# Proof involving complex numbers

I need a little help with this proof. I haven't got any idea how to proceed. The pseudo clue in the question doesn't clarify much.

The real numbers $r$ and $\theta$, where $r > 0$ and $-\pi < \theta < \pi$, are such that, $$rcos\theta + 2r^2cos2\theta + 3r^3cos3\theta = 0$$ $$rsin\theta + 2r^2sin2\theta + 3r^3sin3\theta = 0$$ By writing $z = r(cos\theta + isin\theta)$,

Show that $z = \frac{1}{3}(-1 \pm i\sqrt2)$.

Deduce the value of $r$ and the 2 possible values of $tan\theta$.

The only thing I see is the last part of the question. Deducing $r$ and $tan\theta$ from $z = \frac{1}{3}(-1 \pm i\sqrt2)$ would be simple. I calculated $r = \frac{1}{\sqrt3}$ and $tan\theta = \pm\sqrt2$.

A more elaborate hint ... start with $z=r(\cos\theta + i\sin\theta)$, then write out $z^2$ in terms of $r$ and $\theta$. Can you then see what to do next? – GEdgar Jun 8 '11 at 13:33
Multiply the second equation by $i$ and add the two together. You should see terms like $\cos n\theta + i\sin n\theta$ appear. Now if you think about what $z^2$ and $z^3$ are you should find a cubic in $z$ (which has three solutions-the problem is missing one).
Alternatively, start with $z=r(\cos\theta+i\sin\theta)$ and calculate $z^2$ and $z^3$, and rewrite the results in terms of $\cos2\theta$, etc., using standard trig identities, then see what your two displayed equations tell you about $z$. – Gerry Myerson Jun 8 '11 at 13:32
Got it, Thank you! Adding the 2 equations gave the cubic $(z + 2z^2 + 3z^3) = 0.$ And since $z \ne 0$, solving the quadratic $3z^2 + 2z + 1 = 0$ proves $z = \frac{1}{3}(-1 \pm i\sqrt2)$ as required. – mathguy80 Jun 8 '11 at 14:51