Index of Refraction in Metal: Approximating Complex Perturbation If you consider waves in a metal, you can write the index of refraction for the metal as,
$$ n^2 = 1 - \frac{\omega_p^2}{\omega^2} $$
I am interested in what will happen if the index is perturbed by some small complex quantity,
$$ n^2 = 1 - \frac{\omega_p^2}{\omega^2} - i\epsilon $$
In the low frequency limit, this usually corresponds to an attenuation. To compute quantities that depend on $n$ (such as absorption), I would like to expand $n$ since $\epsilon$ is small.
The function is,
$$ n = \sqrt{1 - \frac{\omega_p^2}{\omega^2}-i\epsilon} $$
So I write,
$$ n = \sqrt{1 - \frac{\omega_p^2}{\omega^2}}\sqrt{1 - i \frac{\epsilon}{1 - \omega_p^2/\omega^2}} $$
For this problem $\omega_p > \omega$, so I rewrite,
$$ n = i \sqrt{\frac{\omega_p^2}{\omega^2}-1}\sqrt{1 + i \frac{\epsilon}{\omega_p^2/\omega^2 - 1}} $$
And, since $\epsilon << 1$, I tried to expand this:
$$ \approx i \sqrt{\frac{\omega_p^2}{\omega^2}-1}\left(1 + i \frac{\epsilon}{2(\omega_p^2/\omega^2-1)} \right) = \sqrt{\frac{\omega_p^2}{\omega^2}-1}\left(i - \frac{\epsilon}{2(\omega_p^2/\omega^2-1)} \right)$$
As a sanity check, I tried plugging in some numbers. $\omega_p=2$, $\omega=1$, and $\epsilon=0.01$. The expansion is off by a factor of $-1$. I believe it is due to the branch cut of the complex square root. If I place $i$ back into the square root,
$$ \sqrt{-1 - i \frac{\epsilon}{\omega_p^2/\omega^2-1}} $$
is in the third quadrant of the complex plane. This means that the square root should be in the fourth quadrant, where I have a positive real part and negative imaginary part (in contrast to my expansion above, where the opposite is the case).
I spent a long time hunting for this negative, so my question is, is there a cleaner way to expand the function where I won't have this branch cut issue?
 A: I'm not sure I understand your question correctly, but let me know if the below helps. 
Let  \begin{array}$ n^2(\omega) &= 1 - \frac{\omega_p^2}{\omega^2} - i\epsilon\\
&=-\Omega-i\epsilon\end{array}
where $\Omega:=-\left(1 - \frac{\omega_p^2}{\omega^2}\right)>0$ by assumption. 
Then, expanding $n(\omega)$ for small $\epsilon$ we find (using some algebra-software: is this OK with you?)
$$\sqrt{-\Omega-i\epsilon} = -i\cdot\mathrm{csgn}(i(-\Omega-i\epsilon))\sqrt{\Omega}+\frac{1}{2}\frac{\mathrm{csgn}(i(-\Omega-i\epsilon))}{\sqrt{\Omega}}\epsilon+\mathcal{O}(\epsilon^2)$$
where the csgn is defined on this page as:  

In our case we get: 
$$\sqrt{-\Omega-i\epsilon} = -i\sqrt{\Omega}+\frac{1}{2}\frac{1}{\sqrt{\Omega}}\epsilon +\mathcal{O}(\epsilon^2)$$ which lies in the fourth quadrant. 
Does this answer your question? 
A: There is a formula, which is checked by squaring the right side $$\sqrt{a+ib}=\pm \sqrt{(\sqrt{a^2+b^2}+a)/2}\pm i\sqrt{(\sqrt{a^2+b^2}-a)/2}$$, the branches of this formula must be chosen from the condition $$\sqrt{a+ib}=(a^2+b^2)^{0.25}exp(iarg(a+bi)/2+i\pi k),k=0,1$$. Depending on the value $arg(a+bi)/2$, choose the root sign.The exact formula is determined up to a multiplier $\pm 1$.
Therefore we have
 $$\sqrt{1-i\frac{\epsilon}{\omega_p^2/\omega^2-1}}=\pm\sqrt{(\sqrt{1+(\frac{\epsilon}{\omega_p^2/\omega^2-1})^2}+1)/2}\pm i\sqrt{(\sqrt{1+(\frac{\epsilon}{\omega_p^2/\omega^2-1})^2}-1)/2}=\pm [1+(\frac{\epsilon}{\omega_p^2/\omega^2-1})^2]^{0.25}(1- i\frac{\epsilon}{\omega_p^2/\omega^2-1}/2+O(\epsilon^2))$$
A: I think you're hitting something equivalent to the classic high school math class riddle: Find the flaw in the equation
$$1 = \sqrt{1} = \sqrt{-1\times-1}=\sqrt{-1}\times\sqrt{-1}=i^2=-1$$
You do a step near the beginning involving $\sqrt{AB} \rightarrow \sqrt{A}\sqrt{B}$, which is not always true, instead you should replace $\sqrt{AB}\rightarrow\pm\sqrt{A}\sqrt{B}$.
At the very end, you should get an equation that looks like $n=\pm(\text{something})$, and then you pick the sign based on what quadrant you know $n$ is supposed to be in.
