Why is the log of a negative number the positive number plus pi * i?

Question

Does it wrap around or something? If so, in which space?

Answer 1

It does wrap around. It actually wraps around the complex plane.

Think of this like this, the natural log (becuase other logs can be derived out of this one) is the inverse of the exponential. The exponential in the real numbers is one-to-one but the exponential of a real number is always positive.

This is why log of nonzero positive real numbers so nice.

Now, if we take the exponential of complex numbers in general, it is not one-to-one anymore.

For example, exp(0)=exp(2*pi*i)=exp(4*pi*i)...=1.

So when we take the log, you can expect multiple answers back. In order to even talk about the log, it first must be well-defined. So one way to do it is to restrict the domain and the range of the exponential so that we can take its inverse.

Another way to do it is then to give all answers.

For example, log(1)=2*pi*i*z where z is any integer, if I allow complex answers. Notice if I let z=0, I get back the real answer that I would expect of zero.

Now to be specific, lets say you want log(-111).
We say that log(-111)=i*pi+log(111) because when I exponentiate it again, I get -111.

exp(i*pi+log(111))=
exp(i*pi)exp(log(111))=

Since exp(i*pi)=-1 and exp(log(111))=111, we get -111 back.

Now I already told you that in the complex plane, the exponential is not one-to-one so
exp(i*pi)=exp(3*i*pi)=exp(5*i*pi)=...=-1

(Multiply i*pi with any odd integer).

So a more complete answer would be
log(-111)=(2z+1)*i*pi+log(111) where z is any integer. If you let z=0, you get the answer that you are familiar with.

I don't know what level of mathematics you are familiar with, but I assure you that this is not as simple as it seems. It is a very technical matter to define log of negative or complex numbers. I had to wait all the way up until I took complex analysis (upper division math) but then even then my professor just brushed the matter.

Answer 2

This comes from the theory of complex numbers. If you look at the Euler's formula link below you see that there is a polar coordinate system on the complex plane of the form (r, theta) = r*e^(i*theta). This is for 0<= theta < 2*Pi, r>0.

Thus -r = r*e^(i*Pi). Taking the natural log of both side.

ln(-r) = ln(r) + ln(e^(i*Pi)) = ln(r) + i*Pi.

I am assuming you are familiar with polar coordinates, the radian scale of angle measures, and complex numbers.

Answer 3

It is not quite true. What is true is that e^{ x + i pi} = - e^x. But it is also true that e^{ x - i pi} = - e^x, so there is no unique definition of the log for negative numbers.If you want to get a better picture of mutivalued complex functions, you have to look into Riemann surfaces... Good luck.

Answer 4

Is ip i*.