<s>[INST] 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?
 [/INST] 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 $\angle AOE$.)
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