Simplifying $6e^{i\pi/6}\times6e^{i\pi/4}$ to $a+bi$ form I am asked to express
$$6e^{i\pi/6}\times6e^{i\pi/4}$$
in the form $a + bi$.
If I use Euler's theorem, then $e^{iz}=\cos(z)+i \sin(z)$
$$6e^{i\pi/6}\times6e^{i\pi/4}=\left[\,6\cos\left(\frac{\pi}{6}\right)+i6\sin\left(\frac{\pi}{6}\right)\,\right]
\times \left[6\cos\left(\frac{\pi}{4}\right)+i6\sin\left(\frac{\pi}{4}\right)\right] \tag1$$
when I simplify I get
$$\frac{18\sqrt3}{\sqrt2}+\frac{18i\sqrt3}{2}+\frac{18i}{\sqrt2}-\frac{18}{\sqrt2} \tag2$$
which simplifies to
$$\frac{18\sqrt3-18}{\sqrt2}+i\left(\frac{18\sqrt3}{2}+\frac{18}{\sqrt2}\right) \tag3$$
Is this correct, or did I do something wrong?

Always make sure to consult a special trig ratios chart; there shouldn't be an irrational denominator anywhere in there. Your answer might be technically correct, but it's sloppy and not fully simplified.
$$\sin\left(\frac{\pi}{6}\right)=\frac{1}{2} \\
\cos\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\\
\sin\left(\frac{\pi}{4}
\right)=\cos\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}$$
We can also clean this up by taking care of the real multiplication first. $$6e^{i\pi/6}\cdot6e^{i\pi/4}=36(e^{i\pi/6}\cdot e^{i\pi/4})$$ Then we have $$36(e^{i\pi/6}\cdot e^{i\pi/4})=36\left(\frac{\sqrt3+i}{2}\right)\left(\frac{\sqrt2+\sqrt2i}{2}\right)=36\left(\frac{\sqrt6-\sqrt2}{4}+\frac{\sqrt6+\sqrt2}{4}i\right)=9(\sqrt6-\sqrt2)+9(\sqrt6+\sqrt2)i\text{.}$$
Normally, I would tell you to use your laws of exponents first to make it even easier, but $\dfrac{5\pi}{12}$ isn't an angle with a special trig ratio, so you're actually in the right for not simplifying that far right away.
