Integrals of fractions(Complex)

Question

I'm a bit clueless about some (presumably basic) complex integrals.

How would I integrate (over a circle centered at the origin, let's say of radius 2) things like $\frac{1}{z^2+z+1}$ or $\frac {1}{z^2+2z-3}$? I know nothing beyond Cauchy's integration formula.

Answer 1

$z^2+2z-3=(z+3)(z-1)$ So,

$$\frac{1}{z^2+2z-3} = \frac14 \left(\frac{1}{z-1}-\frac{1}{z+3}\right)$$ So, we have $$\int_{C} \frac14 \left(\frac{1}{z-1}-\frac{1}{z+3}\right)dz$$

If you know Cauchy's integral formula, then you know that the second term integrates to zero since it is analytic in, and on, $C$.

Note that we could have integrated the integral of $\frac{1}{z-1}$ directly by using the complex logarithm after making a simple substitution. However, caution needs to be applied as the complex logarithm is a multi-valued function and evaluation of $\log (2e^{i\pi}-1)=\log (3)+i\pi$ and $\log (2e^{-i\pi}-1)=\log (3)-i\pi$ on either side of the integration limits are obviously not the same. For those unfamiliar with the complex logarithm, we proceed as follows.

For the integral of $\frac{1}{z-1}$, parameterize the circle $|z|=2$. Let $z=2e^{i\theta}$, $dz=2ie^{i\theta}d\theta$, and form the integral

$$\frac14 \int_0^{2\pi} \frac{2ie^{i\theta}}{2e^{i\theta}-1}d\theta=\frac14 \int_{-\pi}^{\pi} \frac{2ie^{i\theta}(2e^{-i\theta}-1)}{(2e^{i\theta}-1)(2e^{-i\theta}-1)}d\theta=\frac14 \int_{-\pi}^{\pi} \frac{2i(2-\cos \theta)}{5-4\cos \theta}d\theta$$

where the sine term integrated to zero because that integrand was an odd function of $\theta$.

Now, let's work on the integrand a bit of the last term. It is easy to show that,

$$\frac{2-\cos \theta}{5-4\cos \theta}=\frac14 \left(1+\frac{3}{5-4\cos \theta}\right)$$

So, we now have $$\frac14 \int_{-\pi}^{\pi} \frac{2ie^{i\theta}}{2e^{i\theta}-1}d\theta=i\frac{\pi}{4}+ i\frac{3}{8}\int_{-\pi}^{\pi} \frac {d\theta}{5-4\cos \theta}$$

The remaining integral has an integrand with a useful antiderivative, $\frac23 \arctan \left(3 \tan \left(\frac{\theta}{2}\right)\right)$ with $-\pi\le \theta \le \pi$. Evaluating the integral yields

$$\frac14 \int_0^{2\pi} \frac{2ie^{i\theta}}{2e^{i\theta}-1}d\theta=i\frac{\pi}{4}+i\frac{\pi}{4}=i\frac{\pi}{2}$$

Answer 2

$$ z^2+z+1 = \left(z^2+z+\frac 1 4 \right) + \frac 3 4 = \left(z + \frac 1 2 \right)^2 +\frac 3 4 = \left(z+\frac 1 2 + i\frac{\sqrt 3} 2 \right)\left( z+\frac 1 2 - i\frac{\sqrt 3} 2 \right) $$

The question now is whether the roots $\dfrac 1 2 \pm i \dfrac{\sqrt 3} 2$ are inside or outside the circle along which you're integrating. Since $\left|\dfrac 1 2 \pm i\dfrac{\sqrt 3} 2\right| = 1 < 2$ and your circle has radius $2$, They're inside it. So you have $\displaystyle\int_\gamma \cdots = (2\pi i\cdot\text{sum of residues})$.