This is for an assignment, describing the procedure is most beneficial for me, rather that solely computing the result.

I want to evaluate the following integral:

$$\int_C\text{Re}\;z\,dz\,\text{ from }-4\text{ to } 4$$ Along the line segments from $-4$ to $-4-4i$ to $4-4i$ to $4$, e.g:

Bad drawing of the path described above

So I want to evaluate three integrals of the above three paths, and add them together, is this correct?

Parametrisation for each curve respectively: \begin{align} z(t)=-4-4it,\quad &0\leq t \leq 1\\ z(t)=-4-4i+8t,\quad &0\leq t \leq 1\\ z(t)=4-4i+4it,\quad &0\leq t \leq 1 \end{align}

$$\int_0^1 (-4-4it)(-4i)dt + \int_0^1 (-4-4i+8t)(8)dt+\int_0^1 (4-4i+4it)(4i)dt$$ $$=16\int_0^1 (i-t)dt+32\int_0^1(-1-i+2t)dt+16\int_0^1 (i+1-4t)dt$$ $$=16\left[it-\frac{t^2}{2}\right]_0^1 + 32\left[-t-it+t^2\right]_0^1+16\left[it+t-2t^2\right]_0^1$$ $$=16(i-\frac12)+32(-1-i+1)+16(i+1-2)$$ $$16i-8-32-32i+32+16i+16-32$$ $$=-24$$

  • $\begingroup$ Yes. Your next step is to parametrize the line segments. $\endgroup$ – whacka May 8 '15 at 4:20
  • $\begingroup$ Well you probably don't want to substitute $x$ just yet in the last equation, but you have the right idea. On the left vertical leg, $\operatorname{Re}z$ is constant and $dz = i\,dy$, similarly for the other legs. $\endgroup$ – Cameron Williams May 8 '15 at 4:21
  • $\begingroup$ @SkiesBurn Please let me know how I can improve my answer. I just want to give you the best answer I can. $\endgroup$ – Mark Viola May 8 '15 at 4:28
  • $\begingroup$ Correct, but the notation $$\int_\alpha^\beta x\,dz$$ is not advisable, because you need to emphasize that we are talking about a straight line from $\alpha$ to $\beta$. If we took a different route from $\alpha$ to $\beta$, the result might be different. $\endgroup$ – David May 8 '15 at 4:31
  • $\begingroup$ On the first path, $\operatorname{Re}{z}=-4$, $dz = i dy$. On the second, $\operatorname{Re}{z}=x$, $dz =dx$. On the third, $\operatorname{Re}{z}=4$, $dz = i dy$. $\endgroup$ – Ron Gordon May 8 '15 at 7:41

$$\int_C xdz = \int_0^{-4} (-4) \,i\,dy+\int_{-4}^{4}x\,dx+\int_{-4}^0 (4) \,i\,dy=i32$$

  • $\begingroup$ So I can write $z=-4-4xi, 0\leq x \leq 1$ for the parametrisation of the first curve. $z=-4i-12+8x, 1\leq x \leq 2$ etc. Are those strange symbols for $y=\pm4$ actually $\pm $ signs? $\endgroup$ – Skies burn May 8 '15 at 4:39
  • $\begingroup$ Yes that is the starting point. $\endgroup$ – Skies burn May 8 '15 at 5:03
  • $\begingroup$ What's wrong with my attempt? $\endgroup$ – Skies burn May 8 '15 at 7:00
  • $\begingroup$ How on earth is there a real part in the answer? The integral over $x$ should be zero. The nonzero pieces of the integral are imaginary. $\endgroup$ – Ron Gordon May 8 '15 at 7:04
  • $\begingroup$ @RonGordon So how do I do it right? Why is my attempt above all real? $\endgroup$ – Skies burn May 8 '15 at 7:38

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