# Integrating using residue theorem

I have to evaluate the following integral using the residue theorem. $$\int_{|z-i|=3} \frac {dz}{(z^2+2)(z+1)}.$$ First I found that there are three singularities, all 1st order poles. These are at $-1$, $\sqrt2i$ and $-\sqrt2i$.

My path is a circle centered at $i$ with a radius of 3. This means that I should sum up the residue of all three residues and none of them will be ignored correct?

I'm learning the residue theorem now. When I'm applying it to one singularity at a time, will I be using the whole polynomial $\frac 1 {(z^2+2)(z+1)}$ and applying the shortcut where I take the derivative of the bottom and leave the numerator unchanged and then plug in my singularity value to calculate the residue at that point? Thanks!

Yes, you are correct. However, by recalling that the Sum of residues at singularities plus residue at infinity is zero, you may also compute just one residue, the residue at infinity, $$\text{Res}(f,\infty)=-\text{Res}(f(1/z)/z^2,0)=-\text{Res}\left(\frac{z}{(1+2z^2)(1+z)},0\right)=0.$$

• I calculated each residue independently and verified that it does in fact come out to 0. I don't quite understand when this is applicable though. Lets assume I arrive at a problem where I have found my singularities and they all fall within the integral path, what question does one ask oneself to know if one can just calculate one residue at infinity instead? – J. Paredes Feb 22 '18 at 18:46
• If $f$ is analytic in $\mathbb{C}$ except at a finite number of singularities then the sum of the residues at the singularities plus the residue at infinity is zero. So if we have to evaluate a line integral of $f$ along a closed curve (counter-clockwise), we can 1) evaluate the sum of the residues INSIDE the curve or 2) evaluate minus the sum of the residues OUTSIDE the curve. We choose the option with fewer (or easier ) residues to evaluate. – Robert Z Feb 22 '18 at 18:59

You are correct, the region $|z-i|\leq 3$ encloses all the poles of $\frac{1}{(z^2+2)(z+1)}$. In particular

$$\oint_{|z-i|=3}\frac{dz}{(z^2+2)(z+1)} = \oint_{|z|=R}\frac{dz}{(z^2+2)(z+1)}$$ for any $R$ large enough (both sides are the same sum of residues). On the other hand the length of $|z|=R$ is $2\pi R$ and the function $\frac{1}{(z^2+2)(z+1)}$ is $\ll\frac{1}{R^3}$ on $|z|=R$, hence the RHS is arbitrarily close to $0$. In particular we may state $$\oint_{|z-i|=3}\frac{dz}{(z^2+2)(z+1)} = 0$$ without actually computing the residues of $f(z)$ at $\pm i\sqrt{2}$ and $-1$.

$\frac {1}{(z+i\sqrt 2)(1-i\sqrt 2)(z+1)} = \frac {1}{(-i2\sqrt 2)(i\sqrt 2+1)}\frac {1}{(z+i\sqrt 2)} +\frac {1}{(i2\sqrt 2)(-i\sqrt 2+1)}\frac {1}{(z-i\sqrt 2)} + \frac {1}{(-1+i\sqrt2)(-1-i\sqrt 2)}\frac {1}{(z+1)}$

Since all three poles are inside the contour:

$\oint_{|z+i| =3} \frac {1}{(z^2 + 2)(z+1)} \ dz = 2\pi i\left(\frac {1}{(-i2\sqrt 2)(i\sqrt 2+1)} +\frac {1}{(i2\sqrt 2)(-i\sqrt 2+1)} + \frac {1}{(-1+i\sqrt2)(-1-i\sqrt 2)}\right)$