# Show that the Cauchy integral formula implies the Cauchy-Goursat Theorem

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.

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

Let $z_0$ in the interior of $C$ and $g(z)=(z-z_0)f(z)$. Then Cauchy Integral Formula provides
$$0=g(z_0) = \frac{1}{2\pi i} \int_{c}\frac{g(z)}{z-z_0}\,dz= \frac{1}{2\pi i} \int_{c}\frac{f(z)(z-z_0)}{z-z_0}\,dz=\frac{1}{2\pi i} \int_{c}f(z)\,dz.$$
Hint: what if you take $g(z) = (z - z_0) f(z)$?
By the way, your integral formula has a missing $i$.
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