Let $f(z)=\frac{z^2}{z-(1-i)}$. Compute the integral over $C$ where $C=\{z: |z+3i-5|=1\}$.

Question

Let $f(z)=\dfrac{z^2}{z-(1-i)}$. Compute the integral over $C$ where $C=\{z: |z+3i-5|=1\}$.

Is it correct to say that the function is analytic on and everywhere inside $C$ therefore by Cauchy's Theorem the function integrated over $C$ is equal to $0$.

Answer 1

Yes. $$f(z)=\frac{z^2}{z-(1-i)}$$ and since $$|1-i+3i-5|=|-4+2i|>1$$ the point $1-i$ lies outside $C$, so $$\int_Cf\,dz=0.$$