Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The following text are screenshots from my textbook:

enter image description here

enter image description here

I am not sure how 14.56 is derived. Since if z-a is a term in the denominator, it is not continuous there if z = a. How can we just apply Cauchy's theorem and say it equals to 0 there? Shouldn't it be handled like case c?

Detailed explanations are appreciated!


share|cite|improve this question
up vote 2 down vote accepted

The point is you're integrating around a small circle around the point $a$, not going through it, as in real analysis $\frac{1}{x^n}$ is differentiable away from 0 with antiderivative $\frac{1}{-(n-1)} \frac{1}{x^{n-1}}$ for $n > 1$.

Edit: let's recall how to compute a contour integral, over $\gamma$ with endpoints $z_0, z_1$ as in the picture. By definition, if $f = f_1 + i f_2$, where $f_1, f_2: \mathbb{C} \rightarrow \mathbb{R}$, then formally set $dz = dx + i dy$ and mash out: $$\int_\gamma (f_1 + i f_2)(dx + i dy) = \int_\gamma f_1dx - f_2dy + i \int_\gamma f_1 dy + f_2 dx$$So a complex contour is just 2 regular line integrals.

Now, I claim that because of this, we can import the following theorem from line integrals: if one is trying to integrate $P dx + Q dy$ over $\gamma$, now let's say $\gamma$ joins points $a$ and $b$, and in any neighborhood of the image of $\gamma$, we have some function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ such that $\partial_x f = P, \partial_y f = Q$, then $$\int_\gamma Pdx + Qdy = f(b) - f(a)$$This is pretty much the fundamental theorem of calculus.

Since we saw a complex contour integral is just 2 line integrals, you might hope some form of this theorem can apply, and indeed, it does, suppose we have in any neighborhood of the image of $\gamma$ a holomorphic function $F$ satisfying $F'(z) = f(z)$, unwinding this and writing $F = F_1 + i F_2$, this means in real analysis terms that: $$\partial_x F_1 = f_1 \quad \quad \partial_x F_2 = f_2$$ Since the first column of the Jacobian shows where $(1,0)$ goes under the action of $dF$, i.e. multiplication by $f(z)$.

With that, our contour integral above is just $$\int_\gamma f_1dx - f_2dy + i \int_\gamma f_1 dy + f_2 dx = \int_\gamma \partial_x F_1 dx - \partial_x F_2 dy + i \int_\gamma \partial_x F_1 dy + \partial_x F_2 dx$$Now we use the Cauchy-Riemann equations, which just say that $dF$ looks like a scaled rotation matrix, since that's what multiplication by a complex number does, we get: $$\int_\gamma \partial_x F_1 dx + \partial_y F_1 dy + i \int_\gamma \partial_y F_2 dy + \partial_x F_2 dy$$Now this we can straight up apply our real analysis theorem about contour integrals to, we obtain $$ = F_1(b) - F_1(a) + i(F_2(b) - F_2(a)) = F(b) - F(a)$$

Anyways, long story short: we can salvage a version of FTC, and all it cares about is having a local antiderivative. Now, $\frac{1}{z^2}, \frac{1}{z^3}, \ldots$ all have antiderivatives on $\mathbb{C} - 0$, so everything is fine and dandy so long as $\gamma$ doesn't touch 0. In this case since $a = b$ the integral is just 0.

For $\frac{1}{z}$, the issue is that we don't have a well defined antiderivative on $\mathbb{C} - 0$. Indeed, the problem is with $\log z$. In the case of $\mathbb{R^{>0}} \rightarrow \mathbb{R}$, this is just the inverse to $e^x$. But now, $e^z$ isn't a bijection from $\mathbb{C} \rightarrow \mathbb{C} - 0$, but it nearly is: $e^z = e^{z'}$ iff $z - z'$ is an integral multiple of $2\pi i$. In that sense $\log$ will be "multivalued" - when you wind around the origin once, any choice of $\log$ will change by $2\pi i$, so you can't get a well-defined holomorphic log on all of $\mathbb{C} - 0$.

share|cite|improve this answer
But why isn't 1/x differentiable away from 0 with antiderivative log(x)? – Sullivan Dec 13 '12 at 15:30
And also why are we considering it away from 0 for 1/x^n and within a circle surrounding 0 for 1/x? – Sullivan Dec 13 '12 at 15:33
This is extremely helpful! – Sullivan Dec 20 '12 at 3:06
thanks man, glad to help – uncookedfalcon Dec 23 '12 at 17:50

As long as you aren't being pedantic about the contour here (which isn't really the point of the example from the text), you can establish (14.56) by straightforward computation as follows.

Without loss of generality, take the contour $C$ to be a circle about $z=a$ of radius $r$, which can be parametrized as $$z-a=re^{i\theta} \implies dz=ire^{i\theta}\,d\theta \quad\text{ and }\quad {1\over z-a}\,dz=i\,d\theta.$$

Then, for $n=2,3,\dots$,

\begin{align*} \oint_C {1\over (z-a)^n\,dz}&=\oint_C {1\over (z-a)^{n-1}}\cdot {1\over z-a}\,dz\\ &=\int_0^{2\pi} {1\over (re^{i\theta})^{n-1}}\cdot i\,d\theta\\ &={i\over r^{n-1}}\int_0^{2\pi} e^{-i(n-1)\theta}\,d\theta\\ &={i\over r^{n-1}}\left[{e^{-i(n-1)\theta}\over -i(n-1)}\right]_0^{2\pi}\\ &=0. \end{align*}

Hope that helps.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.