OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Chapter Test

Accessing OP Malhotra Maths Class 11 Solutions Chapter 9 Complex Numbers Chapter Test can be a valuable tool for students seeking extra practice.

S Chand Class 11 ICSE Maths Solutions Chapter 9 Complex Numbers Chapter Test

Question 1.
Find the square root of 5 – 12i.
Solution:
Let \(\sqrt{5-12 i}\) = x – iy ; on squaring both sides; we have
5 – 12i = (x – iy)² = x² – y² – 2ixy
On equating real and imaginary parts on both sides ; we have
x² – y² = 5 …(1)
and 2xy = 12 …(2)
Now x² + y² = \(\sqrt{\left(x^2-y^2\right)^2+(2 x y)^2}\)
= \(\sqrt{5^2+12^2}\)
= \(\sqrt{25+144}\) = 13 … (1)
On adding eqn. (1) and (3) ; we have
2x² = 18 ⇒ x² = 9 ⇒ x = ± 3
eqn. (3) – eqn. (1) gives ; 2y² = 8 ⇒ y = ± 2
Since xy be of positive sign ∴ x and y are of same sign.
∴ When x = 3, y – 2 or when x = – 3, y = – 2
Thus, \(\sqrt{5-12 i}\) = (3 – 2i) or – (3 – 2i)
= ± (3 – 2i)

Question 2.
Find the locus of a complex number Z = x + iy, satisfying the relation |z + i| = |z + 2|.
Illustrate the locus of z in the Argand plane.
Solution:
Given |z + i| = |z + 2|; where z = x + iy
⇒ | 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 x
x² + (y + 1)² = (x + 2)² + y²
⇒ x² + y² + 2y + 1 = x² + y² + 4x + 4
⇒ 2y + 1 = 4x + 4
⇒ 4x – 2y + 3 = 0
Thus locus of z is a straight line intersecting coordinate axes at \(\left(-\frac{3}{4}, 0\right) \text { and }\left(0, \frac{3}{2}\right)\)

Question 3.
Express \(\frac{13 i}{2-3 i}\) in the form A + Bi.
Solution:
Let z = \(\frac{13 i}{2-3 i} \times \frac{2+3 i}{2+3 i}=\frac{13 i(2+3 i)}{2^2+3^2}=\frac{26 i-39}{13}\) = 2i – 3 = – 3 + 2i

Question 4.
If z = x + yi and \(\frac{|z-1-i|+4}{3|z-1-i|-2}\) = 1, show that x² +y² – 2x- 2y – 7 = 0.
Solution:
Given z = x + iy and \(\frac{|z-1-i|+4}{3|z-1-i|-2}\) = 1
⇒ |x + iy – 1 – i| + 4 = 3|x + iy – 1 – i | – 2
⇒ | (x – 1) + i(y – 1) | = 3 | (x – 1) + i(y – 1) | – 6
\(\sqrt{(x-1)^2+(y-1)^2}=3 \sqrt{(x-1)^2+(y-1)^2}-6\)
⇒ 2\(\sqrt{(x-1)^2+(y-1)^2}\) = 6
⇒ \(\sqrt{(x-1)^2+(y-1)^2}\) = 3
On squaring both sides ; we have
(x – 1)² + (y – 1)² = 9 ⇒ x² + y² – 2x – 2y – 7 = 0

Question 5.
If ω and ω² are cube roots of unity, prove that (2 – ω + 2ω²) (2 + 2ω – ω²) = 9.
Solution:
(2 – ω + 2ω²) (2 + 2ω – ω²) = [2(1 + ω²) – ω] [2 (1 + ω) – ω²] [∵ 1 + ω + ω² = 0]
= [2 (- ω) – ω] [- 2ω² – ω²]
= (- 3ω) (- 3ω²) = 9ω³ = 9 [∵ ω³ = 1]

Question 6.
If z₁, z₂ ∈ C (set of complex numbers), prove that | z₁ + z₂ | ≤ | z₁ | + | z₂ |.
Solution:

