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How to find this integral?

$\int_c (x^2+iy^3)dz$ when

$c$ is a segment that connects $z=1$ with $z=i$?

I know that $z(t) = (1-t)z_1 + tz_2 = 1 -t+ti$.

How do I use that in this cases?

1 Answers 1

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\int_c (x^2+iy^3) dz = \int_0^1 \left[(1-t)^2+i(t)^3\right]z'(t)dt

because $x=1-t$ and $y=t$ per what you've written, and z'(t)=i-1. Now if you take the scalar factor of z' outside the integral you can evaluate it with real calculus and then multiply to finish.

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    I've now realized that the book I am reading is a tragedy. I couldn't figure this out by myself.2012-01-26