z = 12 /_ 30°

w = √18 /_ 225°

z • w = 36 √2 /_ 255°

Use √(a^2 + b^2) = r and (θ) = arctan(b/a) to get polar forms

z = 12[√(3) + i] = 24e^i(𝛑/6)

w = 3(-1 - i) = 3√(2)e^i(5𝛑/4) because -1 – i is in third sector.

w = 3(-1 - i) = 3√(2)e^i(-3𝛑/4) is equivalent because e^i2𝛑 = 1

Use r₁e^i(θ₁)*r₂e^i(θ₂) = r₁r₂e^i(θ₁+θ₂) to get polar product

zw = 72√(2)e^i[(2 + 15)/12]𝛑 = 72√(2)e^i(17𝛑/12)

zw = 72√(2)e^i(-7𝛑/12) would be equivalent.

The rectangular form can be found from (72√2)cis(17π/12) if required

z = 12(sqrt(3)+ i) = 24(√3/2 + 1/2 i) = 24cis(π/6)

w = −3 − 3i = 3√2 (-1/√2 - 1/√2 i) = 3√2 cis(5π/4)

(r₁cisθ₁)(r₂cisθ₂) = r₁r₂cis(θ₁+θ₂)

zw = (24cis(π/6))((3√2)cis(5π/4)) = (24×3√2)cis(π/6+5π/4) = (72√2)cis(17π/12) ✓

this is my work and i can't find the mistake

First, put z & w into polar form and write then here.

Then we will proceed...

Hopefully no one will spoil you the answer thereby depriving you from your personal enhancement; that would be very inconsiderate of them.