Evaluating contour integrals along given C's

Question

Ok, so I have the following problem that I am working on. It says to evaluate $$\int \frac{z}{(z-1)(z-2)}dz$$ where C are given by \begin{align} a)& \ \ C:\lvert z \rvert=\frac12\\ b)& \ \ C:\vert z+1 \rvert=1\\ c)& \ \ C:\lvert z-1 \rvert=\frac12\\ d)& \ \ C:\lvert z \rvert=4 \end{align}

So, my first thought was to use the Cauchy Integral Formula but after graphing

$a)$, which has zeroes at $1$ and $2$, which both lie outside the graph of the circle given by $a)$ so I believe then that $a)$ would be $0$

For $b)$ I have a circle centered at (-1,0) with radius of 1, which means that I would need to use Cauchy's Formula.

For $c)$ it is a circle centered at (1,0) with radius of $\frac12$

And for $d)$ I have a circle centered at the origin with radis of 4 so both zeroes would work.

Is my thinking correct?? If not, can someone help with understanding the problem.

Answer 1

So, my first thought was to use the Cauchy Integral Formula but after graphing $a)$, which has zeroes at $1$ and $2$, which both lie outside the graph of the circle given by $a)$ so I believe then that $a)$ would be $0$

Correct.

For $b)$ then, I would think that it does not exist since $C:\vert z+1 \rvert=1$n is a point at 0

I don't understand what you're saying, but the integrand function has no singularities in the given circle, the integral exists.

For $c)$ it is a circle centered at the origin, with radius $\frac32$ so I could use $z=1$ but not $z=2$ since it lies outside the circle.

No, it's a circle centered at $1$ with radius $\frac 1 2$. Use Cauchy's integral formula for an appropriate integrand.

And for $d)$ I have a circle centered at the origin with radis of 4 so both zeroes would work.

Again, I don't understand what you mean with 'both zeroes would work', but note that you can't use Cauchy's integral formula (not directly anyway). Try partial fractions and then use Cauchy's integral formula.

Note that for all $z\in \mathbb C\setminus \{1,2\}$ one has $\dfrac z{(z-1)(z-2)}=\dfrac 2{z-2}-\dfrac 1 {z-1}$.

Thus $\displaystyle \int _C\dfrac z{(z-1)(z-2)}=\int _C\dfrac 2{z-2}-\dfrac 1 {z-1}=2\cdot 2\pi i-2\pi i=2\pi i$.

The second equality is a consequence of Cauchy's integral formula and the fact that $z\mapsto 2$ and $z\mapsto 1$ are holomorphic functions inside $C$ (they are so everywhere, really) and $1$ and $2$ are inside $C$.