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My question for my exam on linear algebra was to evaluate the integral $\displaystyle \oint \operatorname{im}(z^2)dz$ , where integration is done under area bounded by vertices $0$, $1$, $i$ in the complex plane.

Since $\operatorname{im}(z^2)=2xy$, is the answer $0$?

*EDIT:- I am posting a picture of the question below enter image description here

Robert Z
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Arun Sunny
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5 Answers5

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On the paths $[0,1]$ and $[i,0]$, the function is $0$, hence its path integral is too.

On $[1,i]$ we have to compute it. This path is parametrised by $\begin{cases}x=1-t\\y=t\end{cases}(0\le t\le 1)$, so we just have to compute $$\int_{[1,i]}\operatorname{im}z^2\,\mathrm d z=\int_0^{1}\!2t(1-t)\, (-1+i)\,\mathrm d t=(-1+i)\biggl[t^2-\frac{2t^3}3\biggr]_0^1=\frac{-1+i}3.$$

Bernard
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No. It is $\int _0^{1}\int_0^{x} 2xy \, dy\, dx=1/4$.

Kavi Rama Murthy
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    I believe you're mistaken; you've computed the integral over the interior of the triangle, but what's asked for is the path integral over the boundary of the triangle. – John Hughes Jul 02 '18 at 10:14
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    @JohnHughes The phrasing in the question is a little confusing since it asks for both the line integral and the integration done under an area. – Michael Burr Jul 02 '18 at 10:14
  • It clearly says integral under the area though the notation does not agree with this. I took his wording as what he wanted. – Kavi Rama Murthy Jul 02 '18 at 10:16
  • Hey, I have now added a picture of the question itself. – Arun Sunny Jul 02 '18 at 10:24
  • I agree it's confusing; it's somewhat less confusing when you see the original (as is so often the case). In this case, it appears that my guess was the right one, but to be honest, I was about 40% believing that yours was. :) – John Hughes Jul 02 '18 at 11:03
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I guess that here we are talking about the complex line integral of $\mbox{Im}(z^2)$ around the triangle of vertices $0$, $1$, $i$ in the complex plane (counterclockwise?).

Hence, since $\mbox{Im}(z^2)=2xy\not=0$ just on the the segment $S$ from $1$ to $i$, it follows that $$\begin{align*} \int_{C}\mbox{Im}(z^2)dz&=\int_{C}2xyd(x+iy)=2\int_{S}xydx+2i\int_{S}xydy \\&=2\int_{x=1}^0x(1-x)dx+2i\int_{y=0}^1(1-y)ydy\\&=2(-1+i)\int_{t=0}^1(t-t^2)dt=\frac{-1+i}{3}.\end{align*}$$

Robert Z
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The phrasing in your question is a little awkward since the question asks for $\oint$, which is a line integral, and states that "where integration is done under area," which sounds like an area integral.

Breaking the triangle into its three edges, we can construct the paths: \begin{align} \gamma_1(t)&=t\\ \gamma_2(t)&=(1-t)+ti\\ \gamma_3(t)&=(1-t)i. \end{align}

Then $\gamma_1$ is the path from $0$ to $1$, $\gamma_2$ is the path from $1$ to $i$, and $\gamma_3$ is the path from $i$ to $0$ (each path has domain $[0,1]$).

Then, since \begin{align} \gamma_1'(t)&=1\\ \gamma_2'(t)&=-1+i\\ \gamma_3'(t)&=-i, \end{align} we can substitute these into a line integral to get \begin{align} \oint\Im(z^2)dz&=\int_0^1 \Im(t^2)\cdot 1dt\\&+\int_0^1 \Im(((1-t)+ti)^2)\cdot(-1+i)dt+\int_0^1\Im(((1-t)i)^2)\cdot(-1)dt \end{align} In the first and third integrals, the argument to $\Im$ is real, so the integrand in these cases is $0$. This leaves the second integral: $$ \int_0^1 \Im(((1-t)+ti)^2)\cdot(-1+i)dt=(-1+i)\int_0^1\Im((1-2t)+2t(1-t)i)dt. $$ We can simplify this to $$ (-1+i)\int_0^12t(1-t)dt=2(-1+i)\left.\left(\frac{t^2}{2}-\frac{t^3}{3}\right)\right|_0^1=2(-1+i)\frac{1}{6}=\frac{1}{3}(-1+i). $$

Michael Burr
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It is general complex integration. Realy strange question for a linear algebra exam !! This answer is right.

Sonu Lamba
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