How to simplify ${(1+2i)}^6$? How to simplify ${(1+2i)}^6$ using De Moivre's formula?
I have found that $r=\sqrt 5$ and $\tan x=2$ but I can't find the exact value of $x$.
 A: Hint You have that $\tan(\theta) = 2$ and you need $\cos(6\theta)$ and $\sin(6\theta).$ Drawing a triangle in the first quadrant you can see that if $\tan(\theta) = 2$ you have $\cos(\theta) = 1/\sqrt{5}$ and $\sin(\theta) = 2/\sqrt{5}.$ Now can you find $\sin(6\theta)$ and $\cos(6\theta)$?
As others have said, it might be easier to use a hybrid approach where you square or cube first and then do the rest with trig.
A: Using the binomial formula is better, but is not what OP asked for. Using De Moivre's formula directly would be difficult, although doable if one has knowledge of Chebyshev's polynomials. I pointed this out in my answer, and was downvoted for it. So I have overwritten this answer.
A: $$|1+2i|=\sqrt5\;,\;\;\arg(1+2i)=\arctan 2\implies (1+2i)^6=5^3e^{6\arctan 2\cdot i} $$
and now:
$$\begin{cases}&5^3=125\\{}\\
&e^{6\arctan2\cdot i}=0.936+0.352\,i\end{cases}\implies(1+2i)^6=125(0.936+0.352\,i)=117+44i$$
But this looks weird and, anyway, is way simpler first calculating the third power and then squaring...
A: With $\tan(\theta)=2$ and the angle addition formula for tangent:
$$
\begin{align}
\tan(6\theta)
&=\tan(3\theta+3\theta)\\
&=\frac{2\tan(3\theta)}{1-\tan^2(3\theta)}
=\frac{2\tan(\theta+2\theta)}{1-\tan^2(\theta+2\theta)}\\
&=\frac{2\frac{2+\tan(2\theta)}{1-2\tan(2\theta)}}{1-\left(\frac{2+\tan(2\theta)}{1-2\tan(2\theta)}\right)^2}\\
&=\frac{2(2+\tan(2\theta))(1-2\tan(2\theta))}{(1-2\tan(2\theta))^2-(2+\tan(2\theta))^2}\\
&=\frac{2(2+\frac{2+2}{1-2\cdot2})(1-2\frac{2+2}{1-2\cdot2})}{(1-2\frac{2+2}{1-2\cdot2})^2-(2+\frac{2+2}{1-2\cdot2})^2}\\
&=\frac{2(2-\frac{4}{3})(1+2\frac{4}{3})}{(1+2\frac{4}{3})^2-(2-\frac{4}{3})^2}\\
&=\frac{2(2/3)(11/3)}{(11/3)^2-(2/3)^2}=\frac{44}{117}\\
\end{align}$$
So that is the tangent of the angular argument you are looking for. Now you know your result has integer coefficients, so possibilities are $\pm(117+44i)$, $\pm(234+88i)$, etc. 
You know the norm squared of your result is $5^6=15625$, and that matches the norm squared of $\pm(117+44i)$. So which of these two possibilities is it? The original angle $\theta$ is clearly between $60$ degrees and $90$ degrees, so $6\theta$ is between $360$ degrees and $540$. That means it is an upper plane angle. So the value must be $117+44i$
