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#2 (permalink) | |
Wehret Den Anfängen!
Location: Ontario, Canada
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Quote:
If I remember correctly there is a family of solutions. Do you know how to raise a complex number to a real power? Do you know how to raise a real number to a complex power?
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Last edited by JHVH : 10-29-4004 BC at 09:00 PM. Reason: Time for a rest. |
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#3 (permalink) |
aka: freakylongname
Location: South of the Great While North
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This might be helpful to you... http://www.math.toronto.edu/mathnet/...omplexexp.html
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"Reality is just a crutch for people who can't cope with drugs." Robin Williams. |
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#4 (permalink) |
Wehret Den Anfängen!
Location: Ontario, Canada
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i = e^(pi*i*(1/2 + 2*k))
i^(a+b*i) = e^(pi*i*(1/2 + 2*k))^(a+b*i) = e^(pi*(1/2 + 2k)*(-b+a*i)) = e^(-pi*b*(1/2+2k) + pi*i*a*(1/2+2k)) = e^(-pi*b*(1/2+2k)) * e^(pi*i*a*(1/2+2k)) If b is not 0, this is a set of points arranged in a spiral. The magnatude of each point is e^(-pi*b*(1/2+2k)) while the angle is e^(pi*i*a*(1/2+2k)) If a is rational, the result takes on finitely many angles. If a is irrational, the angles form a dense net. Ie, if b is 0 and a is irrational, then the set of roots is dense on the complex unit circle. If b is not zero, then the set of roots spirals from 0 out to infinity. This seems less than useful. You could identify a particular root as the primary root.
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Last edited by JHVH : 10-29-4004 BC at 09:00 PM. Reason: Time for a rest. |
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Tags |
complex, exponites, numbers |
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