Return to Answer

There are two branch points:

$$ z=-\frac{1}{2}\pm i \frac{\sqrt{2}}{2} $$

we can set the branch cuts connecting these two points and set up a contour like this (sorry for the poor drawing):

The two small circles in green near the two singularities have no contribution, since the function goes as $\frac{1}{\sqrt{z}}$. And the four blue lines have no contribution too, because they cancel each other in pairs since they are on the same branch. So the only contribution comes from the integration between the branch points in red. We can do this line integral easily:

-2i Integrate[1/Sqrt[4 z^2 + 4 z + 3] /. z -> -1/2 + I y, {y, -Sqrt[2]/2, Sqrt[2]/2}] (* -i π *) 

There are two branch points:

$$ z=-\frac{1}{2}\pm i \frac{\sqrt{2}}{2} $$

we can set the branch cuts connecting these two points and set up a contour like this (sorry for the poor drawing):

The two small circles in green near the two singularities have no contribution, since the function goes as $\frac{1}{\sqrt{z}}$ near the poles. And the four blue lines have no contribution too, because they cancel each other in pairs since they are on the same branch. So the only contribution comes from the integration between the branch points in red. We can do this line integral easily:

-2i Integrate[1/Sqrt[4 z^2 + 4 z + 3] /. z -> -1/2 + I y, {y, -Sqrt[2]/2, Sqrt[2]/2}] (* -i π *) 

There are two branch points:

$$ z=-\frac{1}{2}\pm i \frac{\sqrt{2}}{2} $$

we can set the branch cuts connecting these two points and set up a contour like this(sorry for the poor drawing):

The two small circles in green near the two singularities has no contribution, since the the function goes as $\frac{1}{\sqrt{z}}$. And the four blue lines has no contribution too,because they cancel each other in pairs since they are on the same branch. So the only contribution comes from the integration between the branch points in red. We can do this line integral easily:

-2i Integrate[1/Sqrt[4 z^2 + 4 z + 3] /. z -> -1/2 + I y, {y, -Sqrt[2]/2, Sqrt[2]/2}] (* -iπ *) 

There are two branch points:

$$ z=-\frac{1}{2}\pm i \frac{\sqrt{2}}{2} $$

we can set the branch cuts connecting these two points and set up a contour like this  (sorry for the poor drawing):

The two small circles in green near the two singularities have no contribution, since the function goes as $\frac{1}{\sqrt{z}}$. And the four blue lines have no contribution too, because they cancel each other in pairs since they are on the same branch. So the only contribution comes from the integration between the branch points in red. We can do this line integral easily:

-2i Integrate[1/Sqrt[4 z^2 + 4 z + 3] /. z -> -1/2 + I y, {y, -Sqrt[2]/2, Sqrt[2]/2}] (* -i π *) 

There are two branch points:

$$ z=-\frac{1}{2}\pm i \frac{\sqrt{2}}{2} $$

we can set the branch cuts connects these two points and set up a contour like this:

Since the small circles near the two singularities has no contribution, the only contribution comes from the integration between the branch points:

-2i Integrate[1/Sqrt[4 z^2 + 4 z + 3] /. z -> -1/2 + I y, {y, -Sqrt[2]/2, Sqrt[2]/2}] (* -iπ *) 

There are two branch points:

$$ z=-\frac{1}{2}\pm i \frac{\sqrt{2}}{2} $$

we can set the branch cuts connecting these two points and set up a contour like this(sorry for the poor drawing):

The two small circles in green near the two singularities has no contribution, since the the function goes as $\frac{1}{\sqrt{z}}$. And the four blue lines has no contribution too,because they cancel each other in pairs since they are on the same branch. So the only contribution comes from the integration between the branch points in red. We can do this line integral easily:

-2i Integrate[1/Sqrt[4 z^2 + 4 z + 3] /. z -> -1/2 + I y, {y, -Sqrt[2]/2, Sqrt[2]/2}] (* -iπ *)