# Integral on a half circle

complex-analysisintegrationline-integrals

I was very confused by this integral: $$I=\int_\gamma\frac{1}{z^2} dz$$, where $$\gamma$$ is the upper half of the unit circle. I know that when $$\gamma$$ is the unit circle, this would integrate to 0. Then I thought if we have $$a+bi$$ in the upper half, we also have $$-a-bi$$ in the lower half, and their square is the same. So by symmetry, we would have that the integral along the upper half equals the integral along the lower half. Then each of them would be 0, which obviously isn't the case. So I was wondering what I am missing here.
Thank you!

By definition, if the domain of $$\gamma$$ is $$[a,b]$$, then$$\int_\gamma\frac1{z^2}\,\mathrm dz=\int_a^b\frac{\gamma'(t)}{\gamma^2(t)}\,\mathrm dt$$and it seems to me that you are not taking that $$\gamma'(t)$$ into account.
To be more concrete, take $$\gamma\colon[0,\pi]\longrightarrow\Bbb C$$ defined by $$\gamma(t)=e^{it}$$ and take $$\eta\colon[\pi,2\pi]\longrightarrow\Bbb C$$ also defined by $$\eta(t)=e^{it}$$. Then\begin{align}\int_\eta\frac1{z^2}\,\mathrm dz&=\int_\pi^{2\pi}\frac{ie^{it}}{e^{2it}}\,\mathrm dt\\&=\int_0^\pi\frac{ie^{i(t+\pi)}}{e^{2i(t+\pi)}}\,\mathrm dt\\&=-\int_0^\pi\frac{ie^{it}}{e^{2it}}\,\mathrm dt,\end{align}since $$e^{\pi i}=-1$$ and $$e^{2\pi i}=1$$. But the final integral is just $$\int_\gamma\frac1{z^2}\,\mathrm dz$$.
The same argument applies if we have any even function instead of $$z^2$$.