The following problem arises in E. N. Economou's ''Green's Functions in Quantum Physics'', 3e.
Evaluate the following integral over the unit circle $$ \oint\frac{dw}{(w-w_+)(w-w_-)} $$ where $w_\pm = z\pm\sqrt{z^2-1}$, and $z\in\mathbb{C}$ such that $|w_+|\ne|w_-|$.
Supposing $w_+$ is the root inside the unit circle, the answer is
$$ \frac{1}{w_+-w_-}=\frac{1}{2\sqrt{z^2-1}} $$
I'm fine up to this part. Economou then says
where the square root [in the answer above] is the one that has the same sign of the imaginary part as the sign of $\mbox{Im}\,z$.
Could someone explain what justifies this choice of the principal branch? Forgive me if this is an obvious question.
Thank you!
Let $z=x+iy$ and $s = u+iv$ be a square root of $z^2-1$. ($x, y, u, v$ all real numbers). Then $$ \begin{align} u^2-v^2 &= x^2-y^2-1 \, ,\\ 2uv &= 2xy \, . \end{align} $$ $w_+ = z+s$ and $w_- = w-s$ are the roots of $w^2-2zw+1=0$, so that their product is equal to one. It is also assumed that $|z+s| \ne |z-s|$, which means that exactly one of the roots has modulus less than one. Therefore $$ \begin{align} |w_+| < 1 &\iff |z+s| ^2 < |z-s|^2 \\ &\iff 0 < \operatorname{Re} (z \bar s) \\ & \iff xu + yv > 0 \, . \end{align} $$
If $y \ne 0$ then $x=uv/y$ and $$ xu + yv = \frac vy (u^2 + y^2) $$ is positive if and only if $v/y$ is positive, that is if $v$ has the same sign as $y$.
And if $y=0$ then necessaily $v=0$ because otherwise $2uv = 2xy=0$ would imply $u = 0$ and then $xu + yv = 0$.
So $|w_+| < 1$ exactly if $s = \sqrt{z^2-1}$ where the square root is chosen such that its imaginary part and the imaginary part of $z$ are both zero or have the same sign.