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 – z_{1} | ≤ 3 where z1 = 3 – 2i

Solution:

Given | z – z_{1}| ≤ 3 ;

where z_{1} = 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.