$$2e^{(i×\pi/4)}×3e^{(i×\pi/6)}$$
How would I evaluate and simplify the above, and then express it in polar form?
I understand $re^{i\theta} = r(\cos\theta+i\,\sin\theta)$. The question is to find the product of two polar forms: I've changed them to the form $re^{i\theta}$ under the assumption it's easier to compute. Wolfram gives the answer as $6e^{5i×pi/12}$ but I don't see how it comes to this. How is the exponent evaluated?
Its easiest to see this if you recall that $a^b\times a^c=a^{b+c}$. This makes your problem boil down to: $$2e^{\pi i/4}\times 3e^{\pi i/6}=6(e^{\pi i/4}\times e^{\pi i/6})=6(e^{3\pi i/12}\times e^{2\pi i/12})=6 e^{3\pi i/12+2\pi i/12}=6e^{5\pi i/12}.$$