Question 7.
If z = x + yi, ω = \(\frac{2-i z}{2 z-i}\) and | ω | = 1, find the locus of z and illustrate it in the complex plane.
Solution:
Given z = x + iy and w = \(\frac{2-i z}{2 z-i}\)
and | ω | = 1
⇒ \(\frac{2-i z}{2 z-i}\) = 1 ⇒ \(\frac{2-i z}{2 z-i}\) = 1 [∵ \(\left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|}\) ]
⇒ | 2 – i(x + iy) | = |2(x + iy) – i |
⇒ | (2 + y) – ix | = | 2x + i (2y – 1) |
⇒ \(\sqrt{(2+y)^2+x^2}=\sqrt{(2 x)^2+(2 y-1)^2}\)
On squaring both sides ; we have
(2 + y)² + x² = 4x² + (2y – 1)²
⇒ y² + 4y + 4 + x² = 4x² + 4y² – 4y + 1
⇒ 3x² + 3y² – 8y – 3 = 0
⇒ x² + y² – \(\frac { 8 }{ 3 }\)y – 1 = o
⇒ x² + y² – \(\frac { 8 }{ 3 }\)y + \(\frac { 16 }{ 9 }\) – 1 – \(\frac { 16 }{ 9 }\) = o
⇒ x² + (y – \(\frac { 4 }{ 3 }\))² = \(\frac { 25 }{ 9 }\) = (\(\frac { 5 }{ 3 }\))²
Thus locus of z represents a circle with centre (0, \(\frac { 4 }{ 3 }\)) and radius \(\frac { 5 }{ 3 }\).

Question 8.
Simplify : (1 – 3ω + ω²) (1 + ω – 3ω²).
Solution:
(1 – 3ω + ω²) (1 + ω – 3ω²) = (1 + ω² – 3ω) (1 + ω – 3ω²)
= (- ω – 3ω) (- ω² – 3ω²) [∵ 1 + ω + ω² = 0]
= (- 4ω) (- 4ω²)
= 16ω³
= 16 x 1 = 16

Question 9.
Sketch in the complex plane the set of points z satisfying \(\left|\frac{z-3}{z+1}\right|\) = 3.
Solution:
Given \(\left|\frac{z-3}{z+1}\right|\) = 3 ; where z = x + iy

On squaring both sides ; we have
(x – 3)² + y² = 9[(x + 1)² + y²]
⇒ 9[x² + 2x + 1 + y²] = x ² + y² – 6x + 9
⇒ 8x² + 8y² + 24x = 0
⇒ x² + y² + 3x = 0
which represents the set of points in the circle whose centre (\(\frac { -3 }{ 2 }\), 0) and radius \(\frac { 3 }{ 2 }\) units.

Question 10.
Given that
\(\frac{2 \sqrt{3} \cos 30^{\circ}-2 i \sin 30^{\circ}}{\sqrt{2}\left(\cos 45^{\circ}+i \sin 45^{\circ}\right)}\) = A + Bi, find the values of A and B.
Solution:
Given \(\frac{2 \sqrt{3} \cos 30^{\circ}-2 i \sin 30^{\circ}}{\sqrt{2}\left(\cos 45^{\circ}+i \sin 45^{\circ}\right)}\) = A + Bi

On comparing real and imaginary parts on both sides ; we have A = 1; B = – 2

Question 11.
Simplify :
(1 – ω) (1 – ω²) (1 – ω⁴) (1 – ω⁸).
Solution:
(1 – ω) (1 – ω²) (1 – ω⁴) (1 – ω⁸)
= (1 – ω) (1 – ω²) (1 – ω) (1 – ω²) [∵ ω³ = ω⁶ = 1]
= [(1 – ω)(1 – ω²)]²
= [1 – ω – ω² + ω³]²
= [1 – (ω + ω²) + 1]²
= [1 – (- 1) + 1]² = 9 [∵ 1 + ω + ω² = 0]

