solutions to relative simple expression

I could need a hint on how to find all solutions to this equation

$\displaystyle \left( \frac{1+z^{-1}}{1+z} \right)^N = C$

where C is a real constant, N is a positive integer, and and z is the complex variable. I'm not sure I can just write

$\displaystyle \frac{1+z^{-1}}{1+z} \right = C^{1/N}$. It should probably look more like $\displaystyle \frac{1+z^{-1}}{1+z} \right = C^{1/N}e^{2\pi i k/N}$, for $\displaystyle k=0,1,...N-1$.

But then I'm unsure as how to proceed. I have tried writing z in polar form but it didn't give me a break through.

Re: solutions to relative simple expression

Quote:

Originally Posted by

**niaren** I could need a hint on how to find all solutions to this equation

$\displaystyle \left( \frac{1+z^{-1}}{1+z} \right)^N = C$

where C is a real constant, N is a positive integer, and and z is the complex variable. I'm not sure I can just write

$\displaystyle \frac{1+z^{-1}}{1+z} \right = C^{1/N}$. It should probably look more like $\displaystyle \frac{1+z^{-1}}{1+z} \right = C^{1/N}e^{2\pi i k/N}$, for $\displaystyle k=0,1,...N-1$.

But then I'm unsure as how to proceed. I have tried writing z in polar form but it didn't give me a break through.

$\displaystyle \displaystyle \begin{align*} \left(\frac{1 + z^{-1}}{1 + z}\right)^N &= C \\ \left(\frac{1 + \frac{1}{z}}{1 + z}\right)^N &=C \\ \left(\frac{\frac{z + 1}{z}}{1 + z}\right)^N &= C \\ \left(\frac{1 + z}{z(1 + z)}\right)^N &= C \\ \left(\frac{1}{z}\right)^N &= C \\ z^{-N} &= C \\ z &= C^{-\frac{1}{N}} \end{align*}$

Re: solutions to relative simple expression

Dammit :) Don't know how I could miss that. Thanks.

Re: solutions to relative simple expression

Quote:

Originally Posted by

**niaren** Dammit :) Don't know how I could miss that. Thanks.

I expect there are other solutions too though...

Re: solutions to relative simple expression

Do you mean that there is another way to arrive at the N solutions or do you mean that there are more than N solutions?

Re: solutions to relative simple expression

Quote:

Originally Posted by

**niaren** $\displaystyle \frac{1+z^{-1}}{1+z} \right = C^{1/N}$.

Let p = 1/N

[1 + z^(-1)] / (1 + z) = C^p

(z + 1) / z = C^p + zC^p

z^2 C^p + z(C^p - 1) - 1 = 0

z = {1 + C^p +- SQRT[C^(2p) + 2C^p + 1]} / (2C^p)

I'm sure you can see your way to wrap that up, getting same as Prove It

Re: solutions to relative simple expression

Quote:

Originally Posted by

**Wilmer** Let p = 1/N

z = {1 + C^p +- SQRT[C^(2p) + 2C^p + 1]} / (2C^p)

I actually had that expression written down but gave up because it looked too complicated. I will look at it tomorrow.

Re: solutions to relative simple expression

Quote:

Originally Posted by

**niaren** I actually had that expression written down but gave up because it looked too complicated. I will look at it tomorrow.

Why?

C^(2p) + 2C^p + 1 = (C^p + 1)^2

Re: solutions to relative simple expression

$\displaystyle \displaystyle{ \left(\frac{1}{z}\right)^{N}=C\angle 2k\pi}$

$\displaystyle \frac{1}{Z}=C^{1/N}\angle (2k\pi/N),\qquad k=0,1,2,...(N-1).$

Re: solutions to relative simple expression

Do something similar with this one, Bob? :

[(1 + 1/a) / (1 + a)]^(1/m) + [(1 + b) / (1 + 1/b)]^n = k

Solve for b.

b = [(ka^(1/m) - 1) / a^(1/m)]^(1/n) ... unless I goofed!