Complex Analysis Calculator
Check the Cauchy–Riemann equations, find harmonic conjugates, compute residues and contour integrals of rational functions, and find the Möbius map through three points.
About the Complex Analysis Calculator
f = u + iv is analytic where u_x = v_y and u_y = −v_x (the Cauchy–Riemann equations) with continuous partial derivatives. Then u and v are harmonic (∇²u = ∇²v = 0), and a harmonic u has a harmonic conjugate v, found by integrating v_x = −u_y and v_y = u_x.
The residue of f at a pole is the coefficient of 1/(z − a) in its Laurent series; for a simple pole it is P(a)/Q′(a). The residue theorem gives ∮ f dz = 2πi × (sum of the residues inside the contour). A Möbius (bilinear) map w = (az + b)/(cz + d) is fixed by where it sends three points, through the cross-ratio.
The formula
u_x = v_y, u_y = −v_x; ∮ f dz = 2πi Σ Res; (w − w₁)(w₂ − w₃)/((w − w₃)(w₂ − w₁)) = (z − z₁)(z₂ − z₃)/((z − z₃)(z₂ − z₁))
- Simple pole
- Res = P(a)/Q′(a)
- Pole of order m
- Res = (1/(m − 1)!) lim d^(m−1)/dz^(m−1) [(z − a)^m f(z)]
Worked example
f(z) = (z + 1)/(z²(z − 2))
- Poles: 0 (order 2), 2 (simple)
- Res at 0 = −3/4, Res at 2 = 3/4
- On |z| = 3: ∮ f dz = 2πi(−3/4 + 3/4) = 0
Questions people ask
How are residues computed here?
By integrating f around a small circle round each pole (accurate to about 15 digits), then shown exactly when they are simple fractions or surds; for a simple pole this agrees with P(a)/Q′(a).
Why does a Möbius map need three points?
It has four coefficients but they matter only up to a common factor, leaving three degrees of freedom.