4.4 The Principal Logarithm

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In the last section we computed the path integral of f(z)=1/z over the path γ(t)=eit parameterizing the unit circle and got a non-zero answer γf=02π1eitieitdt=2πi from which we concluded f does not have an antiderivative on C{0}. (If it did have an antiderivative on C{0} the path integral above would have to be zero!)

We can write 1z=11(1z)=n=0(1)n(z1)n for all |z1|<1 using our knowledge of the geometric series. This gives a power series representation of f on B(1,1). Let's define F(z)=n=0(1)nn+1(z1)n+1 which is the term-by-term antiderivative of our power series representing f on B(1,1). We can check (for example, using the ratio test) that its radius of convergence is 1. As a power series in its own right, it is holomorphic on B(1,1) and certainly F(z)=f(z) for all z. Thus f(z)=1/z does have an antiderivative on B(1,1)! We now have two questions.

  1. Why do we have an antiderivative on B(1,1) but not on B(0,1)?
  2. What is the antiderivative of f on B(1,1)?

We will have to wait a little to answer the first question. For the second question, recall first of all from real analysis that (ln(x))=1x so that our antiderivative F has (at least on the real axis) something to do with the natural logarithm. In real analysis the natural logarithm can be defined as the functional inverse of the exponential function. Let's try to define an inverse of our exponential function exp.

Fix wC{0}. We would like to solve the equation w=exp(z) for z. Writing z=x+iy gives w=ex(cos(y)+isin(y)) and we see that |w|=ex and that y is an argument of w. Thus the solutions are z=ln(|w|)+i(Arg(w)+2nπ) for any nZ.

In particular, every non-zero complex number z has a (indeed many) natural logarithms: a complex number w such that exp(z)=w. The quantity Arg(z) was defined as the argument of z lying in the interval (π,π]. The function zArg(z) is not continuous, so we cannot define a continuous function zln(|z|)+iArg(z) on all of C{0}. However, if we exclude the negative real axis (,0]={x+iy:x0 and y=0} where the output of Arg jumps, we do get a continuous function.

Definition (Principal Logarithm)

The principal logarithm of zC(,0] is Log(z)=ln(|z|)+iArg(z) and this defines a function Log:C(,0]C.

Example

What is Log(i)? From Arg(i)=π/2 we calculate Log(i)=ln(|i|)+iArg(i)=iπ/2 which is reasonable because exp(iπ/2)=i.

Having defined a logarithm, we can define complex powers of complex numbers as follows.

Definition (Principal Value)

For b,zC with b0 define bz=exp(zLog(b)) to be the principal value of bz.

Since z|z| is continuous on C and zArg(z) is continuous on C(,0] the principal logarithm is continuous on its domain. In fact, more is true.

Theorem

The principal logarithm is holomorphic on its domain and Log(z)=1z for all zC(,0].

Proof:

Fix zC(,0]. We need to evalue the limit limh0Log(z+h)Log(z)h and will do so using the relationship between Log and exp. Put w=Log(z) and k(h)=Log(z+h)w. Since Log is continuous the limit h0 implies that k(h)0. We may therefore calculate limh0w+k(h)wexp(w+k(h))exp(w)=limh0k(h)exp(w+k(h))exp(w)=1limh0exp(w+k(h))exp(w)k(h)=1exp(w)=1z as desired.

We have established the following about the function f(z)=1/z.

What difference between the domains C{0} and C(,0] is responsible for the two statements above? We will uncover the full answer to this question next week.