Euler's Formula

z = ∣ z ∣ e^{i ϕ} = ∣ z ∣ (cos (ϕ) + i sin (ϕ))

z_{1} z_{2} = (∣ z_{1} ∣ e^{i ϕ_{1}}) (∣ z_{2} ∣ e^{i ϕ_{2}}) = (∣ z_{1} ∣∣ z_{2} ∣ e^{i (ϕ_{1} + ϕ_{2})})

Consequence of Euler’s Identity
Suppose $z_{1}$ has an angle $ϕ_{1}$ and $z_{2}$ has $ϕ_{2}$
The product $z_{1} z_{2}$ has angle $ϕ_{1} + ϕ_{2}$ , and modulus $∣ z_{1} ∣∣ z_{2} ∣$

Notes