How do I solve this complex integral?

Question

$$\displaystyle\int_c\frac{2z-1}{z^2-1}\ dz$$

where $c$ is a circle radius $1$ centred at $1$.

I know it has poles at $i$ and $-i$ but these don't fall within the contour so I'm not sure i can use cauchy's residue theorem. Is the integral just $0$ or is there another way to work it out? Do I work out the path integral?

Answer 1

Hint: From Cauchy's integral formula (CIF), since $1 \in D=\{ z : | z − 1| < 1\}$ and $-1 \notin D$ $$g(1)=\frac{1}{2\pi i}\int\limits_{|z−1|=1} \frac{g(z)}{z-1}dz$$ where $g(z)=\frac{2z-1}{z+1}$ (check that $g(z)$ satisfies the criteria required by CIF). Or $$\int\limits_{|z−1|=1} \frac{g(z)}{z-1}dz=\int\limits_{|z−1|=1} \frac{2z-1}{z^2-1}dz=2\pi \cdot i \cdot g(1)=\pi\cdot i$$

Plenty of examples to understand this technique here, here, here, here and here.

Answer 2

If all else fails, you can always brute-force it.

Let $c:[0,1)\to\Bbb{C}$ be a function mapping the interval $[0,1)$ to your circle.

The usual formula for the contour integral applies

$$\int_c\frac{2z-1}{z^2-1}\ dz=\int_0^1\left(\frac{2c(t)-1}{c(t)^2-1}\cdot\frac{d}{dt}c(t)\right)\ dt$$

Answer 3

The residue at $z=1$ is $\lim_{z\to1}(z-1)\dfrac {2z-1}{z^2-1}=\dfrac 12$.

Thus by the residue theorem, we get $2\pi i\cdot\dfrac 12=\pi \cdot i$.