Cauchy’s Integral Formula: Conditions and Worked Examples

State the formula together with its assumptions

For a positively oriented simple closed contour C, suppose f is analytic on an open region containing the contour and its interior, and a lies inside C. Cauchy’s formula then gives the contour integral of f(z)/(z−a) as 2πi f(a). Positive orientation means counterclockwise. The numerator function f must satisfy the analyticity condition; the full integrand deliberately has the denominator factor z−a. Writing the formula without those hypotheses can lead to applying it to a singular numerator or the wrong region.

Identify the numerator function and the interior point

Consider the integral of (z²+1)/(z−(1+i)) around the counterclockwise circle |z|=2. Here f(z)=z²+1 is a polynomial and therefore analytic everywhere. The point a=1+i has modulus √2, less than 2, so it lies inside the circle. The theorem applies and reduces the contour integral to 2πi times f(1+i). The contour is not represented by ordinary real lower and upper limits; its geometry and orientation are essential parts of the problem.

Evaluate the first example exactly

Since (1+i)²=1+2 i+i²=2 i, the numerator value is f(1+i)=1+2 i. Multiplying by 2πi gives 2πi+4πi²=−4π+2πi. Numerically this is approximately−12.5663706+6.2831853 i. In the local Complex Number Calculator, multiply the number with real part 0 and imaginary part 6.283185307179586 by the number with real part 1 and imaginary part 2. This checks the final product only; the contour and analytic conditions were established separately.

Use the derivative form for higher powers

Under the same analytic and contour assumptions, the integral of f(z)/(z−a)^(n+1) equals 2πi f^(n)(a)/n!, where n is a nonnegative integer. For a squared denominator, n=1 and the formula uses the first derivative. Take f(z)=z³+2 z and a=1 inside |z|=2. Since f′(z)=3 z²+2, f′(1)=5, so the integral of (z³+2 z)/(z−1)² is 10πi. Multiplying 2πi by 5 in the local arithmetic form gives approximately 31.4159265 i.

Do not confuse denominator power with derivative order

A denominator(z−a)^3 corresponds to n=2, so the second derivative is divided by 2!, not by 3!. For f(z)=z³ at a=1, f″(1)=6 and the integral becomes 2πi×6/2=6πi. This example checks both the shift from exponent to derivative order and the factorial. Using the third derivative or forgetting the factorial changes the result even if the contour condition and complex multiplication are otherwise correct.

Check the contour position and orientation

If the point a is outside the contour and the integrand has no other singularities inside, Cauchy’s theorem gives zero instead of the interior-point formula. If a lies on the contour, the ordinary formula cannot be applied as written. Reversing a simple contour to clockwise changes the sign of its integral. Multiple traversals require accounting for the winding number rather than silently assuming one counterclockwise circuit. Draw the contour and mark all potentially singular points before substituting a value into f.

Inspect additional singularities in the numerator

If the proposed numerator is f(z)=1/(z−b), it is not analytic at b. When b also lies inside the contour, the simple application that treats f as analytic fails. You may need partial fractions or a residue theorem calculation that accounts for all enclosed singularities. The denominator's appearance alone does not select the answer. A valid solution explains why the chosen numerator is analytic throughout the required region, rather than merely matching the visible pattern f(z)/(z−a).

Keep the theorem separate from numerical quadrature

A real one-variable integral form cannot accept a complex contour, orientation and analytic-domain assumptions simply by replacing x with z. Parametrizing a contour produces a different calculation involving z(t), its derivative and generally complex values. Cauchy’s formula can avoid that quadrature when its conditions hold, but the final arithmetic checker does not supply a proof. Record the contour, numerator, interior point, derivative order and exact complex result so that every part of the application can be reviewed independently.

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