ADDING SUBTRACTING MULTIPLYING COMPLEX NUMBERS

Complex number will consists of two parts,

(i) Real part

(ii) Imaginary part

General form of complex number is a + ib

Here a is real and b is imaginary.

Simplify :

Problem 1 :

(-3 + 4i) + (-4 + 7i)

Solution :

 = (-3 + 4i) + (-4 + 7i)

= (-3 – 4) + 4i + 7i

= -7 + 11i

Problem 2 :

(3 - 6i) + (7 + 3i)

Solution :

= 3 - 6i + 7 + 3i

= (3 + 7) + 3i - 6i

= 10 - 3i

Problem 3 :

(3 + 8i) + (1 - i)

Solution :

= (3 + 1) + (8i – i)

= 4 + 7i

Problem 4 :

(4 - 4i) + (-4 + 6i)

Solution :

= 4 - 4 - 4i + 6i

= 2i

Problem 5 :

-6i + (3i) - (-7 - 8i)

Solution :

= (-6i) + (3i) - (-7 - 8i)

Distributing negative, we get

= (-6i) + (3i) + 7 + 8i

= -6i + 3i + 8i + 7

= 5i + 7

Problem 6 :

(-5 - 3i) - (8 + i)

Solution :

= (-5 - 3i) - (8 + i)

= -5 - 3i - 8 - i

= -5 - 8 - 3i - i

= -13 – 4i

Problem 7 :

(8 - 5i) - (-4 - 3i)

Solution :

= (8 - 5i) - (-4 - 3i)

= 8 - 5i + 4 + 3i

= 12 – 2i

Problem 8 :

-4 + (7i) - (1 - 5i)

Solution :

-4 + (7i) - (1 - 5i) = (-4 – 1) + (7i – (-5i))

= -5 + 12i

Problem 9 :

(-6 – 6i) – (2 – 2i)

Solution :

= (-6 – 6i) – (2 – 2i)

= -6 - 2 - 6i + 2i

= -8 – 4i

Problem 10 :

(-3i) – (-5 + 7i) + i

Solution :

= (-3i) – (-5 + 7i) + i

= -3i + 5 - 7i + i

= -9i + 5

Problem 11 :

(8i) (-6i) (-3i)

Solution :

(8i) (-6i) (-3i)

= -48i²(-3i)

= 144 i² i

= 144(-1) i

= -144 i

Problem 12 :

(2i) (3i) (-6i)

Solution :

= (2i) (3i) (-6i)

= 6i² (-6i)

= -36 ⋅ (i²) i

= -36(-1) i

= 36i

Find the values of x and y that satisfy the equation.

Problem 13 :

4x + 2i = 8 + yi

Solution :

4x + 2i = 8 + yi

By equating the corresponding terms, we get

So, the values of x and y are 2 and 2 respectively.

Problem 13 :

3x + 6i = 27 + yi

Solution :

Equating the real part,

3x = 27

x = 27/3

x = 9

Equating the imaginary part, we get

y = 6

So, the values of x and y are 9 and 6 respectively.

Problem 14 :

−10x + 12i = 20 + 3yi

Solution :

Equating the real part, we get

-10x = 20

x = -20/10

x = -2

Equating the imaginary part, we get

3y = 12

y = 12/3

y = 4

Problem 15 :

Find the value of the real number y such that (3 + 2i) (1 + iy) is 

a) real   b) Imaginary

Solution :

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

Using distributive property, we get

= 3 + 3yi + 2i + 2yi²

= 3 + 3yi + 2i + 2y(-1)

= 3 + 3yi + 2i - 2y

= (3 - 2y) + i(3y + 2)

So, the real part is 3 - 2y and imaginary part is 3y + 2.

Problem 16 :

Write in the form x + yi

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

Solution :

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

Multiplying both numerator and denominator by the conjugate of the denominator, we get

= [(2 + 3i)/(1 + i)] [(1 - i) / (1 - i)]

= (2 + 3i)(1 - i) / (1 + i) (1 - i)

= (2 - 2i + 3i - 3i²)/(1² - i²)

= (2 - 2i + 3i + 3)/(1² - (-1))

= (5 + i)/(1 + 1)

= (5 + i)/2

= (5/2) + (1/2)i

So, the real part is 5/2 and imaginary part is 1/2.