Strange algebra on the page of wikipedia over contour integration In the wikipedia article about contour integration we can see in example 4 the following algebra
$${\displaystyle {\begin{aligned}\color{red}{\int _{R}^{\varepsilon }{\frac {\sqrt {z}}{z^{2}+6z+8}}\,dz}&=\int _{R}^{\varepsilon }{\frac {e^{{\frac {1}{2}}\operatorname {Log} z}}{z^{2}+6z+8}}\,dz\\&=\int _{R}^{\varepsilon }{\frac {e^{{\frac {1}{2}}(\log |z|+i\arg {z})}}{z^{2}+6z+8}}\,dz\\&=\int _{R}^{\varepsilon }{\frac {e^{{\frac {1}{2}}\log |z|}e^{{\frac {1}{2}}(2\pi i)}}{z^{2}+6z+8}}\,dz\\&=\int _{R}^{\varepsilon }{\frac {e^{{\frac {1}{2}}\log |z|}e^{\pi i}}{z^{2}+6z+8}}\,dz\\&=\color{red}{\int _{R}^{\varepsilon }{\frac {-{\sqrt {z}}}{z^{2}+6z+8}}\,dz}\\&=\int _{\varepsilon }^{R}{\frac {\sqrt {z}}{z^{2}+6z+8}}\,dz.\end{aligned}}}\tag1$$
Both integrals marked in red are opposite, so we can conclude that for $I:=\int _{R}^{\varepsilon }{\frac {\sqrt {z}}{z^{2}+6z+8}}\,dz$ we have that
$$2I=\int _{R}^{\varepsilon }{\frac {\sqrt {z}}{z^{2}+6z+8}}\,dz+\int _{R}^{\varepsilon }{\frac {\sqrt {z}}{z^{2}+6z+8}}\,dz\\=\int _{R}^{\varepsilon }{\frac {\sqrt {z}}{z^{2}+6z+8}}\,dz-\int _{R}^{\varepsilon }{\frac {\sqrt {z}}{z^{2}+6z+8}}\,dz=0\tag2$$
I cant see how $(2)$ could be wrong here. There is a mistake in the wikipedia article? Or $(2)$ (for some strange reason) doesn't holds?
 A: Wikipedia is rather confusingly using the notation $\sqrt{z}$ with two different meanings in (1).  At the start, $\sqrt{z}$ refers to the branch of the square root function which you obtain by going around the keyhole, so $\sqrt{z}$ is actually negative when $z$ is between $R$ and $\varepsilon$ (since $\arg z$ is taken to be $2\pi$ rather than $0$).  But in the second use of $\sqrt{z}$ which you highlighted in red, the other branch of the square root is being used.  So in the resulting final equation $$\int _{R}^{\varepsilon }{\frac {\sqrt {z}}{z^{2}+6z+8}}\,dz=\int ^{R}_{\varepsilon }{\frac {\sqrt {z}}{z^{2}+6z+8}}\,dz,$$ the $\sqrt{z}$ on the left and the $\sqrt{z}$ on the right are not the same (they differ by a sign).  To somewhat justify this rather unfortunate choice of notation, note that in each case $\sqrt{z}$ denotes the branch of the square root that is used on the relevant part of the contour (a different branch is used when going from $R$ to $\varepsilon$ than when going from $\varepsilon$ to $R$).  This really doesn't excuse the second part you highlighted in red, though, where $\sqrt{z}$ is used with a different meaning despite the bounds of the integral still being from $R$ to $\varepsilon$.
