# Evaluating a contour integral using Cauchy's generalised integral formula

Exercise: Evaluate the following integral using the Cauchy Integral Formula (the contour is traversed once anti-clockwise):

$$\int_{|z| = 2} \frac{(z^2 + 1)}{(z - 3) (z^2 - 1)} dz$$

I found the singularities as being:

$$z=3\; and\; z = \pm 1$$

I then plotted these points and the contour on the argand diagram and then rewrote the integral as:

$$\int_{|z| = 2} \frac{\frac{(z^2 + 1)}{(z - 3)}}{(z^2 - 1)} dz$$

because the singularity $z = 3$ is outside the contour $|z| = 2$ and to use Cauchy's generalised integral formula $f(z)$ must not have any singularities within the contour on the plane.

At this point I can't just immediately use the formula because of the form of the denominator. Expanding it into $(z - 1) (z + 1)$ doesn't satisfy the formula either. $(z - 1) (z + 1)$ != $(z - w)^{n-1}$.

How do I evaluate this integral?

• Are you sure they didn't mean using Cauchy's residue theorem? – spaceisdarkgreen Oct 1 '17 at 2:40
• Yeah, this question was under the section "Cauchy's integral formula". The residue related theorem is presented later in the lecture notes. – user486323 Oct 1 '17 at 2:43
• Yeah, I usually think of the integral formula as only handling one singularity at a time. So I guess what you could do is rewrite the contour so that it consists of a contour going around $+1$ and another contour going around $-1.$ This is kind of moving towards just proving the residue theorem, but it'll work. – spaceisdarkgreen Oct 1 '17 at 2:51
• Did you not see how my hint lets you use cauchy integral formula directly? Just write $\frac{1}{z^2-1} = \frac{1}{2}\left(\frac{1}{z+1}-\frac{1}{z-1}\right)$ and then you can split the integral into two integrals each of which only has one singularity. – spaceisdarkgreen Oct 1 '17 at 3:19
• Maybe my algebra is poor but I got a different (incorrect) answer when I substituted that in. I found the other answer easier to follow. – user486323 Oct 1 '17 at 4:45

Hint: \begin{align} \dfrac{1}{2\pi i} \int_{|z| = 2} \frac{\frac{(z^2 + 1)}{(z - 3)}}{(z^2 - 1)} dz &=\dfrac{1}{2\pi i} \int_{C_1} \frac{\frac{(z^2 + 1)}{(z - 3)(z-1)}}{(z + 1)^1} dz +\dfrac{1}{2\pi i} \int_{C_2} \frac{\frac{(z^2 + 1)}{(z - 3)(z+1)}}{(z - 1)^1} dz \\ &=\dfrac{1}{2\pi i} \int_{C_1} \frac{f_1}{(z + 1)^1} dz +\dfrac{1}{2\pi i} \int_{C_2} \frac{f_2}{(z - 1)^1} dz \end{align} where $f_1$ ad $f_2$ are analytic over $C_1$ and $C_2$ respectively. • Why can you break the integral into different parts? – user486323 Oct 1 '17 at 2:51
• You can break an integral into different parts about every singularity over a simple connected domain with your contour!. – Nosrati Oct 1 '17 at 2:54
• Is there a theorem that I can use to support this? I showed my professor that solution for feedback and he said that I can't continue using the integral formula after breaking the integral into two pieces because f(z) won't be holomorphic because of the singularity being in the numerator. Personally, I can see that you're redefining the contour (so f(z) in those cases are actually holomorphic and satisfy the integral formula) for each integral but could you point me to something I can use to rationalise that decision? – user486323 Oct 3 '17 at 13:11
• Can you redefine the contour for each integral and make $C_1$ and $C_2$ about $z=-1$ and $z=1$.? – Nosrati Oct 3 '17 at 13:22
• @user486323 In $|z|<2$ the function is holomorphic except in points $z=\pm1$ so in this circle except in a very small neighborhood about these points we have $\int_{C-{(C_1\cup C_2)}}f=0$ so $\int_{C}f=\int_{C_1\cup C_2}f$. – Nosrati Oct 3 '17 at 14:22

HINT

The best way to do this is to probably use the Residue theorem, or rewrite the contour so that it goes around $z=1$ and $z=-1$ separately (thus "moving toward" the residue theorem). However, here's a trick that will let you use only the Cauchy integral formula without modifying the contour: $$\frac{1}{z^2-1} = \frac{1}{2}\left(\frac{1}{z-1} - \frac{1}{z+1}\right)$$