# Computing $\int_{\gamma} {dz \over (z-3)(z)}$

Compute, using the Cauchy Integral Formula,

$$\int_{\gamma} {dz \over (z-3)(z)}$$

where $\gamma$ is the circle of radius $2$ centered at the origin,
oriented counterclockwise.

Attempt:

1. On the one hand we have that

$$\int_{\gamma} {dz \over (z-3)(z)} = -\int_{-\gamma} {{1 \over z} \over (z-3)}\ dz= -2 \pi i {1 \over 3} = -{2 \pi i \over 3}$$

via Cauchy’s Integral Formula. This seems to answer the question.

2. On the other hand, we have that since $3$ is located outside of the circle of radius $2$ centered at the origin, that therefore

$$n(-\gamma, 3) = 0 = {1 \over 2 \pi i}\int_{-\gamma} {dz \over (z – 3)}$$

This makes me uneasy about the answer to the question being $- \frac{2 \pi i}{3}$ as stated above.

Question: Are these two facts consistent with each other despite my uneasiness?

#### Solutions Collecting From Web of "Computing $\int_{\gamma} {dz \over (z-3)(z)}$"

• The equality $\displaystyle \int _\gamma \dfrac{1}{z(z-3)}\mathrm dz = -\int _{-\gamma} \dfrac{\frac 1 z}{z-3}\mathrm dz$ is correct, but I don’t see why you would revert the direction of the path.
• The equality $\displaystyle -\int _{-\gamma} \dfrac{\frac 1 z}{z-3}\mathrm dz= -2 \pi i {1 \over 3}$ is incorrect. Apparently you used cauchy’s integral formula for this. But to do this you would need for $z\mapsto \dfrac 1 z$ to be holomorphic inside $\gamma$. Not only it is not so, it’s not even defined inside $\gamma$. Furthermore, if the numerator was an holomorphic function $f$, you should get $2\pi if(3)$, not $-2\pi if(3)$, because you’re going in the negative direction.
• Your second point is true, but how does it relate to the integral?
• To solve this I suggest one of two ways. One is to consider $z\mapsto \dfrac 1{z-3}$ which is holomorphic inside $\gamma$. From here follows by cauchy’s integral formula that $$\int _\gamma \dfrac{1}{z(z-3)}\mathrm dz=\int _\gamma \dfrac{\frac1{z-3}}{z-0}\mathrm dz=2\pi i\dfrac {-1}3.$$ This can be confirmed by noting that \begin{align}\int _\gamma \dfrac{1}{z(z-3)}\mathrm dz&=\dfrac 1 3 \left[\int _\gamma \dfrac{1}{z-3}\mathrm dz-\int _\gamma \dfrac{1}{z}\mathrm dz\right]\\ &=\dfrac 1 3 \left[0-2\pi i\right] &\text{(Cauchy’s integral formula)}.\end{align} A third alternative is to break up $\gamma$ in $\gamma _+$ (the upper half of the circle) and $\gamma _-$ (the lower half of the circle). On the upper half, the function $F_+$ below, defined on $\mathbb C$ with exception of the negative imaginary axis, is an antiderivative of the integrand: $$z\mapsto \dfrac 1 3\left(\log\left(|z-3|\right)-\log(|z|)\right)+i\left(\text{arg}(z-3)-\text{arg}(z)\right), \text{where arg takes values in }\left]-\dfrac \pi 2, \dfrac {3\pi} 2\right[.$$ Thus \begin{align}\int _{\gamma _+}\dfrac{1}{z(z-3)}&=F_+(-2)-F_+(2)\\ &=\dfrac 1 3\left[\log\left(\dfrac 52\right)+0\right]-\dfrac 1 3\left[\log \left(\dfrac 1 2\right)+\pi i\right]\\ &= \dfrac 1 3 (\log(5)-\pi i).\end{align} Similarly, on the lower half, the function $F_-$ below, defined on $\mathbb C$ with exception of the positive imaginary axis, is an antiderivative of the integrand: $$z\mapsto \dfrac 1 3\left(\log\left(|z-3|\right)-\log(|z|)\right)+i\left(\text{arg}(z-3)-\text{arg}(z)\right), \text{where arg takes values in }\left]\dfrac \pi 2, \dfrac {5\pi} 2\right[.$$ So \begin{align}\int _{\gamma _-}\dfrac{1}{z(z-3)}&=F_-(2)-F_-(-2)\\ &=\dfrac 1 3\left[\log\left(\dfrac 1 2\right)-\pi i\right]-\dfrac 1 3\left[\log \left(\dfrac 5 2\right)+0\right]\\ &=\dfrac 1 3 \left(-\log(5)-\pi i\right).\end{align} Finally, summing the integrals yields $-\dfrac{2\pi i}{3}$.

The integrand has two poles, as is easy to see: 3 and 0. The curve $\ \gamma$ encloses only one of them, 0. Thus, the function$\ \frac{1}{z -3}$ is holomorphic on the area enclosed by the circle, while$\ \frac{1}{z}$ is not.
So yes, your second statement is correct; due to the fact that$\ \frac{1}{z -3}$ has no pole in the area, this function on its own would leave your curve integral with 0. However, because the pole of$\ \frac{1}{z}$ lies within the circle, the integral doesn’t just vanish. In short: apply the Cauchy Integral Formula again, bearing in mind that$\ \frac{1}{z -3}$is holomorphic in the area and only$\ \frac{1}{z}$ has a pole.

Also: Counterclockwise is actually the positive mathematical direction for your curve, so there’s no need to take the negative of $\ \gamma$. With this, I believe you should get the same result as you have right now, but now it’s for the right reasons!