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?

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:
$RHS=\frac3{\sqrt{10}}-\frac4{10\sqrt{10}}=\frac{26}{10\sqrt{10}}=LHS$
