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?
 A: 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.
A: By using the polar form transformation, $$z=a+ib=re^{i\theta}=r(cos\theta + isin\theta)$$
Here $\theta$ can be defined as $arg(z)=tan^{\prime}(\frac{b}{a})$
Let $z_{1}=6e^{i\pi/6}=r_{1}e^{i\theta_{1}}$
This implies, $r_{1}=6$, and $arg(z_{1})=\theta_{1}=\pi/6$
Likewise, let $z_{2}=6e^{i\pi/4}=r_{2}e^{i\theta_{2}}$
This implies, $r_{2}=6$, and $arg(z_{2})=\theta_{2}=\pi/4$
It's clear that, $$z_{1}z_{2}=r_{1}e^{i\theta_{1}}. r_{2}e^{i\theta_{2}}=r_{1}r_{2}[e^{iarg(z_{1}z_{2})}]$$
It's clear-cut that,  $arg(z_{1}z_{2})=arg(z_{1})+arg(z_{2})=\theta_{1}+\theta_{2}$
This implies that, $$z_{1}z_{2}=r_{1}e^{i\theta_{1}}. r_{2}e^{i\theta_{2}}=r_{1}r_{2}[e^{iarg(z_{1}z_{2})}]=r_{1}r_{2}[e^{i(\theta_{1}+\theta_{2})}]$$
By substitution,  $$z_{1}z_{2}=r_{1}r_{2}[e^{i(\theta_{1}+\theta_{2})}]=36[e^{i(5\pi/12)}]$$
As we known that, $z=a+ib=re^{i\theta}=r(cos\theta + isin\theta)$
So that, $$z=a+ib=re^{i\theta}=36[e^{i(5\pi/12)}]=r(cos\theta + isin\theta)=36(cos(5\pi/12)+isin(5\pi/12))$$$$=9(\sqrt6-\sqrt2)+9(\sqrt6+\sqrt2)i$$
