frac { 5π }{ 6 }$$
∴ from eqn. (1) ; we have
z = r [cos θ + i sin θ]
= 2 [cos $$\frac { 5π }{ 6 }$$ + i sin $$\frac { 5π }{ 6 }$$ ]

Question 3.
– 2 + 2$$\sqrt{3}$$ i
Solution:
Let z = – 2 + 2$$\sqrt{3}$$ i … (1)
put – 2 = r cos θ …(2)
2$$\sqrt{3}$$ = r sin θ …(3)
since cos θ < 0 and sin θ > 0 ∴ θ lies in 2nd quadrant.
On squaring and adding (2) and (3); we have

which is the required polar form.

Question 4.
– 1 – i
Solution:
Let z = – 1 – i = x + iy
Here x = – 1 < 0 and y = – 1 < 0
∴ point (x, y) i.e. (- 1, – 1) lies in IIIrd quadrant.
∴ tan θ = $$\left|\frac{{Im}(z)}{{Re}(z)}\right|=\left|\frac{y}{x}\right|=\left|\frac{-1}{-1}\right|$$ = 1
⇒ α = $$\frac { π }{ 4 }$$
∴ arg (z) = – (π – α) = – (π – $$\frac { π }{ 4 }$$) = – $$\frac { 3π }{ 4 }$$
and r = $$\sqrt{x^2+y^2}=\sqrt{(-1)^2+(-1)^2}=\sqrt{2}$$
Thus required polar form of z is given by
z = r [cos (arg z) + z sin (arg z)]
⇒ z = $$\sqrt{2}\left[\cos \left(\frac{-3 \pi}{4}\right)+i \sin \left(\frac{-3 \pi}{4}\right)\right]$$

Question 5.
– 2i
Solution:
Let z = – 2i × x + iy
∴ |z| = |- 2i| = 2
where x – 0 ; y = – 2
∴ point (x, y) i.e. (0, – 2) lies in IVth quadrant.
∴ tan θ = $$\left|\frac{{Im}(z)}{{Re}(z)}\right|=\left|\frac{-2}{0}\right|$$ → ∞ ⇒ α = $$\frac { π }{ 2 }$$
∴ θ = arg (z) = – α = – $$\frac { π }{ 2 }$$
Thus polar form of z is given by z = | z | (cos θ + z sin θ)
= 2 $$\left[\cos \left(-\frac{\pi}{2}\right)+i \sin \left(-\frac{\pi}{2}\right)\right]$$

Question 6.
– 1 – $$\sqrt{3}$$ i
Solution:
Let z = – 1 – $$\sqrt{3}$$ i …(1)
put – 1 = r cos θ …(2)
– $$\sqrt{3}$$ = r sin θ …(3)
Here cos θ, sin θ < 0
∴ θ lies in 3rd quadrant. On squaring and adding eqn. (2) and eqn. (3); we have
r² (cos² θ + sin² θ) = (- 1)² + (- $$\sqrt{3}$$)² = 1 + 3 = 4
r² = 4 ⇒ r = 2 (∵ r > 0)
On dividing eqn. (3) by eqn. (2); we have
tan θ = $$\frac{-\sqrt{3}}{1}$$ = $$\sqrt{3}$$
= $$\tan \left(\frac{\pi}{3}\right)=\tan \left[-\left(\pi-\frac{\pi}{3}\right)\right]$$
⇒ θ = – $$\frac { 2π }{ 3 }$$
∴ from (1) ; z = r [cos θ + i sin θ]
i.e. z = 2$$\left[\cos \left(-\frac{2 \pi}{3}\right)+i \sin \left(-\frac{2 \pi}{3}\right)\right]$$

Question 7.
– 2
Solution:
Let z = – 2 = x + iy
Here x = – 2 and y = 0
∴ (- 2, 0) lies on negative x-axis.
∴ | z | = | – 2 | = 2
∴ arg (z) = π ;
∴ z = 2 [cos π + i sin π] be the required polar form.

