By using Euler’s formula which state that:
e∧(iΘ) = cos Θ + i sin Θ
So, z₁ = 2 ( cos π/6 + i sin π/6 ) = 2 e∧(i π/6)
And z₂ = 3 ( cos π/4 + i sin π/4 ) = 3 e∧(i π/4)
Then z₁ * z₂ = 2 e∧(i π/6) * 3 e∧(i π/4) = 6 e∧[i (π/6 + π/4)]
= 6 e∧(i 5π/12)
= 6 ( cos 5π/12 + i sin 5π/12 )