Solve for z.

where

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- August 18th 2009, 04:44 AMRandom Variablesolve the equation
Solve for z.

where - August 19th 2009, 06:05 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.

- August 20th 2009, 09:09 PMluobo
- August 20th 2009, 09:33 PMRandom VariableQuote:

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

- August 20th 2009, 09:39 PMCaptainBlack
- August 21st 2009, 04:13 AMRandom Variable
- August 21st 2009, 09:02 AMCaptainBlack
- August 21st 2009, 09:38 AMRandom Variable
OK. Do this instead. Let a be any constant (real or complex).

Is that satisfactory? - August 21st 2009, 07:02 PMluobo
- August 21st 2009, 08:27 PMsimplependulum
But what is ?

- August 21st 2009, 09:34 PMRandom Variable
- August 22nd 2009, 12:24 AMCaptainBlack
It is better in that you eliminate the 's in the final answer but I would still prefer a solution that did not have to worry about which branch of the 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 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 - August 22nd 2009, 07:00 PMluobo
- August 23rd 2009, 12:45 AMCaptainBlack
I like this method:

As where for we have:

or:

so:

.

Hence:

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

CB