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Can somebody help me with this?

is there a difference when the contour that travels horizontally from 0 to 1 and then climbs vertically to 1 + i?

Detailed explanations will be greatly appreciated

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Hint: Recall the definition of a contour integral, choose appropriate parametrizations for both cases, plug in and compare the results. – Your Ad Here Oct 16 '12 at 20:19

For non-analytic function integrals along different curves connecting two points in $\mathbb{C}$-plane, in general are different. If $z=x+iy$ and $\gamma=\gamma_1\cup \gamma_2 ,$ where $\gamma_1=\{z=x+iy\colon \,\, 0\leqslant x\leqslant 1, \,\, y=0 \}, \quad \gamma_2=\{z=x+iy\colon \,\, x=1, \,\, 0\leqslant y\leqslant 1\}$ then $$I=\int\limits_{\gamma}|z|^2 dz=\int\limits_{\gamma}(x^2+y^2) (dx+i\ dy)=\int\limits_{\gamma}(x^2+y^2) dx+i\int\limits_{\gamma}(x^2+y^2) dy=\int\limits_{\gamma_1\cup \gamma_2}(x^2+y^2) dx+i\int\limits_{\gamma_1\cup \gamma_2}(x^2+y^2) dy=\int\limits_{\gamma_1}(x^2+y^2) dx +\int\limits_{\gamma_2}(x^2+y^2) dx+i\left(\int\limits_{\gamma_1}(x^2+y^2) dy+\int\limits_{\gamma_2}(x^2+y^2) dy\right).$$
On $\gamma_1: \quad y=0;\quad dy=0;\,\,$ on $\gamma_2: \quad x=1;\quad dx=0,$ thus $I=\int\limits_{0}^1 {x^2} dx+i\int\limits_{0}^1 (1+y^2) dy.$
Integral over path described in the title can be calculated in the same manner.

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thank you very much – mathnoob Oct 17 '12 at 3:01

In this simple case there is a second method which I shall explain using the paths $\gamma':=\sigma[0,i]$ and $\gamma'':=\sigma[i,1+i]$ described in the title of the question.

Along $\gamma'$ one has $\bar z=-z$ and therefore $|z|^2=z\bar z=-z^2$, and along $\gamma''$ one has $\bar z=z-2i$ and therefore $|z|^2=z^2-2iz$.

This gives $$\int_{\gamma'}|z|^2\ dz=\int_{\gamma'}(-z^2)\ dz= -{z^3\over3}\Biggr|_0^i={i\over3}$$ and $$\int_{\gamma''}|z|^2\ dz=\int_{\gamma''}(z^2-2iz)\ dz=\Bigl({z^3\over3}-iz^2\Bigr)\biggr|_i^{1+i}=\ldots\quad.$$

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