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Math Help - A complex rational equation I'm having trouble with

  1. #1
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    Talking A complex rational equation I'm having trouble with

    I've posted before about my fractal project, and this is something else that is related. The function:
    f(z)=\frac{1-z}{(2-z)^{\frac{3}{2}}} where |z|=1.
    The derivative along this line:
    f'(z)=-\frac{ie^{it}(e^{it}+1)\sqrt{2-e^{it}}}{2(2-e^{it})^{3}} for a new real variable t.
    I'm trying to find when the real and/or imaginary parts equal zero.
    \Re\{f'\}=0
    \Im\{f'\}=0
    Is there any other way to solve these equations without splitting f' into x' and iy' just because it is so messy and long? I know one method of removing the imaginary terms from the bottom by multiplying top and bottom by (\frac{1}{2}-e^{it})^3, factoring e^{3it} out of the denominator, and multiplying numerator and denominator by \frac{1}{e^{3it}} resulting in a denominator of 4\cos t-5. It's still messy with a numerator of -\frac{i}{e^{2it}}(e^{it}+1)(\frac{1}{2}-e^{it})^3\sqrt{2-e^{it}} but now, algebraically, the worst part is \sqrt{2-e^{it}}. I'd appreciate any suggestions.
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  2. #2
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    Quote Originally Posted by saywhat View Post
    I've posted before about my fractal project, and this is something else that is related. The function:
    f(z)=\frac{1-z}{(2-z)^{\frac{3}{2}}} where |z|=1.
    The derivative along this line:
    f'(z)=-\frac{ie^{it}(e^{it}+1)\sqrt{2-e^{it}}}{2(2-e^{it})^{3}} for a new real variable t.
    I'm trying to find when the real and/or imaginary parts equal zero.
    \Re\{f'\}=0
    \Im\{f'\}=0
    Is there any other way to solve these equations without splitting f' into x' and iy' just because it is so messy and long? I know one method of removing the imaginary terms from the bottom by multiplying top and bottom by (\frac{1}{2}-e^{it})^3, factoring e^{3it} out of the denominator, and multiplying numerator and denominator by \frac{1}{e^{3it}} resulting in a denominator of 4\cos t-5. It's still messy with a numerator of -\frac{i}{e^{2it}}(e^{it}+1)(\frac{1}{2}-e^{it})^3\sqrt{2-e^{it}} but now, algebraically, the worst part is \sqrt{2-e^{it}}. I'd appreciate any suggestions.
    The only thing I can think of is this:
    \sqrt{2 - e^{it}} = \sqrt{2 - cos(t) - i~sin(t)} = \sqrt{2}\sqrt{1 - \frac{1}{2}cos(t) - \frac{i}{2}sin(t)}  = \sqrt{2} \left ( 1 - \frac{1}{2}cos(t) - \frac{i}{2}sin(t) \right )^{1/2}

    And then apply
    (cos(\theta) + i~sin(\theta))^n = cos(n\theta) + i~sin(n \theta)

    But it's still going to be pretty messy.

    -Dan
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  3. #3
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    Quote Originally Posted by saywhat View Post
    I've posted before about my fractal project, and this is something else that is related. The function:
    f(z)=\frac{1-z}{(2-z)^{\frac{3}{2}}} where |z|=1.
    The derivative along this line:
    f'(z)=-\frac{ie^{it}(e^{it}+1)\sqrt{2-e^{it}}}{2(2-e^{it})^{3}} for a new real variable t.
    I'm trying to find when the real and/or imaginary parts equal zero.
    \Re\{f'\}=0
    \Im\{f'\}=0
    Is there any other way to solve these equations without splitting f' into x' and iy' just because it is so messy and long? I know one method of removing the imaginary terms from the bottom by multiplying top and bottom by (\frac{1}{2}-e^{it})^3, factoring e^{3it} out of the denominator, and multiplying numerator and denominator by \frac{1}{e^{3it}} resulting in a denominator of 4\cos t-5. It's still messy with a numerator of -\frac{i}{e^{2it}}(e^{it}+1)(\frac{1}{2}-e^{it})^3\sqrt{2-e^{it}} but now, algebraically, the worst part is \sqrt{2-e^{it}}. I'd appreciate any suggestions.
    As when e^{it} goes to 2, \left| f'(z)|_{z=t} \right| goes to \infty, we know that this
    des not correspond to a root, so we can can multiply through by
    2(2-e^{it})^{2.5} to get for a root t:

    <br />
ie^{it}(e^{it}+1)=0<br />

    so we need: e^{it}=-1, or t=\pi (single value as t is parameterizing the unit circle).

    RonL
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  4. #4
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    To Dan, thanks. I have gotten thus far before in other equations and have used a formula to separate into x and y, but it is so large and not worth it.
    To RonL, I should have specified, I'm looking for the intercept of the real axis and the intercept of the imaginary axis. That point is the cusp of the curve f(z). Thanks anyway.

    I stumbled upon the identity
    \Im \{z\}=\frac{z-\bar z}{2}
    and because f'(t) is periodic
    \overline{f'(t)}=-f'(-t)
    leading to \Im \{f'(t)\}=\frac{f'(t)+f'(-t)}{2}=0 and thus I need to solve f'(t)=-f'(-t), which hopefully won't call for me to find x and y.
    The other equation \Re \{f'\}=0 can also be done in the same manor. I'll tell you how this goes and thanks again.

    Oh and a note, I substituted z=e^{it}. I should have said that in the original post.
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