How to calculate this integral: $\int_{C}{} (z^2-2z) dz$ where $C$ is a line segment starting at $1$ and ending in $i$

Question

The task is as follows: $$ \text{Calculate } \int_{C}{}f(z) dz \text{, where } \\ \text{a) } f(z)=z^2-2z \text{, } C \text{ - line segment starting at } 1 \text{ and ending at } i \text{;} \\ \text{b) }f(z)=\frac{\bar{z}}{z+i} \text{, } C \text{ - circle } |z+i| = 3 \text{.} $$

I do not know how to calculate either of the integrals. Any help or push in the right direction would be much appreciated. Thank you.

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Update:

I calculated the $a)$ integral to be equal to $\frac{7}{3}-\frac{i}{3}$. If someone could check if I did good, that would be great.

Answer 1

Hint.

For (a), you can simply use the fundamental theorem of calculus (also true in complex analysis) since you have an entire function: $$ \int_Cf(z)dz=F(B)-F(A) $$ where $A$ is the initial point of $C$ and $B$ the final point, and $F$ is an antiderivative of $f$.

For (b), the first thing you need is to parameterize your path $C$, which means writing $$ z=g(t),\quad t\in[0,1] $$ for some function $g$.

Then $$ \int_Cf(z)dz=\int_0^1 f(g(t))g'(t)dt. $$