OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Ex 9(e)

Well-structured OP Malhotra Maths Class 11 Solutions Chapter 9 Complex Numbers Ex 9(e) facilitate a deeper understanding of mathematical principles.

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

Question 1.
Illustrate in the complex plane, the set of points satisfying the following conditions. Explain your answer:
(i) |z| ≤ 3
(ii) arg (z – 2) = \(\frac { π }{ 3 }\)
(iii) | i – 1 – 2z | > 9
Solution:
(i) Let z = x + iy
∴ | z | ≤ 3 ⇒ | x + iy ≤ 3
⇒ x² + y² ≤ 9
which represents the set of points in interior and on the circle with centre (0, 0) and radius 3.

(ii) arg (z – 2) = \(\frac { π }{ 3 }\)
⇒ arg (x + iy – 2) = \(\frac { π }{ 3 }\)
⇒ tan^-1\(\left(\frac{y}{x-2}\right)=\frac{\pi}{3}\)
⇒ \(\frac{y}{x-2}=\tan \frac{\pi}{3}\) = \(\sqrt{3}\) ⇒ y = \(\sqrt{3}\)(x – 2)
which represents the set of points on line which intersects x-axis at (2, 0) and making an angle of 60° with +ve direction of x-axis.

(iii) Given | i – 1 – 2z | > 9
| i – 1 – 2 (x + iy) | > 9, where z = x + iy

Question 2.
Illustrate and explain the region of the Argand’s plane represented by the inequality |z + i| ≥ |z + 2|.
Solution:
Let z = x + iy
given | z + i | ≥ | z + 2 |
⇒ | x + iy + i | ≥ | x + iy + 2 |
⇒ |x + i (y + 1) | ≥ |x + 2 + iy |
⇒ \(\sqrt{x^2+(y+1)^2}\) ≥ \(\sqrt{(x+2)^2+y^2}\)
On squaring both sides ; we have
x² + (y + 1)² ≥ (x + 2)² + y²
⇒ x² + y² + 2y + 1 ≥ x² + 4x + 4 + y²
⇒ 2y + 1 ≥ 4x + 4
⇒ y ≥ 2x + \(\frac { 3 }{ 2 }\)
which represents the set of points in the region lies above the line PQ intersects coordinate axes at (0, \(\frac { 3 }{ 2 }\)) and (- \(\frac { 3 }{ 4 }\), 0)

Question 3.
Illustrate and explain the set of points z in the Argand diagram, which represents | z – z₁ | ≤ 3 where z1 = 3 – 2i
Solution:
Given | z – z₁| ≤ 3 ;
where z₁ = 3 – 2i
⇒ | z – (3 – 2i) | ≤ 3
⇒ | x + iy – 3 + 2i | ≤ 3
⇒ | (x – 3) + i (y + 2) | ≤ 3
⇒ (x – 3)² + (y + 2)² ≤ 9
which represents the set of points in the interior and on the circle with centre (3, – 2) and radius 3.

Question 4.
If z = x + yi and ω = \(\frac{(1-z i)}{z-i}\) then
| ω | = 1 implies that in the complex plane
(a) z lies on the imaginary axis
(b) z lies on the real axis
(c) z lies on the unit circle
(d) None of these
Solution:
Given z = x + iy and w = \(\frac{1-z i}{z-i}\)
and |w| = 1 ⇒ \(\left|\frac{1-z i}{z-i}\right|\) = 1
⇒ |1 – zi | = |z – i| [∵ \(\left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|}\) ]
⇒ | 1 – (x + iy) i | = | x + iy – i |
⇒ |1 – ix + y| = |x + i(y – 1)|
⇒ 1(1 + y) – ix| = | x + | 0 – 1) |
⇒ \(\sqrt{(1+y)^2+(-x)^2}=\sqrt{x^2+(y-1)^2}\)
On squaring both sides ; we have
(1 + y)² + x² = x² + (y – 1)²
⇒ y² + 2y + 1 – x² = x²+y² – 2y + 1
⇒ 4y = 0
⇒ y = 0
which is the eqn. of x-axis. z lies on real axis i.e. x-axis

Question 5.
Find the locus of a complex number z such that arg \(\left(\frac{z-2}{z+2}\right)\) = \(\frac { π }{ 3 }\).
Solution:

Question 6.
If the amplitude of z – 2 – 3i is \(\frac { π }{ 4 }\), then find the locus of z = x +yi.
Solution:
Given z = x + iy
and amp (z – 2 – 3i) = \(\frac { π }{ 4 }\)
⇒ amp (x + iy – 2 – 3 z) = \(\frac { π }{ 4 }\)
⇒ amp [(x – 2) + i (y – 3)] = \(\frac { π }{ 4 }\)
⇒ tan^-1\(\left(\frac{y-3}{x-2}\right)=\frac{\pi}{4}\)
⇒ \(\frac{y-3}{x-2}=\tan \frac{\pi}{4}=1\)
⇒ y – 3 = x – 2
⇒ x – y + 1 = 0
Hence locus of z represents a straight line intersecting coordinate axes at (- 1, 0) and (0, 1).

Question 7.
Find the locus of z if
ω = \(\frac{z}{z-\frac{1}{3} i}\) = 1.
Solution:

Question 8.
A variable complex number z is such that the amplitude of \(\frac{z-1}{z+1}\) is always equal to \(\frac { π }{ 4 }\).
Illustrate the locus of z in the Argand plane.
Solution:

Question 9.
Find the radius and centre of the circle z\(\bar { z }\)+ (1 – i) z + (1 + i) \(\bar { z }\) – 7 =0.
Solution:
Given eqn. of circle be z\(\bar { z }\)+(1 – i)z + (1 + i)\(\bar { z }\) – 7 = 0 …(1)
where z = x + iy and \(\bar { z }\) = x – iy
∴ eqn. (1) becomes ;
⇒ (x + iy) (x – iy) + (1 – i) (x + iy) + (i + i) (x – iy) – 7 = 0
⇒ x² + y² + x + iy – ix + y + x – iy + ix + y – 7 = 0
⇒ x² + y² + 2x + 2y – 7 = 0
⇒ (x² + 2x) + (y² + 2y) – 7 = 0
⇒ (x² + 2x + 1) + (y² + 2y + 1) – 9 = 0
⇒ (x + 1)² + (y + 1)² = 9
which represents a circle with centre (- 1, – 1) and radius 3.

Question 10.
What is the region represented by the inequality 3 < | z – 2 – 3i|< 4 in the Argand plane.
Solution:
Given 3 < |z – 2 – 3i| < 4;

where z = x + iy
⇒ 3 < | x + iy – 2 – 3i| < 4
⇒ 3 < | (x – 2) + | (y – 3) | < 4
⇒ 3 < \(\sqrt{(x-2)^2+(y-3)^2}\) < 4
⇒ 9 < (x – 2)² + (y – 3)² < 4
which represents the set of points lies in the region between two concentric circles with centre (2, 3) and radius 2 and 3.