Sencilla pregunta acerca de la integración de contorno

Question

Si están integrando $$\int_\gamma y^2\,dz$$ Where $\gamma$ is the line segment from $1$ to $i$. You parameterize the line $$x(t)=1-t$$ $$y(t)=t$$ $$\implies z(t)=1-t+it$$

Ahora, si desea utilizar la fórmula: $$\int_\gamma f(z(t))z'(t)\,dt,$$ would you have the integral $$\int_0^1(t)^2(-1+i)\,dt$$ or would you have $$\int_0^1(it)^2(-1+i)\,dt.$$ I'm assuming its the first integral because you want the imaginary part, which is just $t$ and not $lo$.

Answer 1

Es el primero de ellos: $\int_0^1 t^2(-1 + i)\, dt$. La elección de $f(z)$ es $f(z) = \text{Im}(z)^2$. Así tenemos $z(t) = (1 - t) + it$ $z'(t) = -1 + i$ y así

$$\int_\gamma y^2\, dz = \int_0^1 f(z(t))z'(t)\, dt = \int_0^1 t^2(-1 + i)\, dt = \frac{-1 + i}{3}.$$