Contour Integration over Square, Complex Anaylsis

  • #1
Safder Aree
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Homework Statement


Show that
$$\int_C e^zdz = 0$$

Let C be the perimeter of the square with vertices at the points z = 0, z = 1, z = 1 +i and z = i.

Homework Equations


$$z = x + iy$$

The Attempt at a Solution


I know that if a function is analytic/holomorphic on a domain and the contour lies within the domain lies. Then this integral equals 0. I can prove that e^z is analytic everywhere. (using cauchy-riemann) Is that enough to show it works in this case?
 

Answers and Replies

  • #2
mathwonk
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Let me make the statement more correct. Cauchy's theorem says that an analytic function has integral zero over a closed contour, lying in its region of analyticity, provided that contour does not wind around any point where the function is not analytic. In the case of a simple closed contour like this one, that means the function must be analytic on the contour as well as on the points interior to the contour. Does that apply?

In this extremely simple case, it might be more instructive to actually parametrize the contour and compute the integral over the four sides.

Notice that your less precise statement of the Cauchy theorem, ("I know that..."), would also imply that the integral of 1/z is zero over this contour, which is false. I.e. the function needs to be analytic not just on the contour but also at points "interior to" the contour.

Oh yes, and a contour is normally oriented, but when the integral is zero, it does not matter much for the value, since zero = -zero.
 
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  • #3
Safder Aree
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Let me make the statement more correct. Cauchy's theorem says that an analytic function has integral zero over a closed contour, lying in its region of analyticity, provided that contour does not wind around any point where the function is not analytic. In the case of a simple closed contour like this one, that means the function must be analytic on the contour as well as on the points interior to the contour. Does that apply?

In this extremely simple case, it might be more instructive to actually parametrize the contour and compute the integral over the four sides.

Notice that your less precise statement of the Cauchy theorem, ("I know that..."), would also imply that the integral of 1/z is zero over this contour, which is false. I.e. the function needs to be analytic not just on the contour but also at points "interior to" the contour.

That makes sense. Thank you. I'll just solve the integral.
 
  • #4
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I'm not sure if I am parametrizing this correctly, would you be able to double check where i am going wrong?

So for the path 0 -> 1.

$$z(t) = t, z'(t) = 1$$

$$\int_0^1 e^t dt $$


Path 1-> 1+i
$$ z(t) = 1 + it, z'(t) = 1$$
$$\int_0^1 e^{(1+it)}dt $$

Path 1+i -> i
$$z(t) = 1+i -t, z'(t) = -1$$
$$-\int_0^1 e^{1+i -t}dt$$

Path i -> 0
$$z(t) = i -t, z'(t) = -1$$
$$-\int_0^1 e^{i-t}dt$$
 

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