3
$\begingroup$

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.

Thanks.

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

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 $[0,2\pi)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.