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Given complex numbers $z_1$ and $z_2$, let $[z_1, z_2]$ denote the straight line segment path from $z_1$ to $z_2$. Recall that we can parametrize this by $x(t) = z_1 + t(z_2 - z_1)$ for $t \in [0,1]$. In the case $C=[1-i,1+i]$, sketch $C$ and evaluate the integral of the complex conjugate squared.

So I have an exam tomorrow and this topic is on it, and I have been trying so hard to figure out how to do it, but I cannot figure it out. So do I take the integral of $x(t)x'(t)$ or what. Can someone please explain to me how I do contour integration and how I would go about doing this problem. This is a big part of my exam and I really need to learn it.

Thanks so much.

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  • $\begingroup$ If you do not want to help me, then do not comment in this please. What is the point of even saying that? Anyone who can help, I would really appreciate it. Thanks $\endgroup$ – Brian Feb 12 '15 at 4:29
  • $\begingroup$ Perhaps you can look at the definition of line integral in your notes, try to solve this and post your results. It is not really that different from someone telling you "this is the definition of the line integral of a function over a path" in an answer below. On the other hand, I don't think you'll benefit if someone just spoonfeeds the answer to you. $\endgroup$ – Pedro Tamaroff Feb 12 '15 at 4:31
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    $\begingroup$ You should look at the definition of a complex integral, and what it means to integrate a complex function along a given curve. We are given a curve, in this case a straight line segment connecting $1-i$ and $1+i$, and they've told you (in words) which function to integrate. $\endgroup$ – Ducky Feb 12 '15 at 4:31
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Using the parametrization $\gamma\colon [0,1]\to [z_1,z_2]$ given by: $$ \gamma(t)=z_1+t(z_2-z_1), $$ we may compute the integral as follows: $$ \begin{align*} \int_{[z_1,z_2]}\overline{z^2}\ dz&=\int_{t=0}^1\overline{\left(z_1+t(z_2-z_1)\right)^2}\gamma'(t)\ dt\\ &=(z_2-z_1)\overline{\left.\frac{(z_1+t(z_2-z_1))^3}{3(z_2-z_1)}\right|_{0}^1}\\ &=(z_2-z_1)\overline{\frac{z_2^3-z_1^3}{3(z_2-z_1)}}\\ &=(z_2-z_1)\frac{\overline{z_1}^2+\overline{z_1z_2}+\overline{z_2}^2}{3}. \end{align*} $$

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  • $\begingroup$ There's a $z_2 - z_1$ missing. $\endgroup$ – kobe Feb 12 '15 at 5:28
  • $\begingroup$ Where? Please explain. $\endgroup$ – pre-kidney Feb 12 '15 at 5:34
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    $\begingroup$ @pre-kidney since $\gamma'(t) = z_2 - z_1$, the integral on the right of the first line should have a factor of $z_2 - z_1$, right? $\endgroup$ – kobe Feb 12 '15 at 5:56
  • $\begingroup$ Oh yes, of course. My mistake! $\endgroup$ – pre-kidney Feb 12 '15 at 5:57

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