Question: Evaluate $\int_{0}^{1+i} (x^2+iy) dz$ along the path (i) y = x (ii) y = x$^2$
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Subject: Applied Mathematics 4

Topic: Complex Integration

Difficulty: Medium

m4e(34) • 326 views
 modified 12 months ago  • written 16 months ago by
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(i) y = x

dy = dx; dz = dx + idy = dx + idx = (1+i)dx

$I = \int_0^1(x^2 + ix)(1 + i) \,\,dx \\ = (1+i) \int_0^1 (x^2 + ix) \,\, dx \\ = (1+i) [\frac{x^3}{3} + i \frac{x^2}{2}]_0^1 \\ = (1+i)(\frac{1}{3} + \frac{1}{2}i) \\ = \frac{1}{3} + \frac{1}{2}i + \frac{1}{3}i + \frac{1}{2} i^2 \\ = (\frac{1}{3} - \frac{1}{2}) + (\frac{1}{2} + \frac{1}{3})i \\ = \frac{-1}{6} + \frac{5}{6} i$

(ii) y = x$^2$

dy = 2x dx

dz = dx + idy = dx + i(2x dx) = (1 + 2ix) dx

$I = \int_0^1 (x^2 + ix^2)(1 + 2ix) \,\, dx \\ = \int_0^1 x^2(1+i)(1+2ix) \,\, dx \\ = (1+i) \int_0^1 x^2 (1+2ix) \,\, dx \\ = (1+i) \int_0^1 (x^2 + 2ix^3) \,\, dx \\ = (1+i) [\frac{x^3}{3} 2i \frac{x^4}{4}]_0^1 \\ = (1+i) [\frac{1}{3} + \frac{1}{2}i] \\ = \frac{1}{3} + \frac{1}{2}i + \frac{1}{3}i + \frac{1}{2}i^2 \\ = (\frac{1}{3} - \frac{1}{2}) + (\frac{1}{2} + \frac{1}{3})i = \frac{-1}{6} + \frac{5}{6}i$