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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.


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.

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

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  • $\begingroup$ Okay, so is my $g(t)=i-\text{t}i+\text{t}$? $\endgroup$ – bosendorfer Jan 16 at 21:26
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    $\begingroup$ @bosendorfer: you have different $g$ for (a) and (b).// If the parameterization interval is $[0,1]$, you should have $g(0)=1$ and $g(1)=i$ for (a). // Can you work that out? $\endgroup$ – user9464 Jan 16 at 22:08
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    $\begingroup$ @bosendorfer: that's right. But there is a shortcut for (a), I have edited my answer. You only need to parameterize the path for (b). $\endgroup$ – user9464 Jan 16 at 22:26
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    $\begingroup$ @bosendorfer: do you know how to go on? $\endgroup$ – user9464 Jan 16 at 23:04
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    $\begingroup$ @bosendorfer: yes, that's correct. Then you can go on. $\endgroup$ – user9464 Jan 16 at 23:10

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