What is ln (i)?

[Matheuler’s [/math identitite

[Math\\cos {\\theta} + i\\sin {\\theta} = e ^ {I\\theta} [/math


[Mathi = \\cos {\\frac {π} {2}} + I \\sin {\\frac {π} {2}} [/math

[Math = e ^ {i {\\frac {π} {2}}} [/math


[Math\\ln i = \\log_e{e ^ {i\\frac {π} {2}}} [/math

[math = i {\\frac {π} {2}} [/math


The answer given on the Dutch variant of Quora has been translated, but there are better answers (read more fully) to read answers on the English variant.

Let me take a try.

It is true that it is useful to write first:

[Mathi = e ^ A\\tag {1} [/math

To apply ‘ the ‘ logarithm, after all, [Math\\ln (..) [/math and [mathe ^ {(..)} [/math are each other’s inverse and eliminate each other. The problem is that equation (1) knows more than [math1 [/math Solution and the final answer too!

We have:

[Mathi = e ^ {i \ \ pi/2} \\tag * {} [/math

And that would mean:

[Math\\ln (i) = i \ \ pi/2 \ \ Tag * {} [/math

But we also have:

[Mathi = e ^ {i \ \ pi/2 + 2k \ \ Pi i} \\tag * {} [/math

You can also reach the point [Mathi [/math in the complex plane by always rotating an ‘ extra circle around the origin ‘.What I actually use here is that:

[mathe ^ {2k \ \ Pi i} = 1 \ \ Tag * {} [/math

In both expressions above [MATHK \\In \\mathbb{Z} [/math, or in words [MATHK [/math may be any integer.Now when we apply the logarithm we get:

[Math\\ln (i) = \\pi I (1/2 + 2k) \\tag * {} [/math

The natural logarithm gives more than 1 answer.This function is called multiworthy. There are two solutions to bring this back to the interpretation of a function that is known to us.

The first method limits the number of replies.We define a so-called branch of function. That actually works just like that the solution of [mathx ^ 2 = 2 [/math equals [math\\sqrt {2} [ /math or [math-\\sqrt{2} [/math.The default branch is that we assume that we define [Math\\text {Ln} (z) [/math by limiting the argument in the result of [Math\\ln (z) [/math to [math-\\pi < arg (\\ln z) \\le \\pi [/math.Note, if this is meant it is customary to write: [math\\text {ln} (z) [/math instead of [Math\\ln (z) [/math.(See note from Jos van Kan).

Another solution is that we imagine that the function is not located in a flat plane (the complex plane), but on a smooth surface that moves ‘ up ‘ when the variable [MATHX [/math in [Math\\ln (x) [/math makes a full circle Around the origin.This is called a Riemann surface.

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