Why does the graph of y = ln(x) have an imaginary part?

[rollcast]

I was messing around with wolframalpha and tried to make this graph.

http://www.wolframalpha.com/input/?i=plot+y+=+log+x

Now I understand why the blue real part exists and has that shape but I don't understand why it has an imaginary part?

Thanks
AL

 

[D H]

In terms of the real numbers, the natural logarithm function maps positive reals to the reals. Given some negative real number x, there is no real number y such that e^y=x (i.e., such that y=log(x).) The natural logarithm can be extended to the complex numbers via Euler's equation e^x=cos(x)+i*sin(x). This leads to log(-1)=pi*i (And to a host of other values; natural log is no longer unique.)

 

[rollcast]

Ok, but where does imaginary pi come into it?

I understand that from a bit of messing around that any value to the left of the y-axis is found by taking the natural log and adding imaginary pi, but I don't understand where imaginary pi comes from.

 

[D H]

It comes from Euler's identity: e^{\pi i}+1=0.

 

[AlephZero]

I don't know what math background you have, but the key result here is
e^{ix} = \cos x + i \sin x
If you aren't familar with that, Google for "Euler's identity"

Suppose z = \log(-a) where a > 0. Then e^z = -a. Since e^x > 0 for any real value of x, z must be complex so let z = x + iy.

Then e^{x + iy} = e^x(\cos y + i \sin y) = -a

The imaginary part = 0, so \sin y = 0.

So \cos y = \pm 1 and to make the real parts equal we have to take \cos y = -1 and e^x = a or x = \log a.

Putting it all together

\log(-a) = \log(a) + i \pi + 2k\pi where k is any integer.

The graph shows the "principal value" i.e. \log(a) + i \pi.