# solve the equation

• Aug 18th 2009, 04:44 AM
Random Variable
solve the equation
Solve for z.

$i^{z} = z$ where $z \in \mathbb{C}$
• Aug 19th 2009, 06:05 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, 09:09 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, 09:33 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, 09:39 PM
CaptainBlack
Quote:

Originally Posted by Random Variable
Solve for z.

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

Has closed form solution in terms of Lambert's W

CB
• Aug 21st 2009, 04:13 AM
Random Variable
Quote:

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

CB

I know.

Spoiler:

$i^{z} = z$

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

$z i^{-z} = 1$

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

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

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

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

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

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

• Aug 21st 2009, 09:02 AM
CaptainBlack
Quote:

Originally Posted by Random Variable
I know.

Spoiler:

$i^{z} = z$

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

$z i^{-z} = 1$

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

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

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

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

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

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

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

(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, 09:38 AM
Random Variable
OK. Do this instead. Let a be any constant (real or complex).

$a^{z} = z$

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

$z a^{-z} = 1$

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

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

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

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

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

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

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

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

Is that satisfactory?
• Aug 21st 2009, 07:02 PM
luobo
Quote:

Originally Posted by Random Variable
OK. Do this instead. Let a be any constant (real or complex).

$a^{z} = z$

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

$z a^{-z} = 1$

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

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

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

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

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

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

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

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

Is that satisfactory?

This looks good to me!
• Aug 21st 2009, 08:27 PM
simplependulum
But what is $W(z)$ ?
• Aug 21st 2009, 09:34 PM
Random Variable
Quote:

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

Lambert W-Function

It's the function W(z) that solves $z = W(z)e^{W(z)}$ for a given value of z.
• Aug 22nd 2009, 12:24 AM
CaptainBlack
Quote:

Originally Posted by Random Variable
OK. Do this instead. Let a be any constant (real or complex).

$a^{z} = z$

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

$z a^{-z} = 1$

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

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

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

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

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

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

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

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

Is that satisfactory?

It is better in that you eliminate the $\log$ 's in the final answer but I would still prefer a solution that did not have to worry about which branch of the $\log$ 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 $W$ 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, 07:00 PM
luobo
Quote:

Originally Posted by CaptainBlack
It is better in that you eliminate the $\log$ 's in the final answer but I would still prefer a solution that did not have to worry about which branch of the $\log$ 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 $W$ 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 23rd 2009, 12:45 AM
CaptainBlack
Quote:

Originally Posted by Random Variable
Solve for z.

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

I like this method:

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

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

or:

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

so:

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

Hence:

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

(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 $n$, but the Lambert-W is itself multi-valued)

CB