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Louis Romao
Jun 23, 2019

ln(a+bi)

in Math Problems


4 comments
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Ian Fowler
Nov 05, 2021  ·  Edited: Nov 05, 2021

That works. Well done.


Try this:

Use the polar form of a + bi = R*e^it where

R = sqrt(a^2 + b^2) and t = arctan(b/a)


So ln(a + bi) = ln(R*e^it)

= ln(R) + it

= 1/2*ln(a^2 + b^2) + i* arctan(b/a)


Of course, in the complex numbers, any integer multiple - 2npi - added to t also works. What we both have done here is find the principal value when n= 0.


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Feb 20

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4 comments