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?
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?
\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.