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Math Help - Finding complex roots?

  1. #1
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    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.
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  2. #2
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    \begin{gathered}<br />
  z = 1 + 2i \hfill \\<br />
  \left| z \right| = \sqrt 5  \hfill \\<br />
  \phi  = \arg \left( z \right) = \arctan (2) \hfill \\<br />
  \sqrt[6]{5} e^{i(\phi  + 2k\pi )/3} \,,\,k =  - 1,0,1 \hfill \\ <br />
\end{gathered}
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  3. #3
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    Quote Originally Posted by Plato View Post
    \begin{gathered}<br />
  z = 1 + 2i \hfill \\<br />
  \left| z \right| = \sqrt 5  \hfill \\<br />
  \phi  = \arg \left( z \right) = \arctan (2) \hfill \\<br />
  \sqrt[6]{5} e^{i(\phi  + 2k\pi )/3} \,,\,k =  - 1,0,1 \hfill \\ <br />
\end{gathered}
    Well that doesnt seem right, as z=3*sqrt(1 + 2i) and not z=1 + 2i ? Am I wrong in thinking that?
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  4. #4
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    Quote Originally Posted by LooNiE View Post
    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}<br />
  \sqrt[6]{5}\left( {\cos \left( {\frac{\phi }<br />
{3}} \right) + i\sin \left( {\frac{\phi }<br />
{3}} \right)} \right) \hfill \\<br />
  \sqrt[6]{5}\left( {\cos \left( {\frac{\phi }<br />
{3} - \frac{{2\pi }}<br />
{3}} \right) + i\sin \left( {\frac{\phi }<br />
{3} - \frac{{2\pi }}<br />
{3}} \right)} \right) \hfill \\ <br />
\end{gathered}
    <br />
\sqrt[6]{5}\left( {\cos \left( {\frac{\phi }<br />
{3} + \frac{{2\pi }}<br />
{3}} \right) + i\sin \left( {\frac{\phi }<br />
{3} + \frac{{2\pi }}<br />
{3}} \right)} \right).
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