Find contour integral around the circle $\oint\frac{2z-1}{z(z-1)}dz$

Question

$$\oint\frac{2z-1}{z(z-1)}dz$$ counterclockwise around the circle $|z|=2$

I have substituted $z=2e^{it}$, $dz=2e^{it}dt$...

My integral is now:

$$\int_0^{2\pi}\frac{4e^{it}-1}{4e^{2it}-2e^{it}}$$ Can anyone advice with the next step to take?

Answer 1

Well, no need of the substitution. Note that the integrand isn't analytic at the points $z =0, 1$ both of which lie inside $C$. Writing it as, $$\frac{2z-1}{z(z-1)} = (1-2z)\left[\frac1{z}-\frac1{z-1}\right]$$ and using the Cauchy's integral formula gives us: $$I =2\pi i \left[(1-2(0))-(1-2(1))\right] =4\pi i$$

Answer 2

Notice that $$\int_{|z|=2}\frac{2z-1}{z(z-1)}\,dz = \int_{|z|=2}\frac{1}{z-1}\,dz+\int_{|z|=2}\frac{1}{z-0}\,dz.$$ and, since both $1$ and $0$ are in $|z|<2$, by using Cauchy's integral formula with $f(z)=1$, it follows that $$\int_{|z|=2}\frac{2z-1}{z(z-1)}\,dz=2\pi i\cdot 1+2\pi i\cdot 1=4\pi i.$$

Answer 3

$$\frac{2z-1}{z(z-1)} =\frac{1}{z-1}+\frac{1}{z}.$$ then, by cauchy integral formula

$$\int_{|z|=2}\frac{2z-1}{z(z-1)}\,dz = \int_{|z|=2}\frac{1}{z-1}\,dz+\int_{|z|=2}\frac{1}{z}\,dz =4πi.$$