I'm trying to integrate this as a definite integral. $f(z) = x^2+iy^2$ when $z(t) = t^2+it^2$ and $0 \le t \le 1$.
I know that:
$$\int_{z(a)}^{z(b) }f[z(t)]z'(t) \,dt= \int_C f(z) dz$$
I'm confused on how to parametrize the $x$ and $y$ terms in the $f(z)$ function and complete the integration:
$$\int_0^{(1+it)} f[z(t)]z'(t) \,dt= \int_C x^2+iy^2 dz$$
EDIT:
So this is what I have so far. From, $z(t) = t^2+it^2$ and $0 \le t \le 1$,
$x(t) = t^2$ an $y(t) = t^2$
I think we need to substitute these values in for $x^2+iy^2$
So we would have $$f(z(t)) = [(t^2)^2+i(t^2)^2] $$ $$z'(t) = 4t^3+ 4it^3$$ $$\int_C x^2+iy^2 = \int_0^1 f(z(t))(z'(t))dt = \int_0^1 [(t^2)^2+i(t^2)^2][4t^3+ 4it^3]dt $$
Am I on the right track?
Probably the confusing thing is the definition of $f$, it is the map $x+iy\mapsto x^2+iy^2$, where $x=\operatorname{Re}(z)$ and $y=\operatorname{Im}(z)$, then you have that $(f\circ z)(t)=(t^2)^2+i(t^2)^2=t^4(1+i)$, and $$ dz(t)=d(t^2+it^2)=2tdt+2itdt=2t(1+i)dt $$ Then, assuming that $C$ is the curve parametrized by $z$, we have that $$ \int_{C}f(z)dz=\int_{0}^1 (f\circ z)(t)dz(t)=\int_0^{1}2t^5(1+i)^2dt=(1+i)^2\cdot \frac1{3}=i\frac2{3} $$