Calculator test

[Albert Chan]

(Yesterday 08:12 PM)Idnarn Wrote:  \[\ln(re^{i\theta}) = \ln r + i\theta + 2k\pi i\]

(Yesterday 09:22 PM)Commie Wrote:  I'm sorry, but I don't understand wikipedia's interpretation of k, it seems to be quantized?

Perhaps it is better to rewrite LHS with k included, instead of from thin air.

e^(pi * i) = -1

Multiply LHS ln argument by 1 = e^(2*k*pi * i), integer k, for all solutions.
Note: ln(z) may mean only principal value, not all solutions.

\[\ln(r\;e^{\theta \,i}) = \ln(r\; e^{\theta\,i}\;e^{2k\pi\,i}) = \ln r + \theta\,i + 2k\pi\,i \]