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Evaluate ∮(z+1)dz over a square

Short answer

Define ∮_ C f(z)dz = ∫(u dx−v dy)+i∫(v dx+u dy). For f(z)=z+1, entire and analytic on and inside the closed square with vertices 0,1,1+i,i. By Cauchy integral theorem, the integral is zero. Direct antiderivative F(z)=z^2/2+z gives same: F(end)−F(start)=0.

Define complex integral. Evaluate: ∮_C (z + 1) dz where C is the square with vertices at z = 0, z = 1, z = 1 + i and z = i.

Complex integral

For continuous f(z) on a curve C:

∫_C f(z) dz = ∫_C (u dx − v dy) + i ∫_C (v dx + u dy).

Solution

f(z) = z + 1 is a polynomial, so it is analytic everywhere, in particular inside and on the square C with vertices 0, 1, 1 + i, i.

By Cauchy's integral theorem, ∮_C f(z) dz = 0.

Verification by direct integration

An antiderivative is F(z) = z²/2 + z. Along a closed path the start and end points coincide, so

∮_C (z + 1) dz = F(end) − F(start) = 0.

Direct check, one side at a time (counter-clockwise):

0 → 1      : ∫₀¹ (x + 1) dx           = 3/2
1 → 1+i    : z = 1 + iy, dz = i dy
             ∫₀¹ (2 + iy) i dy        = 2i − 1/2
1+i → i    : z = x + i, x: 1→0
             ∫₁⁰ (x + 1 + i) dx       = −(1/2 + 1 + i) = −3/2 − i
i → 0      : z = iy, y: 1→0
             ∫₁⁰ (iy + 1) i dy        = −(−1/2 + i) = 1/2 − i

Real parts: 3/2 − 1/2 − 3/2 + 1/2 = 0. Imaginary parts: 2 − 1 − 1 = 0.

Answer: ∮_C (z + 1) dz = 0

✓ Answer checked by Claude Opus 5.5 (high effort)

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