# A complex rational equation I'm having trouble with

• Oct 18th 2007, 09:41 PM
saywhat
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:
$\displaystyle f(z)=\frac{1-z}{(2-z)^{\frac{3}{2}}}$ where $\displaystyle |z|=1$.
The derivative along this line:
$\displaystyle f'(z)=-\frac{ie^{it}(e^{it}+1)\sqrt{2-e^{it}}}{2(2-e^{it})^{3}}$ for a new real variable $\displaystyle t$.
I'm trying to find when the real and/or imaginary parts equal zero.
$\displaystyle \Re\{f'\}=0$
$\displaystyle \Im\{f'\}=0$
Is there any other way to solve these equations without splitting $\displaystyle f'$ into $\displaystyle x'$ and $\displaystyle iy'$ 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 $\displaystyle (\frac{1}{2}-e^{it})^3$, factoring $\displaystyle e^{3it}$ out of the denominator, and multiplying numerator and denominator by $\displaystyle \frac{1}{e^{3it}}$ resulting in a denominator of $\displaystyle 4\cos t-5$. It's still messy with a numerator of $\displaystyle -\frac{i}{e^{2it}}(e^{it}+1)(\frac{1}{2}-e^{it})^3\sqrt{2-e^{it}}$ but now, algebraically, the worst part is $\displaystyle \sqrt{2-e^{it}}$. I'd appreciate any suggestions.
• Oct 19th 2007, 04:46 AM
topsquark
Quote:

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

The only thing I can think of is this:
$\displaystyle \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)}$$\displaystyle = \sqrt{2} \left ( 1 - \frac{1}{2}cos(t) - \frac{i}{2}sin(t) \right )^{1/2}$

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

But it's still going to be pretty messy.

-Dan
• Oct 19th 2007, 08:37 AM
CaptainBlack
Quote:

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

As when $\displaystyle e^{it}$ goes to $\displaystyle 2$, $\displaystyle \left| f'(z)|_{z=t} \right|$ goes to $\displaystyle \infty$, we know that this
des not correspond to a root, so we can can multiply through by
$\displaystyle 2(2-e^{it})^{2.5}$ to get for a root $\displaystyle t$:

$\displaystyle ie^{it}(e^{it}+1)=0$

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

RonL
• Oct 19th 2007, 02:59 PM
saywhat
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 $\displaystyle f(z)$. Thanks anyway.

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

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