Question 12.
Find the locus of a complex number z = x + yi, satisfying the relation | 2z + 3i | ≥ | 2z + 5 |.
Illustrate the locus in the Argand plane.
Solution:
Given | 2z + 3i | ∈ | 2z + 5 |;
where z = x + iy
⇒ | 2 (x + iy) + 3i | ≥ | 2 (x + iy) + 5 |
⇒ | 2x + i (2y + 3) | ≥ (2x + 5) + 2iy |
\(\sqrt{4 x^2+(2 y+3)^2} \geq \sqrt{(2 x+5)^2+4 y^2}\)
On squaring both sides ; we have
4x² + 4y² + 12y + 9 ≥ 4x² + 20x + 4y² + 25
⇒ 12y – 20x ≥ 16
⇒ 3y – 5x ≥ 4
⇒ 3y ≥ 5x + 4
Hence locus be the set of points in the region not containing origin lying on the line intersecting coordinate axes at (-\(\frac { 4 }{ 5 }\), 0) and (0, \(\frac { 4 }{ 3 }\).

Question 13.
Find the real values of x and y satisfying the equality \(\frac{x-2+(y-3) i}{1+i}\) = 1 – 3i.
Solution:
Given \(\frac{(x-2)+(y-3) i}{1+i}\) = 1 – 3i
⇒ (x – 2) + (y – 3)i = (1 – 3i)(1 + i)
⇒ (x – 2) + (y – 3) i = 1 + i – 3i + 3 = 4 – 2i
On comparing real and imaginary parts on both sides; we have
x – 2 = 4 ⇒ x = 6
and y – 3 = – 2 ⇒ y = 1

Question 14.
If i = \(\sqrt{-1}\), prove the following :
(x + 1 + i) (x + 1 – i) (x – 1 – i) (x – 1 + i) = x⁴ + 4.
Solution:
L.H.S = (x + 1 + i) (x + 1 – i) (x – 1 – i) (x – 1 + i)
= [(x+ 1 )² – i²] [(x – 1 )² – i²]
= [x² + 2x + 2] [x² – 2x + 2]
= (x² + 2 + 2x) (x² + 2 – 2x)
= (x² + 2)² – (2x)²
= x⁴ + 4x² + 4 – 4x²
= x⁴ + 4 = R.H.S

Question 15.
If z = x + yi and | 2z + 1 | = | z – 2i |, show that 3 (x² + y²) + 4 (x – y) = 3.
Solution:
Given z = x + iy
Also, | 2z + 1 | = | z – 2i |
⇒ | 2 (x + iy) + 1 | = | x + iy – 2i |
⇒ | 2x + 1 + 2iy | = | x + i (y – 2) |
⇒ \(\sqrt{(2 x+1)^2+(2 y)^2}=\sqrt{x^2+(y-2)^2}\)
On squaring both sides ; we have
4x² + 4x + 1 + 4y² = x² + y² – 4y + 4
⇒ 3x² + 3y² + 4x + 4y – 3 = 0

Question 16.
Find the amplitude of the complex number \(\sin \frac{6 \pi}{5}+i\left(1-\cos \frac{6 \pi}{5}\right)\).
Solution:

Question 17.
Express \(\frac{1-2 i}{2+i}+\frac{3+i}{2-i}\) in the form a + bi.
Solution:
Let z = \(\frac{(1-2 i)}{2+i}+\frac{3+i}{2-i}=\frac{(1-2 i)(2-i)+(3+i)(2+i)}{(2+i)(2-i)}=\frac{2-5 i-2+6+5 i-1}{2^2-i^2}\)
= \(\frac{5}{2^2+1}=\frac{5}{5}\) = 1 = 1 + 0i

Question 18.
Find the value of x and y given that (x + yi) (2 – 3i) = 4 + i.
Solution:
Given (x + iy) (2 – 3i) = 4 + i
⇒ x + iy = \(\frac{4+i}{2-3 i} \times \frac{2+3 i}{2+3 i}=\frac{8+12 i+2 i-3}{2^2-(3 i)^2}=\frac{5+14 i}{4+9}\)
⇒ x + iy = \(\frac{5}{13}+\frac{14}{13} i\)
On equating real and imaginary parts on both sides ; we have
x = \(\frac { 5 }{ 13 }\) and y = \(\frac { 14 }{ 13 }\)

