# solve the equation

• Aug 18th 2009, 03:44 AM
Random Variable
solve the equation
Solve for z.

$\displaystyle i^{z} = z$ where $\displaystyle z \in \mathbb{C}$
• Aug 19th 2009, 05:05 PM
Random 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 PM
luobo
Quote:

Originally Posted by Random 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.

The complex power of a complex number is multi-valued. So there could be an infinite number of solutions. Today is too late, we can discuss this sometime later. ;-)
• Aug 20th 2009, 08:33 PM
Random Variable
Quote:

The complex power of a complex number is multi-valued. So there could be an infinite number of solutions.
True. It depends which branch you pick.
• Aug 20th 2009, 08:39 PM
CaptainBlack
Quote:

Originally Posted by Random Variable
Solve for z.

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

Has closed form solution in terms of Lambert's W

CB
• Aug 21st 2009, 03:13 AM
Random Variable
Quote:

Originally Posted by CaptainBlack
Has closed form solution in terms of Lambert's W

CB

I know.

Spoiler:

$\displaystyle i^{z} = z$

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

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

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

$\displaystyle -ze^{-z \ln i} = -1$

$\displaystyle -z \ln i \ e^{-z \ln i} = - \ln i$

$\displaystyle W(-z \ln i e^{-z \ln i}) = W (- \ln i)$

$\displaystyle -z \ln i = W(-\ln i)$

$\displaystyle z = \frac{W(-\ln i)}{-\ln i}$

• Aug 21st 2009, 08:02 AM
CaptainBlack
Quote:

Originally Posted by Random Variable
I know.

Spoiler:

$\displaystyle i^{z} = z$

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

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

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

$\displaystyle -ze^{-z \ln i} = -1$

$\displaystyle -z \ln i \ e^{-z \ln i} = - \ln i$

$\displaystyle W(-z \ln i e^{-z \ln i}) = W (- \ln i)$

$\displaystyle -z \ln i = W(-\ln i)$

$\displaystyle z = \frac{W(-\ln i)}{-\ln i}$

Horrible, try putting $\displaystyle i=e^{\pi i/2}$

(the term horrible here is not intended to be insulting, but is a term often used to describe an aesthetically unpleasing piece of mathematics)

CB
• Aug 21st 2009, 08:38 AM
Random 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 PM
luobo
Quote:

Originally Posted by Random 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?

This looks good to me!
• Aug 21st 2009, 07:27 PM
simplependulum
But what is $\displaystyle W(z)$ ?
• Aug 21st 2009, 08:34 PM
Random Variable
Quote:

Originally Posted by simplependulum
But what is $\displaystyle W(z)$ ?

Lambert W-Function

It's the function W(z) that solves $\displaystyle z = W(z)e^{W(z)}$ for a given value of z.
• Aug 21st 2009, 11:24 PM
CaptainBlack
Quote:

Originally Posted by Random 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?

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 PM
luobo
Quote:

Originally Posted by CaptainBlack
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

The multi-valued-ness needs to be discussed and dealt with, because that is the key here.
• Aug 22nd 2009, 11:45 PM
CaptainBlack
Quote:

Originally Posted by Random Variable
Solve for z.

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

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