If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Precalculus

### Course: Precalculus>Unit 3

Lesson 7: Graphically multiplying complex numbers

# Multiplying complex numbers graphically example: -1-i

We can multiply complex numbers graphically on the complex plane. We rotate an amount equal to the argument and scale by the modulus of the complex number by which we're multiplying. Created by Sal Khan.

## Want to join the conversation?

• Okay I get that when you multiply by a complex number, you are going to rotate by the argument of that complex number, scale the modulus of Z by the modulus of this complex number. But I don't get why? Sal hasn't proven why this works
• Statement to prove:
The modulus and argument of the product of any two complex numbers is the product of the moduli of the two complex numbers and the sum of the arguments of the two complex numbers, respectively.

Givens:
Let z₁ = r₁ ⋅ e^(iθ₁), z₂ = r₂ ⋅ e^(iθ₂), and z₃ = r₃ ⋅ e^(iθ₃)
Let z₃ = z₁ ⋅ z₂
The moduli and arguments of z₁, z₂, and z₃ are r₁ and θ₁, r₂ and θ₂, and r₃ and θ₃, respectively.
z₁ and z₂ can be any two complex numbers.
z₃ is the product of z₁ and z₂.

Proof:
z₃ = z₁ ⋅ z₂ ; given
= [r₁ ⋅ e^(iθ₁)] ⋅ [r₂ ⋅ e^(iθ₂)] ; substitution
= r₁ ⋅ e^(iθ₁) ⋅ r₂ ⋅ e^(iθ₂) ; associative property
= r₁ ⋅ r₂ ⋅ e^(iθ₁) ⋅ e^(iθ₂) ; commutative property
= r₁ ⋅ r₂ ⋅ e^(iθ₁ + iθ₂) ; product rule of exponents
z₃ = r₁ ⋅ r₂ ⋅ e^[i(θ₁ + θ₂)] ; distributive property
z₃ = r₃ ⋅ e^(iθ₃) ; given
r₃ = r₁ ⋅ r₂ and θ₃ = θ₁ + θ₂ ; comparison of coefficients and powers
The modulus and argument of z₃ is the product of the moduli of z₁ and z₂ and the sum of the arguments of z₁ and z₂, respectively.
∵ z₃ is the product of z₁ and z₂, and z₁ and z₂ can be any two complex numbers
∴ The modulus and argument of the product of any two complex numbers is the product of the moduli of the two complex numbers and the sum of the arguments of the two complex numbers, respectively.
• Could an alternative solution be just to multiply the 2 complex numbers? (0-3i)(-1-i). Graphically seams to be correct