Explain why $\displaystyle \int_{C_1(0)} f(z) dz =0$ for the function $\dfrac{z^2}{z-3}$. In case there's some confusion with the notation, $C_1(0)=$ circle of radius $1$ centered at $0$ in $\mathbb{C}$.
We were given several theorems in class for this, but here's one that I think might be applicable:
Assume $f(z)$ analytic in domain $\Omega$ and $\Gamma \subset \Omega$ a closed Jordan curve whose interior is contained in $\Omega$ so that $f(z)$ is analytic on and inside $\Gamma$. Then $$\int_{\Gamma} f(z) dz = 0.$$
since $\Gamma=C_1(0)$ is a Jordan curve sitting inside the disk $\Omega=D_2(0)$. The function $f(z)$ has discontinuity at the point $z=3$, but that's okay since $\left|3-0 \right|^2=9 > 4$, so $3 \not\in D_2(0)$. Thus it is left to show that $f(z)$ is analytic on the disk. Is everything correct so far? Am I using the right theorem here?
Now I want to prove that $\dfrac{z^2}{z-3}$ is analytic, but I'm having trouble breaking it into real and complex parts so that I can check that the first order partial derivatives are continuous and that the Cauchy Riemann equations are satisfied. After expanding the $z^2$ term it just gets ugly. Assuming $z=x+iy$, I get $$\frac{x^2-y^2+i(2xy)}{(x+iy)-3}.$$ How do I show that this function is analytic?
Best Answer
If you have some theorem, proposition, or lemma that the sum, difference, quotient, and product of analytic functions are analytic (except at points where the denominator is zero), then $\frac{z^2}{z-3}$ is analytic because the function $z\mapsto z$ is.