Find $a$, $b$ for $z = \frac{1}{(a + ib)^3}$ to be on negative real axis 
For what values of $a$ and $b$ does
  \begin{equation}
z = \frac{1}{(a + ib)^3}
\end{equation}
  lie on the negative real axis. Hence, or otherwise, find an expression in terms of $a$ only for $\lvert z \rvert$ when $\operatorname{arg}(z) = - \pi$.

How would I approach this question?

For $z$ to lie on the negative real axis, we have 
$$z = \frac{1}{(a + ib)^3}=-r=re^{i(1+2n)\pi}$$
where $r>0$. Rearrange to get 
$$a+bi= \frac1{\sqrt[3]r} e^{-i\frac{(1+2n)\pi}3}
=\frac1{\sqrt[3]r} \left( \cos\frac{1+2n}3\pi - i\sin\frac{1+2n}3\pi \right)
$$
with $n=0,1,2$. Then, 
$$a= \frac1{\sqrt[3]r} \cos\frac{1+2n}3\pi,\>\>\>\>\>
b= -\frac1{\sqrt[3]r} \sin\frac{1+2n}3\pi$$
Evalute at $n=0,1,2$ to obtain the values for $(a,b)$
$$(a.b)=( -c,0),\>(c,\pm\sqrt3c)$$
where $c$ represents any positive real number.
