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

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?

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## 1 Answer

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

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. –  tamakisnen Jan 26 '12 at 20:51