derive: r1/r2*[cos(u-v) + isin(u-v)]?

Question

I have to derive this equation using the trigonometic identities, it should relate to the "sum and difference" identity

Answer 1

Let z₁ = r₁[cos u + i.sin u] and z₂ = r₂[cos v + i.sin v]

dividing: z₁/z₂ = r₁/r₂.[cos u + i.sin u]/[cos v + i.sin v]

the trick is to multiply top and bottom by the conjugate of [cos v + i.sin v]
so multiply top and bottom by [cos v - i.sin v] .

The top bit is:
[cos u + i.sin u][cos v - i.sin v]
= (cos u.cos v + sin u.sin v) + i(sin u.cos v - cos u.sin v]
= cos(u - v) + i.sin(u - v)

The bottom bit is:
[cos v + i.sin v][cos v - i.sin v]
= cos² v + sin² v
= 1

Hence z₁/z₂ = r₁/r₂{cos(u - v) + i.sin(u - v)}