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Complex exponentiation 
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#1
May2309, 05:51 AM

P: 117

Mathworld seems to be the only place which explicitly defines complex exponentiation. I'll be even more explicit than Wolfram and write out the full formula (I'm guessing there's no shorter way for the general case):
(a+bi)^(c+di) = (a^{2}+b^{2})^(c/2) * e^(d*atan2(b,a)) * cos(c*atan2(b,a)+0.5*d*ln(a^{2}+b^{2}) ) + (a^{2}+b^{2})^(c/2) * e^(d*atan2(b,a)) * sin (c*atan2(b,a)+0.5*d*ln(a^{2}+b^{2}) ) i My question is this: Wikipedia says that the result for complex exponentiation is multivalued when the exponent is irrational  in other words, there's more than one possible solution. How would I change the above formula to produce the other solution/s? If anyone could give the other answers to say (2.2+3.3i)^(4.4+5.5i) to clarify the answer to the above, then that would be a great example too. Or are there an infinite number of possible answers? Finally, if complex exponentiation is supposed to give multiple answers when the exponent is irrational, then why do certain calculator applets on the internet that deal with complex arithmetic only give one answer? 


#2
May2309, 05:58 AM

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Um, normally one defines x^y as exp{y log x} which is multivalued as log is multivalued. This is a lot more compact than your answer, though I don't dispute it.
What does atan2(b,a) mean? That might well be multivalued as everything else there is single valued. I don't know why you want such a complicated example, and I will do i^i instead. i^i=exp{i log i} so we need to know what log(i) is. i=exp{i pi/2 +2npi} for any integer n, so log{i} = {i pi/2 +2npi} hence i^i = exp{pi/2 2npi} where n is an integer. Normally, one takes a canonical choice of argument, i.e. Arg(i)=pi/2, so we usually define i^i=exp{pi/2}. 


#3
May2309, 06:37 AM

P: 117

Atan2 is a very useful function often used in computer science to automatically sort out the quadrant business as well as 'degenerate' cases such as atan(1/0) which would throw up an error. It's really handy  be nice for it to be a standard everywhere really. 


#4
May2309, 06:57 AM

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P: 9,396

Complex exponentiation
Right, so I'm going to guess that Atan means inverse tan. Tan is periodic, hence its inverse is a multivalued 'function', but you can pick a principal branch. Which is exactly what you have done.
Your ln in the context above is not the mutlivalued bit  its argument is a positive real number  so it is just the ordinary logarithm. The reason why (online) calculators give a single answer for complex exponentiation is exactly because your Atan function chooses a single value, and the genuine multivaluedness comes from the fact that actually there are infinitely many choices for it. 


#5
May2309, 08:31 AM

P: 117

Aha, thank you  it's all starting to piece together now! Am I right in saying then that the DFT definition for example (which uses a complex exponent) would use this "principal branch" thing? (I'm sure the exponent term in that DFT formula can be reduced regardless).
Just one more question now (as long as there's a relatively simple answer  don't worry otherwise): Why does the exponent have to be irrational to produce the multivalue thing? It makes sense to me that if it was merely noninteger (e.g. 2.35), or even just integer, that the multivalue thing would come into play. 


#6
May2309, 03:27 PM

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P: 9,396

No  that is no a principal branch or anything  exp is a power series and is unique.
Taking rational powers i.e. 1/n results in n possible answers, not infinitely many  that is the difference. 


#7
May2309, 06:03 PM

P: 117

Okay, many thanks for your help again! If you have Paypal, I'd be happy to donate a little if you like.



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