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The contour of this integration is a square form by (1,i),(1,-i),(-1,i),(-1,-i)

How to show that $\int_c dz/z$ = $2\pi i$ without deformation of contour or parametize by $z=e^{it}$?

i 've try to use the formula $\int_c f(z)dz=\int ^b _a f(z(t))z'(t)dt$ but everything doesn't seems too neat!

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Are you familiar with ? – mathmath8128 Oct 11 '11 at 2:12
The way you've currently ordered the corners of your square, your integral becomes zero. You should be ordering the points anticlockwise to get the $2\pi i$ thing. – J. M. Oct 11 '11 at 2:16
@J.M. - why did you say that? – Victor Oct 11 '11 at 2:43
The order of your points matter. In any event: you know how to construct the parametric equations of a line joining two points, yes? – J. M. Oct 11 '11 at 2:53
up vote 2 down vote accepted

On the right half-plane the function $z\mapsto{1\over z}$ is the derivative of ${\rm Log}$, the principal value of the logarithm. Therefore $$\int_{1-i}^{1+i}{1\over z}\>dz={\rm Log}(1+i)-{\rm Log}(1-i)=\Bigl({1\over2}\log2+{i\pi\over4}\Bigr)-\Bigl({1\over2}\log2-{i\pi\over4}\Bigr)={i\pi\over2}\ .$$ Replacing $z$ by $i^kz$, $\>k=1,2,3$, in the integral one sees that all four sides of the square contribute the same amount to the final result, which is therefore $2\pi i$.

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thanks you for your help! – Victor Oct 11 '11 at 21:23

Split the square up into its four sides (in order): Right, Top, Left, Bottom. It's imperative that we make the paths go anti-clockwise. Let $x$ vary from $-1$ to $+1$; we find the parametrizations of each side in terms of $x$, as well as their signed direction (e.g. going up is $+i$).

  • Right: This side goes up from $1-i$ to $1+i$, so has sign $+i$. Its parametrization is $z=1+ix$.
  • Top: This side goes left from $1+i$ to $-1+i$, so has sign $-1$. Parametrization: $z=-x+i$.
  • Left: This side goes down from $-1+i$ to $-1-i$, so sign is $-i$. Parametrization: $z=-1-ix$.
  • Bottom: Goes right from $-1-i$ to $1-i$, so sign is $+1$. Parametrization: $z=x-i$.

Putting these facts together, we must compute the reparametrized integral $$\int_{-1}^{+1}\frac{+i}{1+ix}dx+\int_{-1}^{+1}\frac{-1}{-x+i}dx+\int_{-1}^{+1}\frac{-i}{-1-ix}dx+\int_{-1}^{+1}\frac{+1}{x-i}dx.$$ By linearity of integration, we can put these integrals together. Now:

  1. How can these four fractions be combined together? Hint: multiply numerator and denominator of the first by $-i$, the second by $-1$, the third by $i$... you should end up with four copies of the same fraction!
  2. Put the fraction into the form $a+ib$ and you should be able to integrate using facts from real-variable calculus. Note for the real part that integrating an odd function over $-1,+1$ will return zero, and for the imaginary part keep in mind your trigonometric antiderivatives.
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Hmm... somehow I had read his question to be taking the corners of the square as $1,i,-1,-i$. In any event: as a variation, one could use the line segment parametrization $(1-t)z_1+tz_2$ as well, and the calculations are similar. – J. M. Oct 11 '11 at 3:48

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