Complex number proof involving angles I need to show that (3 + i)^3 = 18 + 26i and use this to show that the angle AOC = 3AOB, where O, A, B, C are points in the plane given by O = (0, 0), A = (1, 0), B = (3, 1) and C = (18, 26).
This is what I have done:
Expand (3 + i)^3
(3 + i)^3 = (3 + i)(3 + i)(3 + i)
      = 9 + 3i + 3i + i^2 * (3 + i)

      = 9 + 6i - 1 * (3 + i)

      = (8 + 6i) * (3 + i)

      = 24 + 8i + 18i + 6i^2

      = 24 + 26i - 6

      = 18 + 26i

Find the magnitude of OB = 3 + i and OC = 18 + 26i to determine ->OB and ->OC
|OB| = sqrt(3^2 + 1^2) = sqrt(10)
|OC| = sqrt(18^2 + 26^2) = 10 * sqrt(10)
Therefore,
->OB = 3 + i = sqrt(10) * cis(AOB)
->OC = 18 + 26i = 10 * sqrt(10) * cis(AOC)
Cube ->OB and show to show AOC = 3AOB
->OB^3 = (3 + i)^3 = (sqrt(10))^3 * cis^3(AOB)
   = 18 + 26i = 10 * sqrt(10) * cis^3(AOB)

We found that (3 + i)^3 = 18 + 26i and by cubing ->OB we find the magnitude is the same as ->OC. Since this is the case, this shows that the angle AOC = 3AOB.
Would this be the correct way to solve this question?
 A: It is actually easier to use polar co-ordinates on this one.
Let's say that $B$ had the polar co-ordinates $(r_1, \theta_1)$ and $C$ had $(r_2, \theta_2)$. Then by the multiplication all you have to show is this:
$r_2=r_1^3 \implies r_2^2=r_1^6 \ldots (1)$
$\theta_2=3\theta_1 \ldots (2)$
Calculating $r_2$ would be the distance $OC$, which can be easily done with the Pythagorean theorem and the coordinates of $C$
$r_2^2=18^2+26^2=10^3=(3^2+1^2)^3=(r_1^2)^3=r_1^6$
For the next one we'll have to use this:
$\tan 3x=\dfrac{3\tan x -\tan^3 x}{1-3\tan^2 x}$
So, what's the tangent of $\theta_2$
$\tan \theta_2=\dfrac{26}{18}$
And by the way $\tan \theta_1=\dfrac{1}{3}$
$=\dfrac{26}{27 \cdot \dfrac{2}{3}}=\dfrac{\dfrac{26}{27}}{\dfrac{2}{3}}$
$=\dfrac{1-\dfrac{1}{27}}{1-\dfrac{1}{3}}=\dfrac{3 \cdot \dfrac{1}{3}-(\dfrac{1}{3})^3}{1-3 \cdot (\dfrac{1}{9})^2}$
$=\dfrac{3\tan \theta_1-\tan^3 \theta_1}{1-3\tan^2 \theta_1}=\tan 3\theta_1$
Therefore $3\theta_1=\theta_2$
To be honest I would've never done the identity part practically. That was just for fun. The point was that $\dfrac{\tan^{-1} \dfrac{26}{18}}{\tan^{-1} \dfrac{1}{3}}=3$
Anyway, peace!
Edit: If you show that $(1)$ is true then the one for angles falls in perfectly going by how polar co-ordinates in the complex plane are used in multiplication. But a little algebra doesn't hurt.
A: Use $\sin 3\theta = 3\sin\theta - 4\sin^3 \theta$ (Mathworld [16])
$\sin\angle AOB=\frac1{\sqrt{10}}$ and $\sin\angle AOC=\frac{26}{10\sqrt{10}}$
so we have:
$RHS=\frac3{\sqrt{10}}-\frac4{10\sqrt{10}}=\frac{26}{10\sqrt{10}}=LHS$
A: The method works. You might be able to do it a little easier.
To cube $3+i$ you can use the binomial formula:
$$ (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3,$$
so
\begin{align}
(3+i)^3 &= 3^3 + 3\cdot 3^2 i + 3\cdot 3 i^2 + i^3 \\
 &= 27 + 27 i + 9 (-1) + (-i) \\
 &= 18 + 26 i.
\end{align}
You found that 
$\DeclareMathOperator\cis{cis}|OB|^3 \cis(3\angle AOB) = |OC| \cis(\angle AOC)$,
and if you have studied the $r\cis\theta$ form sufficiently already,
this is enough to prove that $\cis(3\angle AOB) = \cis(\angle AOC)$.
The tricky part is to show that this really means $3\angle AOB = \angle AOC$
and not something like $3\angle AOB = \angle AOC+2\pi$ or
$3\angle AOB = \angle AOC - 2\pi.$
(Consider the fact that $(-3+2i)^3 = 9 + 46 i$; plot $D = (-3,2)$ and $E = (9,46)$
and consider what this says about the ratio of $\angle AOD$ to $\angle AOE$.)
