# How do I show that $\int_C e^z dz = 0$ using the parametrizations $\gamma 1$, $\gamma 2$, $\gamma 3$, and $\gamma 4$?

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.

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