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Thread: Complex Mapping

  1. #1
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    Complex Mapping

    Find the image of the following region




    under the action of $\displaystyle f(z)=z^5$.

    I am horrible with mappings. I wrote $\displaystyle z^5=r^5 \cdot e^{5i\theta_0}$, and tried points in both regions, and then I ended up with the whole complex plane as the output and I don't think this is right. How do I find the image of this region? Thanks for the help.
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  2. #2
    MHF Contributor

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    The line y= 3x corresponds to the set of complex number a+ 3ai= a(1+ 3i) for a any real number (do you see why?). $\displaystyle z^5= (a+ 3ai)^5= a^5(1+ 3i)^5$. Now you could, if you wanted, multiply that out directly or use the "polar form". It will, in any case, be a constant, say u+ vi, so $\displaystyle z^5= a^5(u+ vi)$ which gives "parametric" equations [/tex]y= va^5[/tex] and $\displaystyle x= ua^5$. $\displaystyle a^5= \frac{x}{u}$ (as long as u is not 0) so $\displaystyle y= v\frac{x}{u}= \frac{v}{u} x$ another straight line. Similarly for y= x/2 which corresponds to the set of complex numbers $\displaystyle a+ \frac{1}{2}ai$.

    That is, $\displaystyle z^5$ maps that region, bounded by straight lines, into another region bounded by straight lines. You will need to compute $\displaystyle (1+ 3i)^5$ and $\displaystyle (1+ \frac{1}{2})^5$ to determine exactly which straight lines.
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