Evaluate $\int(z^2-z+2) dz$ from $i$ to $1$ along the contour $C$ given in the figure. The figure shown is the line $y=1-x^2$ from i to 1.

I'm having trouble parameterizing this curve. If someone can help me find the curve itself, z, then I can figure it out from there. Thanks in advance

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    $\begingroup$ $\gamma(t)=t+(1-t^{2})i,0\leq t\leq 1$? $\endgroup$ – DY88 Mar 23 '16 at 2:22

Functions are always easy to parameterize, as they "come parameterized" by default. For instance, if you feel comfortable calling your parameter $t$, then the curve of the function $y = 1 - x^2$ has natural parameterization $(t, 1-t^2)$. Or perhaps you'll prefer it in the form $t + (1-t^2)i$.

Then you choose $t$ accordingly to give the correct beginning and ending points.

  • $\begingroup$ "... as they "come parameterized" by default." wise words, I never thought of that. $\endgroup$ – Larara Mar 23 '16 at 2:26

Since the curve is $y=1-x^2$, we can easily parameterize with $x=t$ and $y=1-t^2$. Since $z=x+iy$, we therefore have $z=t +i(1-t^2)$. Use this to find $dz$ in terms of $dt$ and then use both of these expressions to express the integral in terms of $t$ and $dt$.

So what about the limits of integration? Note that when $t=0$ we have $z=i$ and when $t=1$ we have $z=1$. So the integral is taken from 0 to 1.


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