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- Newsgroups: sci.math
- Path: sparky!uunet!mnemosyne.cs.du.edu!nyx!lmitchel
- From: lmitchel@nyx.cs.du.edu (lloyd mitchell)
- Subject: Re: Raising Complex Numbers To A Complex Power...
- Message-ID: <1992Oct15.011901.21838@mnemosyne.cs.du.edu>
- Sender: usenet@mnemosyne.cs.du.edu (netnews admin account)
- Organization: University of Denver, Dept. of Math & Comp. Sci.
- Date: Thu, 15 Oct 92 01:19:01 GMT
- Lines: 22
-
- In article <wer8=RK00UhBQ4C1JT@andrew.cmu.edu> cd2d+@andrew.cmu.edu (Christopher Paul Diehl) writes:
- >
- >Could someone explain how to compute (a+bi)^(c+di)? Thanks for your help...
-
- This is how I do it. (Note: this probably lacks some rigor, but works in most
- cases.)
-
- Let r = |a+bi| and t = arg(a+bi). Thus, a+bi = r*exp(ti) and ln(a+bi) =
- ln(r)+ti. By virtue of the natural log function being the inverse of the
- exponential function, a+bi = exp(ln(a+bi)), so (a+bi)^(c+di) =
-
- [exp(ln(a+bi)]^(c+di) =
- [exp(ln(r)+ti)]^(c+di) =
- exp[(ln(r)+ti) * (c+di)] =
- exp(c*ln(r) - d*t) * exp[i*(d*ln(r) + c*t)]
-
- The first factor is the magnitude of the result, and the argument of the second
- factor is the new polar angle.
-
- Hope this helps.
-
- Kerry Mitchell
-