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Evaluate ∫(x²+iy)dz along y = x

Short answer

Parametrise path y=x: z=(1+i)x, dz=(1+i)dx, x∈[0,1], and f = x^2+iy = x^2+ix. Integral = (1+i)∫_0^1 (x^2+ix)dx = (1+i)(1/3 + i/2) = −1/6 + (5/6)i.

Evaluate ∫₀^{1+i} (x² + iy) dz along the path y = x.

Complex integral

If f(z) is continuous on a curve C, the complex integral is

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

or ∫ f(z(t)) z′(t) dt for a parameterisation z = z(t).

Solution

f = x² + iy, path y = x from z = 0 to z = 1 + i.

Parameterise: z = x + ix = (1 + i)x, dz = (1 + i) dx, with x from 0 to 1. On the path, x² + iy = x² + ix.

I = ∫₀¹ (x² + i x)(1 + i) dx
  = (1 + i) [ x³/3 + i x²/2 ]₀¹
  = (1 + i)(1/3 + i/2)
  = 1/3 + i/2 + i/3 − 1/2
  = −1/6 + 5i/6

Answer: −1/6 + (5/6) i

(The value happens to coincide with the result along y = x². This is not general, because x² + iy is not analytic, so the integral is normally path dependent.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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