# Thread: Complex Analysis: Simplifying Polar Form

1. ## Complex Analysis: Simplifying Polar Form

I am having trouble simplifying the following equations which are the solutions to the equation $(z+1)^5=z^5$.
$\frac{1}{e^{i\frac{2}{5}\pi}-1}$

$\frac{1}{e^{i\frac{4}{5}\pi}-1}$

$\frac{1}{e^{i\frac{6}{5}\pi}-1}$

$\frac{1}{e^{i\frac{8}{5}\pi}-1}$

2. ## Re: Complex Analysis: Simplifying Polar Form

You could expand the fractions with $\exp\left[-i \frac{1}{5} \pi\right]$, $\exp\left[-i \frac{2}{5} \pi\right]$ and so on.

3. ## Re: Complex Analysis: Simplifying Polar Form

I am afraid I really don't understand how this would help.

4. ## Re: Complex Analysis: Simplifying Polar Form

After expanding, the demoninator will be a 2 i Sin[~], which is a multiple of Pi. The numerator is Exp[i~], which could you rewrite in a+ ib if you want to do.

Just try to expand the first solution with $\exp\left[-i \frac{1}{5} \pi\right]$ and you will see it works.

5. ## Re: Complex Analysis: Simplifying Polar Form

Sorry... I really appreciate your help, but I am just not seeing it. I still have both these items in the denominator and am back to 1 in the numerator.

6. ## Re: Complex Analysis: Simplifying Polar Form

I think it might be as simple as it gets for my purpose! Thank you.

7. ## Re: Complex Analysis: Simplifying Polar Form

Originally Posted by tarheelborn
I am having trouble simplifying the following equations which are the solutions to the equation $(z+1)^5=z^5$.
$\frac{1}{e^{i\frac{2}{5}\pi}-1}$
What is the original statement of the problem?
Where do you get those answers?

It might help to know that
$\frac{1}{\exp\left(i\frac{2}{5}\pi}\right)-1}=\frac{\exp\left(-i\frac{2}{5}\pi\right)-1}{|\exp\left(i\frac{2}{5}\pi}\right)-1|^2}$

8. ## Re: Complex Analysis: Simplifying Polar Form

The original problem statement was to find all the solutions of $(z+1)^5=z^5$. I simplified this equation to equal $(1+\frac{1}{z})^5=1$ and then wrote $\frac{1}{z}$ in polar form as $(\frac{1}{r}e^{-i\theta})^5$.

I then simplified to the equation $e^{i(2/5)n\pi}$ for $n=0,\cdots,4$.

Obviously, $n=0$ isn't a solution, so I came up with the other 4 using that format, but I wasn't able to simplify them. But then I wondered if they need to be simplified. Thanks for your help.

9. ## Re: Complex Analysis: Simplifying Polar Form

Okay, I will explain again

$\frac{1}{e^{i\frac{2}{5}\pi}-1} = \frac{1}{e^{i\frac{2}{5}\pi}-1} \cdot \frac{e^{-i\frac{1}{5}\pi}}{e^{-i\frac{1}{5}\pi}}= \frac{e^{-i\frac{1}{5}\pi}}{e^{i\frac{1}{5}\pi}- e^{-i\frac{1}{5}\pi}} = \frac{e^{-i\frac{1}{5}\pi}}{2 i \sin\left(\frac{1}{5}\pi\right)} = \frac{ e^{-i\frac{1}{10}\pi}}{2 \sin\left(\frac{1}{5}\pi\right)}$

10. ## Re: Complex Analysis: Simplifying Polar Form

Oh, sorry, yes I see that. Thank you.