Is the Singularity at Zero Removable for the Function g(z) = f(z)/z?

[futurebird]

Context: this is complex anaylsis II and I can use:

Cauchy's integral theorem
Liouville's Theorem
Taylor's Theorem
Morera's Theorem

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Let's say you have a function like g(z) = f(z)/z

And you know that f(z) is entire. But, then you find out that

$$\displaystyle\oint_{c} g(w) dw =0$$

c is a closed curve

that implies that g(z) is analytic in the region where this occurs, the region includes zero. But, how could g(z) be analytic? there is some kind of singularity at z=0. Could it just be removable? I'm not happy with this.

 

[Mystic998]

I'm pretty sure that what you have is

$$\displaystyle\oint_{c} g(w) dw = \displaystyle\oint_{c} \frac{f(w)}{w} dw = 2 \pi i * n(c, 0)*f(0)$$

So for that to be zero for all closed curves, f(0) = 0. Dunno if that helps any. And, as always, I could be wrong.

 

[futurebird]

I'm pretty sure that what you have is

$$\displaystyle\oint_{c} g(w) dw = \displaystyle\oint_{c} \frac{f(w)}{w} dw = 2 \pi i * n(c, 0)*f(0)$$

So for that to be zero for all closed curves, f(0) = 0. Dunno if that helps any. And, as always, I could be wrong.

thanks!

That happens to be how I got to this point. I want to say g(z) is entire, but I'm uset by the singularity at zero. If g(x) was something like...

3z/z can one say that it is "entire" even though it has no defined value at zero?

 

[Mystic998]

Wow, I'm really slow today. Okay, g(z) is not entire because, as you said, it's not analytic at 0. Why does your argument fail? Because the integral $\displaystyle\oint_{c} g(w) dw$ is not defined if c passes through 0, so it's not 0 for every closed curve in the complex plane. However, since z*g(z) goes to 0 as z goes to 0 (because f(0) = 0 and f is continuous), the singularity at 0 is removable. So you could extend g to an analytic function on the whole plane, and that's probably good enough.