# Complex Numbers and Quadratic Equations – Class XI – Exercise 5.1

1: Express the given complex number in the form a + ib: (5i)(-3/5 i)

Solution:

(5i)(-3/5 i) = -5 x 3/5 x i x i

= -3i2

= -3(-1) = 3

2: Express the given complex number in the form a + ib: i9 + i 19

Solution:

i9 + i 19 = i4×2+1 + i4×4+3

= (i4)2.i + (i4)4.i3

[i4 = 1 and i3 = -1]

= 1 x i+1 x (-i)

= i + (-i)

= 0

1. Express the given complex number on the form a + ib: i-39

Solution:

i-39 = i-4×9-3

=(i4)-9 x i-3

= (1)-9 x i-3

= 1/i3

= 1/-I x i/i  = -i/i2 x i/i = -i/i2 = -i/-i = i

1. Express the given complex number in the form a + ib: 3(7 + 7i)+i(7+7i)

Solution:

3(7 + 7i)+i(7+7i) = 21 + 21i + 7i +7i2

= 21 + 28i + 7 x (-1)

= 14 + 28i

1. Express the given comples number in the form a + ib: (1-i)-(-1+6i)

Solution:

(1 – i )-(-1 + 6i) = 1 – i + 1 – 6i

= 2 – 7i

1. Express the given complex number in the form a + ib: (1/5 + i2/5)-(4 + i5/2)

Solution:

(1/5 + i 2/5)-(4 + I 5/2) = 1/5 + 2/5i – 4 – i 5/2

=(1/5 – 4) + i(2/55/2)

=-19/5 + i(-21/10)

= –19/521/10 i

1. Express the given complex number in the form a + ib:

[(1/3 + i  7/3) + (4 + i 1/3)] – (-4/3 + i)

Solution:

[(1/3 + i  7/3) + (4 + i 1/3)] – (-4/3 + i) = 1/3 +  7/3 i + 4 +  1/3i + 4/3 – i

= ( 1/3 + 4 + 4/3) + i(7/3 + 1/3 -1)

= 17/3 + i 5/3

1. Express the given complex number in the form a + ib: (1 – i)4

Solution:

(1 – i)4 = [(1 – i)2]2

= [12 + i2 – 2i] = [1 – 1 – 2i]2 = (-2i)2 = (-2i) x (-2i) = 4i2 = -4

1. Express the given complex number in the form a + ib: (1/3 + 3i)2

Solution:

(1/3 + 3i)2 = (1/3)3 + (3i)3 + 3(1/3)(3i)(1/3 + 3i)

= (1/27) + 27i3 + 3(1/3)(3i)(1/3 + 3i)

= 1/27 + 27(-i) + i + 9i2

= 1/27 – 27i + i  – 9

= (1/27 – 9) + i(-27 + 1)

= ‑242/27 – 26i

1. express he given complex number in the form a + ib: (-2 – 1/3i)3

Solution:

(-2 – 1/3i)3 = (-1)3 (2 + 1/3 i)3

= – [ 23 + (1/3)3 + 3(2)(i/3)(2 + i/3)]

= – [ 8 + (i)3/27 + 2i(2 + i/3)]

= – [ 8 – i/27 + 4i + 2i2/3]

= – [8 – i/27 + 4i – 2/3]

= – [22/3 + 107/27 i ]

= – 22/3107/27 i

1. Find the multiplicative inverse of the complex number 4 – 3i

Solution:

Let z = 4 – 3i

Then ź = 4 + 3i and |z|2 = 42 + (-3)2 = 16 + 9 = 25

Therefore the multiplicative inverse of 4 – 3i is given by

z-1 = ź/|z|2 = 4 + 3i/25  = 4/25 + 3/25 i

1. Find the multiplicative inverse of the complex number √5 + 3i

Solution:

Let z = √5 + 3i

Then ź = √5 – 3i and |z|2 = (√5)2 + (3i)2 = 5 + 9 = 14

Therefore the multiplicative inverse of √5 + 3i

z-1 = ź/|z|2 = √5 + 3i/14 = √5/14 + 3/14 i

1. Find the multiplicative inverse of the complex number –i

Solution:

Let z = -i

Then ź = I and |z|2 = (i)2 = 1

Therefore the multiplicative inverse of –I is given by

z-1 = ź/|z|2 = i/I = i

1. Express the following expression in the form of a + ib.

(3 + i√5)(3 – i√5)/(√3 + √2i) – (√3 – √2i)

Solution:

(3 + i√5)(3 – i√5)/(√3 + √2i) – (√3 – √2i)  = [32 – (i√5)2] /(√3 + √2i – √3 + √2i )

= (9 – 5i2)/2√2i

= 9+5/2√2i x i/i

= 14i/2√2i2

= 14i/2√2i2

= 14i/2√2(-1)

= -7i/√2 x √2/√2 = -7√2i/2