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Math Help - Roots and Complex Numbers

  1. #1
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    Roots and Complex Numbers

    Hi folks

    Got myself some dodgey questions to get through and Im bit clouded in what to do since the book did not help...

    Find the real and imaginary parts of (1 + i)^35

    Find all complex solutions to the equation (z + 1)^4 = 16z^4

    And

    Find the complex logarithm of [sqrt]3 + i, find a complex number z=a+bi (where a and b are real), such that e^z = [sqrt]3 +i

    Any help is appreciated
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  2. #2
    Senior Member Peritus's Avatar
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    Quote Originally Posted by NeloAngelo View Post
    Find the real and imaginary parts of (1 + i)^35
    <br />
1 + i = \left| {1 + i} \right|e^{i\arg (1 + i)}  = \sqrt 2 e^{i\frac{\pi }<br />
{4}} <br />

    therefore:

    <br />
(1 + i)^{35}  = 2^{\frac{{35}}<br />
{2}} e^{i\frac{{35\pi }}<br />
{4}}

    <br /> <br />
\Re e\left\{ {(1 + i)^{35} } \right\} = \Re e\left\{ {2^{\frac{{35}}<br />
{2}} e^{i\frac{{35\pi }}<br />
{4}} } \right\} = 2^{\frac{{35}}<br />
{2}} \cos \left( {\frac{{35\pi }}<br />
{4}} \right)<br /> <br />

    <br />
\Im m\left\{ {(1 + i)^{35} } \right\} = \Im m\left\{ {2^{\frac{{35}}<br />
{2}} e^{i\frac{{35\pi }}<br />
{4}} } \right\} = 2^{\frac{{35}}<br />
{2}} \sin \left( {\frac{{35\pi }}<br />
{4}} \right)<br />
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  3. #3
    Senior Member Peritus's Avatar
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    Quote Originally Posted by NeloAngelo View Post
    Hi folks
    Find all complex solutions to the equation (z + 1)^4 = 16z^4
    <br />
\begin{gathered}<br />
  \left( {z + 1} \right)^4  = 16z^4  \hfill \\<br />
   \Leftrightarrow \left( {z + 1} \right)^4  = \left( {2z} \right)^4  \hfill \\<br />
   \Leftrightarrow \left( {z + 1} \right)^4  - \left( {2z} \right)^4  = 0 \hfill \\ <br />
\end{gathered}
    <br /> <br />
\begin{gathered}<br />
   \Leftrightarrow \left[ {\left( {z + 1} \right)^2  - \left( {2z} \right)^2 } \right]\left[ {\left( {z + 1} \right)^2  + \left( {2z} \right)^2 } \right] = 0 \hfill \\<br />
   \Leftrightarrow \left( {3z + 1} \right)\left( {1 - z} \right)\left( {5z^2  + 2z + 1} \right) = 0 \hfill \\ <br />
\end{gathered} <br />

    can you continue?
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  4. #4
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    Hi Thx for the feedback
    I presume when the question states Complex Solutions they mean Complex Roots therefore after using the binomial expansion I use the formula involving arg, sin and cos. Is that correct?
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  5. #5
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    Quote Originally Posted by NeloAngelo View Post
    Hi Thx for the feedback
    I presume when the question states Complex Solutions they mean Complex Roots therefore after using the binomial expansion I use the formula involving arg, sin and cos. Is that correct?
    Where do you figure the binomial expansion gets used?? Clearly not in the two Peritus has answered. And not in the one I'm about to help with .....

    Quote Originally Posted by NeloAngelo View Post
    [snip]
    Find the complex logarithm of [sqrt]3 + i, find a complex number z=a+bi (where a and b are real), such that e^z = [sqrt]3 +i

    Any help is appreciated
    You should know that \ln w = \ln |w| + i \theta where \theta = \arg w ......

    Now note that \ln (\sqrt{3} + i) = z \Rightarrow \sqrt{3} + i = e^z .....
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  6. #6
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    Sorry my bad. So is it just the quadratic equations I got to use along with Natural Logs and de Moivre Equation?

    Thx in Advance
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  7. #7
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    Quote Originally Posted by NeloAngelo View Post
    Sorry my bad. So is it just the quadratic equations I got to use along with Natural Logs and de Moivre Equation?

    Thx in Advance
    Yes.
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