# What is ln (i)?

[Matheuler’s [/math identitite

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

So

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

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

So

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

[math = i {\\frac {π} {2}} [/math

**Cheers!**

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.