Solve for z.

$\displaystyle i^{z} = z$ where $\displaystyle z \in \mathbb{C} $

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- Aug 18th 2009, 03:44 AMRandom Variablesolve the equation
Solve for z.

$\displaystyle i^{z} = z$ where $\displaystyle z \in \mathbb{C} $ - Aug 19th 2009, 05:05 PMRandom Variable
I'm not sure if a hint is needed, or if this problem is not challenging at all. I'm leaning towards the latter.

- Aug 20th 2009, 08:09 PMluobo
- Aug 20th 2009, 08:33 PMRandom VariableQuote:

The complex power of a complex number is multi-valued. So there could be an infinite number of solutions.

- Aug 20th 2009, 08:39 PMCaptainBlack
- Aug 21st 2009, 03:13 AMRandom Variable
- Aug 21st 2009, 08:02 AMCaptainBlack
- Aug 21st 2009, 08:38 AMRandom Variable
OK. Do this instead. Let a be any constant (real or complex).

$\displaystyle a^{z} = z $

$\displaystyle a^{-z} = \frac{1}{z} $

$\displaystyle z a^{-z} = 1 $

$\displaystyle -z a^{-z} = -1 $

$\displaystyle -ze^{-z \ \text{Log a}} = -1 $

$\displaystyle -z \ (\text{Log a}) \ e^{-z \ \text{Log a}}= - \text{Log a} $

$\displaystyle W(-z \ (\text{Log a}) \ e^{-z \ \text{Log a}}) = W (- \text{Log a}) $

$\displaystyle -z \text{Log a} = W(-\text{Log a}) $

$\displaystyle z = \frac{W(-\text{Log a})}{-\text{Log a}} $

$\displaystyle \text{Log i} = \ln|i| + i \frac{\pi}{2} = i\frac{\pi}{2}$

$\displaystyle \frac {W(-i \frac{\pi}{2})}{-i\frac{\pi}{2}} \approx 0.4382829366+i0.3605924718$

Is that satisfactory? - Aug 21st 2009, 06:02 PMluobo
- Aug 21st 2009, 07:27 PMsimplependulum
But what is $\displaystyle W(z) $ ?

- Aug 21st 2009, 08:34 PMRandom Variable
- Aug 21st 2009, 11:24 PMCaptainBlack
It is better in that you eliminate the $\displaystyle \log$ 's in the final answer but I would still prefer a solution that did not have to worry about which branch of the $\displaystyle \log$ was in use where. That can be done by the the first step that I proposed (which, while that is not the way I think of it, absorbs the taking of logarithms while effectively fixing the branch).

Also as $\displaystyle W$ is multi-valued the right hand side of final line is just one of the solutions that correspond to the left hand side (as you know, and I know you know).

CB - Aug 22nd 2009, 06:00 PMluobo
- Aug 22nd 2009, 11:45 PMCaptainBlack
I like this method:

As $\displaystyle i=e^{(2n+1/2)\pi i}=e^{-\mu_n}$ where $\displaystyle \mu_n=-(2n+1/2)\pi i$ for $\displaystyle n\in\mathbb{Z}$ we have:

$\displaystyle e^{-\mu_n z}=z$

or:

$\displaystyle \mu_n z\; e^{\mu_n z}=\mu_n$

so:

$\displaystyle \mu_n z=W(\mu_n)$.

Hence:

$\displaystyle z=\frac{1}{\mu_n}W(\mu_n); \ \ \mu_n= -(2n+1/2)\pi i,\ n \in \mathbb{Z}$

(notes:

1. this is at least the third refinement of the original idea scribbled in my note book

2. the multi-valuedness of the solution is not only dependent on $\displaystyle n$, but the Lambert-W is itself multi-valued)

CB