Could somebody help me please? I've got part one solved.

1) Solve the equation: $z^3=i$

I can do this bit: $$ z = \exp \left( \frac{i\pi}{6} + \frac{2k\pi}{3} \right) $$ so $$ z = \exp \left( \frac{i\pi}{6} \right) \qquad \text{or} \qquad z = \exp \left( \frac{5i\pi}{6} \right) \qquad \text{or} \qquad z = \exp \left( -\frac{i\pi}{2} \right) $$

2) Hence find the values for the argument of a complex number $w$ which is such that $$ w^3 = i \cdot \overline{w}^3, $$ where $\overline{w}$ is the complex conjugate.


  • 1
    $\begingroup$ $w/w^*$ is a solution of the first equation. $\endgroup$ – GEdgar Feb 27 '14 at 19:08

Hint on 1):

From $z^3=i$ it follows that $|z|^3=|i|=1$ hence $|z|=1$. If $z=e^{i\phi}$ then $z^3=i$ is equivalent to $e^{i3\phi}=e^{i\frac{1}{2}\pi}$.

So to be solved is: $3\phi=\frac{1}{2}\pi+2k\pi$.

Hint on 2):

If $w=re^{i\phi}$ then $w^3=r^3e^{i3\phi}$, its complex conjugate is $re^{-i\phi}$ and $i(re^{-i\phi})^3=e^{\frac{1}{2}\pi i}r^3e^{-3i\phi}=r^3e^{i(\frac{1}{2}\pi-3\phi)}$.

So to be solved is: $3\phi=\frac{1}{2}\pi-3\phi+2k\pi$.

Here $k\in\mathbb Z$ and in both cases you can restrict to solutions in $[0,2\pi)$.


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