**(I) Select the correct answer from the given alternatives.**

**Question 1.If n is an odd positive integer, then the value of 1 + is:(A) -4i(B) 0(C) 4i(D) 4Answer:**

(B) 0

Hint:

= 1 – 1 + 1 – 1 …..(n odd positive integer)

= 0

**Question 2.(A) -2(B) 1(C) 0(D) -1Answer:**

(D) -1

Hint:

**Question 3.√-3 √-6 is equal to(A) -3√2(B) 3√2(C) 3√2 i(D) -3√2 iAnswer:**

(A) -3√2

Hint:

√-3 √-6

= (√3 i) (√6 i)

= 3√2 (-1)

= -3√2

**Question 4.If ω is a complex cube root of unity, then the value of is:(A) -1(B) 1(C) 0(D) 3Answer:**(C) 0

Hint:

**Question 5.(A) cos 2θ(B) 2cos 2θ(C) 2cos θ(D) 2sin θAnswer:**

(B) 2cos 2θ

Hint:

**Question 6.If ω(≠1) is a cube root of unity and (1 + ω = A + Bω, then A and B are respectively the numbers(A) 0, 1(B) 1, 1(C) 1, 0(D) -1, 1Answer:**

(B) 1, 1

Hint:

= 1 + ω

A = 1, B = 1

**Question 7.**

Hint:

**Question 8.(A) -θ(B) θ(C) π – θ(D) π + θAnswer:**

(A) -θ

Hint:

**Question 9.**

Hint:

**Question 10.If z = x + iy and |z – zi| = 1, then(A) z lies on X-axis(B) z lies on Y-axis(C) z lies on a rectangle(D) z lies on a circleAnswer:**

(D) z lies on a circle

Hint:

|z – zi | = |z| |1 – i| = 1

**(II) Answer the following:**

**Question 1.Simplify the following and express in the form a + ib.(i) 3 + √-64Solution:**

3 + √-64

= 3 + √64 √-1

= 3 + 8i

**(iii) (2 + 3i) (1 – 4i)Solution:**

(2 + 3i)(1 – 4i)

= 2 – 8i + 3i – 12

= 2 – 5i – 12(-1) …..[∵ = -1]

= 14 – 5i

Solution:

= (-8 + 6i)(3 + i)

= -24 – 8i + 18i + 6i^{2}

= -24 + 10i + 6(-1)

= -24 + 10i – 6

= -30 + 10i

Solution:

Solution:

Solution:

Solution:

Solution:

**Question 2.Solve the following equations for x, y ∈ R(i) (4 – 5i)x + (2 + 3i)y = 10 – 7iSolution:**

(4 – 5i)x + (2 + 3i)y = 10 – 7i

(4x + 2y) + (3y – 5x) i = 10 – 7i

Equating real and imaginary parts, we get

4x + 2y= 10 i.e., 2x + y = 5 ……(i)

and 3y – 5x = -7 ……(ii)

Equation (i) × 3 – equation (ii) gives

11x = 22

∴ x = 2

Putting x = 2 in (i), we get

2(2) + y = 5

∴ y = 1

∴ x = 2 and y = 1

x + iy = (7 – i)(2 + 3i)

x + iy = 14 + 21i – 2i – 3

x + iy = 14 + 19i – 3(-1)

x + iy = 17 + 19i

Equating real and imaginary parts, we get

∴ x = 17 and y = 19

**(iii) (x + iy) (5 + 6i) = 2 + 3iSolution:**

2x + 2yi – y + xi = 10

(2x – y) + (x + 2y)i = 10 + 0 . i

Equating real and imaginary parts, we get

2x – y = 10 ……(i)

and x + 2y = 0 ……..(ii)

Equation (i) × 2 + equation (ii) gives, we get

5x = 20

∴ x = 4

Putting x = 4 in (i), we get

2(4) – y = 10

y = 8 – 10

∴ y = -2

∴ x = 4 and y = -2

**Question 3.Evaluate(i) (1 – i +Solution:**

**Question 4.Find the value of(i) + 2 – 3x + 21, if x = 1 + 2iSolution:**

**(ii) + 9 + 35 – x + 164, if x = -5 + 4iSolution:**

**Question 5.Find the square roots of(i) -16 + 30iSolution:**

**(ii) 15 – 8iSolution:**

**(iv) 18iSolution:**

**(v) 3 – 4iSolution:**

**(vi) 6 + 8iSolution:**

**Question 6.Find the modulus and argument of each complex number and express it in the polar form.(i) 8 + 15iSolution:**

**(ii) 6 – iSolution:**

Solution:

Solution:

**(v) 2iSolution:**

**(vi) -3iSolution:**

Solution:

**Question 7.Represent 1 + 21, 2 – i, -3 – 2i, -2 + 3i by points in Argand’s diagram.Solution:**

The complex numbers 1 + 2i, 2 – i, -3 – 2i, -2 + 3i will be represented by the points A(1, 2), B(2, -1), C(-3, -2), D(-2, 3) respectively as shown below:

**Question 8.Solution:**

**Question 9.Find the real numbers x and y such that**

(3x + y) + 2(x + y)i = 5 + 6i

Equating real and imaginary parts, we get

3x + y = 5 ……(i)

and 2(x + y) = 6

i.e., x + y = 3 …….(ii)

Subtracting (ii) from (i), we get

2x = 2

∴ x = 1

Putting x = 1 in (ii), we get

1 + y = 3

∴ y = 2

∴ x = 1, y = 2

**Question 10.Solution:**

**Question 11.Solution:**

**Question 12.Convert the complex numbers in polar form and also in exponential form.Solution:**

**(ii) z = -6 + √2 iSolution:**

z = -6 + √2 i

∴ a = -6, b = √2

i.e. a < 0, b > 0

Solution:

**Question 13.Solution:**

**Question 14.Solution:**

**Question 15.Solution:**

**Question 16.SimplifySolution:**

Solution:

Solution:

**Question 17.Solution:**

**Question 18.If α and β are complex cube roots of unity, prove that (1 – α) (1 – β) (1 – ) (1 –) = 9.Solution:**

α and β are the complex cube roots of unity.

**Question 19.**

**Question 20.If ω is the cube root of unity, then find the value of **

Solution:

If ω is the complex cube root of unity, then