Let $f(z)=3y$, where $z=x+iy\in\mathbb C$, and let $\gamma(t)=t+it^2$, for $t\in[0,1]$. Find $\int f(z)\ dz$.
If the problem read "Let $f(z) = 3z$" for example, I would first find $\gamma'(t)= 1+2it$.
Then, I would find $\int f(\gamma(t))(\gamma'(t))\ dt$, which would become $\int3(t+it^2)(1+2it)\ dt$.
And I would just integrate as normal.
But, given $f(z)=3y$... I missing on what to do. I'm sure it's something simple, but I can't capture it. Can anyone help?
On the curve $\gamma, f(z)=3y=3t^2$ (which is $\it{real}$).