Question 19.
If the ratio \(\frac{z-i}{z-1}\) is purely imaginary, prove that the point z lies on the circle whose centre is the point \(\frac { 1 }{ 2 }\)(1 + i) and radius is \(\frac{1}{\sqrt{2}}\).
Solution:

Question 20.
If (- 2 + \(\sqrt{-3}\))(- 3 + 2\(\sqrt{-3}\)) = a + bi, find the real numbers a and b. With these values of a and b, also find the modulus of a + bi.
Solution:
(- 2 + \(\sqrt{3}\))(- 3 + 2\(\sqrt{-3}\)) = a + ib
⇒ (- 2 + \(\sqrt{3}\)i)(- 3 + 2\(\sqrt{3}\) i) = a + ib
⇒ (6 – 4\(\sqrt{3}\) i – 3\(\sqrt{3}\) i – 6) = a + ib
⇒ – 7\(\sqrt{3}\) i = a + ib
∴ a = 0; b = – 7\(\sqrt{3}\)
Thus |a + ib| = | – 7\(\sqrt{3}\)i| = 7\(\sqrt{3}\)

Question 21.
if 1, ω, ω² are the three cube roots of unity, then simplify : (3 + 5ω + 3ω²)² (1 + 2ω + ω²)
Solution:

Question 22.
Find the locus of a complex number z = x + iy, satisfying the relation | 3z – 4i | < | 3z + 2 |. Illustrate the locus in the Argand plane.
Solution:
Given z = x + iy

Also, | 3z – 4i | ≤ | 3z + 2 |
⇒ | 3 (x + iy) | ≤ | 3 (x + yi) + 2|
⇒ | 3x + iy (3y – 4) | ≤ | (3x + 2) + 3iy|
⇒ \(\sqrt{(3 x)^2+(3 y-4)^2} \leq \sqrt{(3 x+2)^2+(3 y)^2}\)
On squaring both sides ; we have
⇒ 9x² + 9y² – 24y + 16 ≤ 9x² + 12x + 9y² + 4
⇒ 12x + 24y – 12 ≥ 0
⇒ x + 2
Clearly the line x + 2y – 1 = 0 intersecting coordinate axes at (1, 0) and (0, \(\frac { 1 }{ 2 }\))
Hence locus of z consisting of set of points lying in region not containing (0, 0) satisfying x + 2y – 1 ≥ 0.

Question 23.
Find the modulus and argument of the complex number \(\frac{2+i}{4 i+(1+i)^2}\).
Solution:

Question 24.
If | z – 3 + i | = 4, then the locus of z is
(i) x² + y² – 6 = 0
(ii) x² + y² – 3x + y = 0
(iii) x² + y² – 6x – 2 = 0
(iv) x² + y² – 6x + 2y – 6 = 0
Solution:
Given | z – 3 + i | = 4 ; where z = x + iy
⇒ | x + iy – 3 + i | = 34
⇒ \(|(x-3)+i(y+1)|\) = 4
⇒ \(\sqrt{(x-3)^2+(y+1)^2}\) = 4
⇒ (x – 3)² + (y + 1)² = 16
⇒ x² + y² – 6x + 2y – 6 = 0

Question 25.
The locus of the point z is the Argand plane for which | z + 1 |² + | z – 1|² = 4 is a
(a) Straight line
(b) Pair of straight lines
(c) Parabola
(d) Circle
Solution:
Given | z + 1 |² + | z – 1 |² = 4, where z = x + iy ⇒
⇒ | x + 1 + iy |² + | x – 1 + iy |² = 4
⇒ (x + 1)² + y² + (x – 1)² + y² = 4
⇒ 2x² + 2y² = 2
⇒ x² + y² = 1
which clearly represents a circle with centre (0, 0) and radius unity.