1. ## Finding complex roots?

The equation I have is:
z^3 = 1+2i
I need to find the complex roots and give the answer in Cartesian form. From looking at the function, I believe that there should be three roots to the equation. If you can just start it off I think I might be able to do it.

2. $\begin{gathered}
z = 1 + 2i \hfill \\
\left| z \right| = \sqrt 5 \hfill \\
\phi = \arg \left( z \right) = \arctan (2) \hfill \\
\sqrt[6]{5} e^{i(\phi + 2k\pi )/3} \,,\,k = - 1,0,1 \hfill \\
\end{gathered}$

3. Originally Posted by Plato
$\begin{gathered}
z = 1 + 2i \hfill \\
\left| z \right| = \sqrt 5 \hfill \\
\phi = \arg \left( z \right) = \arctan (2) \hfill \\
\sqrt[6]{5} e^{i(\phi + 2k\pi )/3} \,,\,k = - 1,0,1 \hfill \\
\end{gathered}$
Well that doesnt seem right, as z=3*sqrt(1 + 2i) and not z=1 + 2i ? Am I wrong in thinking that?

4. Originally Posted by LooNiE
Well that doesnt seem right, as z=3*sqrt(1 + 2i) and not z=1 + 2i ? Am I wrong in thinking that?
Written in polar form $1 + 2i = \sqrt 5 \left( {\cos (\phi ) + i\sin (\phi )} \right)\;,\;\phi = \arctan (2)$.
Recall that $\left( {\sqrt 5 } \right)^{\frac{1}{3}} = \sqrt[6]{5}$.

Now you are correct. There are three cube roots:
$\begin{gathered}
\sqrt[6]{5}\left( {\cos \left( {\frac{\phi }
{3}} \right) + i\sin \left( {\frac{\phi }
{3}} \right)} \right) \hfill \\
\sqrt[6]{5}\left( {\cos \left( {\frac{\phi }
{3} - \frac{{2\pi }}
{3}} \right) + i\sin \left( {\frac{\phi }
{3} - \frac{{2\pi }}
{3}} \right)} \right) \hfill \\
\end{gathered}$

$
\sqrt[6]{5}\left( {\cos \left( {\frac{\phi }
{3} + \frac{{2\pi }}
{3}} \right) + i\sin \left( {\frac{\phi }
{3} + \frac{{2\pi }}
{3}} \right)} \right)$
.