Skip to main content

Evaluate ∫(x²+iy)dz along y = x²

Short answer

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

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

Complex integral

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

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

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

Solution

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

Parameterise: z = x + i x², dz = (1 + 2ix) dx. On the path, x² + iy = x² + i x² = (1 + i)x².

I = ∫₀¹ (1 + i) x² (1 + 2ix) dx
  = (1 + i) ∫₀¹ (x² + 2i x³) dx
  = (1 + i) [ 1/3 + 2i/4 ]
  = (1 + i)(1/3 + i/2)
  = 1/3 + i/2 + i/3 − 1/2
  = −1/6 + 5i/6

Answer: −1/6 + (5/6) i

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Applied Mathematics chapter 1 questions →