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## Precalculus

### Course: Precalculus>Unit 3

Lesson 8: Multiplying and dividing complex numbers in polar form

# Multiplying complex numbers in polar form

We can multiply two complex numbers in polar form by multiplying their moduli and adding their arguments. Created by Sal Khan.

## Want to join the conversation?

• Thank You for the video, Sal!

I'll show here the algebraic demonstration of the multiplication and division in polar form, using the trigonometric identities, because not everyone looks at the tips and thanks tab.

But I also would like to know if it is really correct.

w1 = a*(cos(x) + i*sin(x))
w2 = b*(cos(y) + i*sin(y))

Multiplication

w1 * w2

a*(cos(x) + i*sin(x)) * b*(cos(y) + i*sin(y))

a*b*(cos(x) + i*sin(x))*(cos(y) + i*sin(y))

ab*(cos(x) + i*sin(x))*(cos(y) + i*sin(y))

ab*(cos(x)cos(y) + cos(x)*sin(y)*i + cos(y)*sin(x)*i + sin(x)*sin(y)*i^2)

i^2 = -1

ab
(cos(x)cos(y) - sin(x)*sin(y) + i(cos(x)sin(y) + cos(y)*sin(x))

ab
(cos(x + y) + i*sin(x + y))

Division

w1/w2

numerator: a*(cos(x) + i*sin(x))
denominator: b*(cos(y) + i*sin(y))

multiply both the numerator and denominator by (cos(y) - i*sin(y))

numerator: a*(cos(x) + i*sin(x))*(cos(y) - i*sin(y))

a*(cos(x)cos(y) - cos(x)*sin(y)*i + sin(x)*cos(y)*i - sin(x)*sin(y)*i^2)

i^2 = -1

a
(cos(x)cos(y) + sin(x)*sin(y) + i(sin(x)cos(y) - cos(x)*sin(y)))

a
(cos(x - y) + i*sin(x - y))

denominator: b*(cos(y) + i*sin(y))*(cos(y) - i*sin(y))

Difference of squares

b*((cos(y))^2 - i^2*(sin(y))^2)

b*((cos(y))^2 + (sin(y))^2)

The fundamental relation of trigonometry

b*1

numerator: a*(cos(x - y) + i*sin(x - y))
denominator: b

a*(cos(x - y) + i*sin(x - y))/b

(a/b)*(cos(x - y) + i*sin(x - y))  