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Let $C$ be the perimeter of the square with vertices at the points $z=0, z=1, z = 1+i$ and $z=i$ traversed once in that order. Show that

$\int_C e^z dz = 0$.

I get the following parametrizations for the square:

$\gamma 1 = t, 0 \le t \le 1$

$\gamma 2 = it + 1, 0 \le t \le 1 $

$\gamma 3 = (1+i)-t, 0 \le t \le 1$

and

$\gamma 4 = i(t+1), 0 \le t \le 1$

Since $\int_{\Gamma} f(z)dz = \int_{a}^b f(z(t))z'(t)dt$, where $a \le t \le b$ I get that

$\int_{\gamma_1} e^tdt + \int_{\gamma_2} e^{it + 1}i dt + \int_{\gamma_3} (-1)(e^{1+i-t})dt + \int_{\gamma_4} e^{i(1+t)}i dt$ and so I get

$e^t \big|_0^1$ + $\frac{e^{it + 1}i}{i} \Big|_0^1$ + $\frac{(-1)e^{1+i-t}}{-1}\Big|_0^1$ + $\frac{ie^{i(1+t)}}{i}\Big|_0^1$ which is equal to

$e-e^0 + e^{i+1}-e + e^{i}-e^{1+i}+e^{2i}-e^i$, but then I get

that the answer is $e^{2i}-1$, not $0$, so I'm not sure where I'm going wrong.

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    $\begingroup$ $\gamma_4$ is moving from $z=i$ to $z=2i$ ? $\endgroup$ – FormerMath Mar 5 at 22:51
  • 1
    $\begingroup$ Use $\gamma_4:t\to i(1-t)$, $t\in[0,1]$. $\endgroup$ – dan_fulea Mar 5 at 23:47

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