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)
lies in first quadrant.
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.