Cauchy's Integral Formula

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Let $f$ be analytic in a simply connected domain $D$ , and let $C$ be a positively oriented simple closed contour lying in $D$ . If $z_0$ lies inside $C$ ,

f(z_0) = \frac{1}{2\pi i}\int_C \frac{f(z)}{z - z_0}\,\text{d}z

Equivalently,

\int_C \frac{f(z)}{z - z_0}\,\text{d}z = 2\pi i f(z_0)

For Derivatives

Let $f$ be analytic in a simply connected domain $D$ , and let $C$ be a positively oriented simple closed contour lying in $D$ . If $z_0$ lies inside $C$ , for $n = 1, 2, 3, \ldots$ ,

f^{(n)}(z_0) = \frac{n!}{2\pi i}\int_C \frac{f(z)}{(z-z_0)^{n+1}}\,\text{d}z

Equivalently,

\int_C \frac{f(z)}{(z-z_0)^{n+1}}\,\text{d}z = \frac{2\pi i}{n!}f^{(n)}(z_0)

Written by

Sahithyan K.

September 13, 2026 1 min read

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