3
$\begingroup$

I'm struggling with this question, the integral formula states:

$$f(z_0) = \frac{1}{2\pi i} \int_{C}\frac{f(z)}{z-z_0}\,dz$$

and the Cauchy-Goursat theorem states:

If $f$ is holomorphic in a simply connected domain $D$ and $C$ contained in $D$ is a closed curve, then $\int_{C}f(z)\,dz =0$

To be perfectly honest I'm not at all sure where to start with question.

$\endgroup$
  • $\begingroup$ I believe you need Cauchy's Integral theorem to prove the integral formula? $\endgroup$ – Ayesha Jan 12 '14 at 20:20
  • $\begingroup$ Hint : apply Cauchy's formula to the function $g(z)=(z-z_0) f(z)$. $\endgroup$ – Glougloubarbaki Jan 12 '14 at 20:23
  • $\begingroup$ Can you show the integral formula implies the theorem? $\endgroup$ – Sam Houston Aug 26 '14 at 8:12
4
$\begingroup$

Let $z_0$ in the interior of $C$ and $g(z)=(z-z_0)f(z)$. Then Cauchy Integral Formula provides

$$ 0=g(z_0) = \frac{1}{2\pi i} \int_{c}\frac{g(z)}{z-z_0}\,dz= \frac{1}{2\pi i} \int_{c}\frac{f(z)(z-z_0)}{z-z_0}\,dz=\frac{1}{2\pi i} \int_{c}f(z)\,dz. $$

$\endgroup$
3
$\begingroup$

Hint: what if you take $g(z) = (z - z_0) f(z)$?

By the way, your integral formula has a missing $i$.

$\endgroup$
  • $\begingroup$ I feel like a complete idiot, this is so simple, thank you Robert. $\endgroup$ – Joe John Jan 12 '14 at 20:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.