Calculate using Cauchy formula $\oint\limits_{\gamma} \frac{d z}{z^{2}+4}$, where $\gamma = |z-2i|=1$ – a closed loop counterclockwise.

Question

How can I calculate that using Cauchy formula $\oint\limits_{\gamma} \frac{d z}{z^{2}+4}$, where $\gamma = |z-2i|=1$ – a closed loop counterclockwise? However, for Cauchy formula I need the denominator to be in the form $(z−z_0)^{n+1}$, how can I rewrite this function so that I can apply the formula?

Answer 1

The key idea here, and one worth remembering for the future, is that of partial fraction decomposition. We can factor $z^2 + 4 = (z - 2i) (z + 2i)$. Now, the idea is to write $\frac{1}{z^2 + 4} = \frac{A}{z - 2i} + \frac{B}{z - 2i}$ for some complex numbers $A, B$. There's a general theorem proving that these exist, but for this concrete example we'll work backwards to find these constants. Indeed, if this equality $\frac{1}{z^2 + 4} = \frac{A}{z - 2i} + \frac{B}{z - 2i}$ held, then we could multiply by $z^2 + 4$ to get $1 = A (z + 2i) + B (z - 2i)$. Hence, the right hand side is equal to $1$ for all $z$. In particular, for $z = 2i$ we get $4iA = 1$ so $A = -\frac{i}{4}$. Similarly, by plugging in $z = -2i$ we get $B = \frac{i}{4}$. You'd be right to be a bit concerned about the rigor, so you could go look at the general proof of partial fraction decomposition or just manually verify that these choices of $A, B$ work.

Now, we have $\frac{1}{z^2 + 4} = \frac{i}{4} \left( \frac{1}{z + 2i} - \frac{1}{z - 2i} \right)$. Plugging this into the path integral yields $\oint_\gamma \frac{1}{z^2 + 4} = \oint_{\gamma} \frac{i}{4} \left( \frac{1}{z + 2i} - \frac{1}{z - 2i} \right)$, so by linearity we need only compute $\oint_\gamma \frac{1}{z - 2i}$ and $\oint_\gamma \frac{1}{z + 2i}$. Now, $\gamma$ loops around $2i$ but $-2i$ is on the outside of the loop. Hence, $\oint_\gamma \frac{1}{z + 2i} = 0$ as $\frac{1}{z + 2i}$ has no singularities encircled by $\gamma$. On the other hand, Cauchy's theorem tells us that $\oint_\gamma \frac{1}{z - 2i} = 2 \pi i$.

Putting this all together yields $\oint_\gamma \frac{1}{z^2 + 4} = \frac{i}{4} (-2 \pi i) = \frac{\pi}{2}$.

Answer 2

If you want to write the integrand in the form $$\frac{f(z)}{z-2i}$$ just set $$\frac1{z^2+4}=\frac{f(z)}{z-2i}$$ and solve for $f(z)$. In this case, we find $$f(z)=\frac1{z+2i}$$ $f$ is analytic on and within the given contour, so Cauchy's integral theorem applies.