Toronto Math Forum

Cauchy's integral formula is $$ f(z) = \frac{1}{2𝜋i} \int_{Γ}\frac{f(z)}{z-a} dz$$
given formula is $$ \int_{Γ}\frac{dz}{z^2-6z+25} = \int_{Γ}\frac{dz}{(z-(3+4i))(z-(3-4i))}$$
For (a), as 3+4i is inside and 3-4i is outside
Let $$f(z)=\frac{1}{z-(3-4i)}$$
by Cauchy theorem $$ \int_{Γ}\frac{f(z) dz}{(z-(3+4i))}=2𝜋i f(3+4i) = \frac{2𝜋i}{8i} = \frac{𝜋}{4}$$
For (b), as 3-4i is inside and 3+4i is outside
Let $$f(z)=\frac{1}{z-(3+4i)}$$
by Cauchy theorem $$ \int_{Γ}\frac{f(z) dz}{(z-(3-4i))}=2𝜋i f(3-4i) = \frac{2𝜋i}{-8i} = \frac{-𝜋}{4}$$
For (c), as both points 3±4i are inside, we can use residue theorem
$$\int_{Γ}\frac{dz}{(z-(3+4i))(z-(3-4i))}$$
$$=2𝜋i(Res(f, 3+-4i) + Res(f, 3+4i))$$
$$=2𝜋i(\frac{1}{8i}+\frac{-1}{8i})=0$$