Question 8.
$$\frac{(1+i)^{13}}{(1-i)^7}$$
Solution:
Let

Further x = – 8 ; y = 0 i.e. point (- 8, 0) lies on negative real axis.
∴ arg (z) = π
∴ z = 8 [cos π + i sin π] be the required polar form

Question 9.
(3 + i) (4 + i)
Solution:
Let z = (3+ i) (4 + i)
= 12 + 3i + 4i – 1 = 11 + 7i …(1)
∴ | z | = $$\sqrt{11^2+7^2}$$
= $$\sqrt{121+49}=\sqrt{170}$$
put 11 = r cos θ …(2)
7 = r sin θ …(3)
On dividing (3) by (2); we have
tan θ = $$\frac { 7 }{ 11 }$$
⇒ θ = tan-1$$\frac { 7 }{ 11 }$$ = tan-1(0.636)
⇒ θ = 32°29′
[since cos θ, sin θ > 0
∴ θ lies in first quadrant]
∴ arg (z) = θ = 32° 29′
∴ from (1); z = r [cos θ + i sin θ]
⇒ z = $$\sqrt{170}$$ [cos (32° 29′) + i sin (32° 29′)]
[On squaring and adding eqn. (1) and (2); r = $$\sqrt{170}$$ ]

Question 10.
$$\frac{(1+i)(2+i)}{(3+i)}$$
Solution:
Let z = $$\frac{(1+i)(2+i)}{3+i}=\frac{2+i+2 i-1}{3+i}$$ = $$\frac{1+3 i}{3+i} \times \frac{3-i}{3-i}$$
⇒ z = $$\frac{3-i+9 i+3}{3^2-i^2}$$
⇒ z = $$\frac{6+8 i}{9+1}=\frac{3+4 i}{5}=\frac{3}{5}+\frac{4}{5} i$$ … (1)
∴ | z | = $$\frac{\sqrt{3^2+4^2}}{5}=\frac{5}{5}$$ = 1
Put $$\frac { 3 }{ 5 }$$ = r cos θ … (2)
and $$\frac { 4 }{ 5 }$$ = r sin θ … (2)
On squaring and adding eqn. (2) and (3);
($$\frac { 3 }{ 5 }$$)² = r² (cos² θ + sin² θ)
⇒ r² = $$\frac{9+16}{25}$$ 1 (∵ r > 0)
On dividing eqn. (3) by eqn. (2); we have
tan θ = $$\frac{4}{5} \times \frac{5}{3}=\frac{4}{3}$$ = 1.3333
⇒ θ = tan-1 (1.3333) = 53° 8′ [∵ θ lies in first quadrant]
∴ from (1); z = r [cos θ + i sin θ]
⇒ z = 1 [cos (53° 8′) + i sin (53° 8′)]
which is the required polar form.

Question 11.
$$\frac{5-i}{2-3 i}$$
Solution:

Question 12.
$$\frac{(3+4 i)(4+5 i)}{(4+3 i)(6+7 i)}$$
Solution:
Let z = $$\frac{(3+4 i)(4+5 i)}{(4+3 i)(6+7 i)}$$

⇒ α = 18° 12′ and arg (z) = a = 18° 12′
∴ required polar form of z be given by z = | z | [cos α + i sin α]
⇒ z = $$\sqrt{\frac{41}{85}}$$ [cos (18° 12′)+ i sin (18° 12′)]

Question 13.
Change the following complex numbers into polar form.
(i) – 4 + 4$$\sqrt{3}$$ i
(ii) $$\frac{1+3 i}{1-2 i}$$
(iii) $$\frac{1+2 i}{1-(1-i)^2}$$
(iv) $$\frac{(3+4 i)(4+5 i)}{(4+3 i)(6+7 i)}$$
Solution:
(i) Let z = – 4 + 4$$\sqrt{3}$$i …(1)
put – 4 = r cos θ …(2)
4$$\sqrt{3}$$ = r sin θ …(3)
On squaring and adding eqn. (2) and eqn. (3); we have
r² = (- 4)² + (4$$\sqrt{3}$$)² = 16 + 48 = 64
⇒ r = 8 (∵ r > 0)
since cos θ < 0 and sin θ > 0 ∴ θ lies in 2nd quadrant.
On dividing eqn. (3) by eqn. (2); we have
tan θ = – $$\sqrt{3}$$ = – tan$$\frac { π }{ 3 }$$ = tan(π – $$\frac { π }{ 3 }$$)
= tan$$\frac { 2π }{ 3 }$$
∴ θ = $$\frac { 2π }{ 3 }$$
∴ from (1); z = r [cos θ + i sin θ]
⇒ z = 8$$\left[\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right]$$
which is the required polar form

