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Topic: Complex numbers (Read 604 times) |
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LZJ
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Complex numbers
« on: Mar 30th, 2003, 4:01am » |
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Here's an interesting maths question:
What is a^i? 'a' is a real number that isn't 0,1 or e.
Express a ^i in terms of x + iy.
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towr
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Re: Complex numbers
« Reply #1 on: Mar 30th, 2003, 6:42am » |
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ai = (elog(a))i = ei*log(a) = cos(log(a)) + i sin(log(a)) (hidden)
if memory serves me right..
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Icarus
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Re: Complex numbers
« Reply #2 on: Mar 30th, 2003, 10:42am » |
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One additional clarification: towr's formula is good for a>0. For a<=0, ai is multivalued.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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LZJ
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Re: Complex numbers
« Reply #3 on: Mar 30th, 2003, 11:02pm » |
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Agreed, cos for negative logarithms, the solution has infinite values.
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towr
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Re: Complex numbers
« Reply #4 on: Mar 31st, 2003, 6:04am » |
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for a < 0 isn't it:
ai = ei*log(a) = ei*(i*pi + log(-a)) = e-pi * ei*log(-a) = e-pi * (cos(log(-a)) + i sin(log(-a))) ?
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| « Last Edit: Mar 31st, 2003, 6:09am by towr » |
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towr
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Re: Complex numbers
« Reply #5 on: Mar 31st, 2003, 12:24pm » |
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hmm.. of course, ei(2k+1)pi = -1 for all integer k, right? which gives the multiple solutions..
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