solve $\int_{c}(1-z)^2dz$; given that $c$ is the contour defined by$x=3\cos\theta$ and $y=3+3\sin\theta$; from $\theta=0$ to $\theta=\pi/2$

Question

As the title says I need to solve this integral, but this is the first problem I have to solve of this type and the example given by our professor is almost entirely different, and he is yet to answers my messages asking about that...

So if you can give me some insight on how can I solve this problem, I would appreciate very much.

Answer 1

Here we can define the curve C as a set of z $\in \mathbb C$ . Note that C is not a closed curve, and so the Cauchy-Goursat theorem you mentioned in some comments is not relevant.

Let $C := \{z \in \mathbb C| z = (3cos(\theta))+(3+3sin(\theta))i , \theta \in [0,\pi/2]\}$. Then, by definition of line integrals, we have that if C is parameterized by $(x,y)=(x(t),y(t)), z=x+yi$, the following holds: $$\int_{C} f(z)dz = \int_a^b f(z(t))\cdot z'(t)dt$$ for $t \in [a,b]$.

This is a basic property of line integrals, and I'll leave the calculations up to you.