(ii) Let z = $$\frac{1+3 i}{1-2 i} \times \frac{1+2 i}{1+2 i}$$
⇒ z = $$\frac{1+2 i+3 i-6}{1^2-(2 i)^2}=\frac{5 i-5}{1+4}$$ = i – 1
⇒ z = – 1 + i …(1)
put – 1 = r cos θ …(2)
and 1 = r sin θ …(3)
since cos θ < 0 and sin θ > 0
∴ θ lies in 2nd quadrant.
On squaring and adding eqn. (1) and (2); we have
r² (cos² θ + sin² θ) = (- 1)² + (1)²
⇒ r² = 2 ⇒ r = $$\sqrt{2}$$ (∵ r > 0)
On dividing eqn. (3) by eqn. (2); we have
tan θ = – 1 = – tan $$\frac { π }{ 4 }$$ = tan (π – $$\frac { π }{ 4 }$$)
⇒ θ = π – $$\frac { π }{ 4 }$$ = $$\frac { 3π }{ 4 }$$
∴ from (1); z = r [cos θ + 2 sin θ]
⇒ z = $$\sqrt{2}\left[\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right]$$
which is the required polar form.

(iii) Let z = $$\frac{1+2 i}{1-(1-i)^2}=\frac{1+2 i}{1-(1-1-2 i)}$$
⇒ z = $$\frac{1+2 i}{1+2 i}$$ = x + iy
Here x = 1 ; y = 0
∴ point (x, y) i.e. (1,0)
lies on positive real axes.
∴ z = 1 = 1 + i0 = cos 0° + 2 sin 0°
which is the required polar form.

(iv) Let z = $$\frac{1+7 i}{(2-i)^2}=\frac{1+7 i}{4-1-4 i}$$ = $$\frac{1+7 i}{3-4 i} \times \frac{3+4 i}{3+4 i}$$
⇒ z = $$\frac{3+4 i+21 i-28}{3^2-(4 i)^2}=\frac{-25+25 i}{9+16}$$
⇒ z = – 1 + i … (1)
put – 1 = r cos θ …(2)
and 1 = r sin θ … (3)
since cos θ < 0 and sin θ > 0 ∴ lies in 2nd quadrant.
On squaring and adding eqn. (1) and (2); we have
r² (cos² θ + sin² θ) = (- l)² + (l)²
⇒ r² = 2 ⇒ r = $$\sqrt{2}$$ (∵ r > 0)
On dividing eqn. (3) by eqn. (2); we have
tan θ = – 1 = – tan $$\tan \frac{\pi}{4}=\tan \left(\pi-\frac{\pi}{4}\right)$$
⇒ θ = π – $$\frac { π }{ 4 }$$ = $$\frac { π }{ 4 }$$
∴ from (1) ; z = r [cos θ + i sin θ]
⇒ z = $$\sqrt{2}\left[\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right]$$
which is the required polar form.

Question 14.
Given the complex number z = $$\frac{-1+\sqrt{3} i}{2}$$ and w = $$\frac{-1-\sqrt{3} i}{2}$$ (where i = $$\sqrt{-1}$$)
(i) Prove that each of these complex numbers is the square of the other.
(ii) Calculate the modulus and argument of w and z.
(iii) Calculate the modulus and argument of $$\frac { w }{ z }$$.
(iv) Represent z and w accurately on the complex plane.
Solution:
Given z = $$\frac{-1+\sqrt{3} i}{2}$$ and w = $$\frac{-1-\sqrt{3} i}{2}$$

Thus each of the complex numbers is the square of the other.

(ii) z = – $$\frac{1}{2}+\frac{\sqrt{3}}{2}$$ = x + iy

(iv) Clearly the complex number z is represented by point P$$\frac{-1-\sqrt{3} i}{2}$$ and the complex number is represented by point Q$$\frac{-1-\sqrt{3} i}{2}$$ in complex plane.