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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?

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    $\begingroup$ The fastest and most natural way to show that $z\mapsto \dfrac{z^2}{z-3}$ is analytic on $C_1(0)$ is, I feel, to just use the definition of analyticity: it is given by power series. $\endgroup$
    – Git Gud
    Commented May 5, 2016 at 5:56
  • $\begingroup$ @GitGud Hmmm, we haven't really talked about power series yet (the professor briefly mentioned it when we were first introduced to the definition of analytic) but I'll do a quick google search and see what I can find $\endgroup$
    – user167857
    Commented May 5, 2016 at 6:15

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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.

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  • $\begingroup$ Yep, proved this in a previous homework assignment actually. Thanks for reminding me about it, just the result I needed. Thank you $\endgroup$
    – user167857
    Commented May 5, 2016 at 6:57

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