The Definition of a Logarithm is:

If xy=a

Then logxa=yGiven this definition, since eiπ=−1

Then shouldn’t ln(−1)=iπ? What is wrong with it?

**Answer**

It is not undefined.

In fact, Ahlfors treatment of complex analysis begins with exponentiation and logarithms. As the other comments point out, once you accept complex values, there is an inherent problem as uniqueness fails: If t=ez, then also t=ez+2nπi for any integer n, so one has to choose which of these infinitely many values z+2nπi will be distinguished as “the” logarithm of t.

We can, however, choose the values of logarithm in a way that the resulting function is continuous (in fact, analytic) at the cost of not having it be defined everywhere on C∖{0}: We need to remove at least a line from the *punctured* plane line so that the domain of logarithm does not contain a loop about the origin; continuity imposes this restriction; think of eiθ for 0≤θ≤2π, we get a discontinuity once we have completed traveling along the loop. We refer to this choice of domain as a choice of *branch*. This question further discusses why this is needed.

When one chooses as domain C∖{x+0i∣x≤0} with log(a+ib) being the number ln√a2+b2+iarctan(b/a), with the arctangent θ chosen so −π<θ<π, we call this the *Principal branch*, usually denoted Log(z).

**Attribution***Source : Link , Question Author : YYC , Answer Author : Community*