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  1. $\int_C \hat{z} dz$ where $C$ is the straight line from $i$ to $2-i$.

  2. $\int_C \frac{dz}{z}$ where $C$ is the straight line from $3$ to $4i$

  3. $\int_C (z-z_0)^{n-1}dz $ for any integer $n$, where C is the contour once around the circle $|z-z_0|=1$.

I know that $z=x+iy$, so $\hat{z}=x-iy$. so for the first part, $\int (x-iy)dz$ from $i$ to $2-i$, but I don't know if to integrate with respect to $dx$ or $dy$?

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up vote 1 down vote accepted

What you are missing is the description of the contour on which you are integrating. For example, the straight line from $z=i$ to $z=2-i$ is described by

$$z=i (1-t) + (2-i) t$$

where $t \in [0,1]$. Then equating real and imaginary parts, $x=2 t$, $y=1-2 t$. Also, $dz = 2 (1-i) dt$. The first integral is then

$$\int_0^1 (2 t - i (1-2 t)) (2 (1-i)) \, dt$$

You can evaluate this.

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how did you get dz=2(1-i)dt? – xfitpi Mar 29 '13 at 1:39
on #2 I've got that x=3(1-t) and y=4t. so when I set up the integral for 1/z dz I'll have 1/(3(1-t)+i(4t)(...)dt and the (...) is for dz, but I don't know how to get that. – xfitpi Mar 29 '13 at 1:49
@xfitpi The derivative of $at + b$ for constants $a$, $b$ is $a$. – Ayman Hourieh Mar 29 '13 at 1:51
@AymanHourieh I still do not understand your abbreviated step on the derivative – xfitpi Mar 29 '13 at 1:54
@xfitpi $dz = \frac{dz}{dt} dt$. And $\frac{dz}{dt} = \frac{d}{dt}\left( i + t(2-2i) \right) = 2-2i$. Does it make sense now? – Ayman Hourieh Mar 29 '13 at 1:57

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