Stimulated emission in laser can be described by the classical oscillator theory (A.E. Siegman Lasers chapter 2 University Science Books 1986).
According to Siegman, the overall susceptibility is $$\tilde{\chi}={Ne^2\over m\epsilon}{1\over \omega_a^2-\omega^2+j\omega\Delta\omega_a}$$
The exact form of the imaginary part: $${\chi}_e={-Ne^2\over m\epsilon}{j\omega\Delta\omega_a\over (\omega_a^2-\omega^2)^2+\omega^2{\Delta\omega_a}^2}$$
When $\omega\approx \omega_a$, $\omega_a^2-\omega^2 \approx 2\omega_a(\omega_a-\omega)$, the exact form of the imaginary part can be approximated by: $${\chi}_a={-Ne^2\over m\epsilon}{j\omega_a\Delta\omega_a\over 4\omega_a^2(\omega_a-\omega)^2+\omega_a^2{\Delta\omega_a}^2}$$
In exercise question 5 of section 2.4, the author asks how far $\omega$ can deviate from $\omega_a$, so that
$${\left\vert {\chi}_e - {\chi}_a\right\vert\over {\chi}_e}=0.1$$
If I do it in a straight forward way, I will get a fourth order polynomial of $\omega$, which doesn't seem possible to get the solution. What is the correct approach here?
For this question, both $\omega_a$ and $\Delta\omega_a$ are known quantities.
This is the original question. Original question
Well, expanding in one-term Taylor series in $\omega$ centered at $\omega_a$, we have (cases are to make the absolute value pick a sign) \begin{align*} \omega &> \omega_a &&: & \frac{|\chi_e - \chi_a|}{\chi_e} &\approx \frac{\omega - \omega_a}{\omega_a} \\ \omega &< \omega_a &&: & \frac{|\chi_e - \chi_a|}{\chi_e} &\approx -\frac{\omega - \omega_a}{\omega_a} \end{align*} From these, the range of $\omega$ is $-[0.9, 1.1]\omega_a$. The minus sign is because $\chi_e$ in the denominator is negative. It is far more common to write the relative error as either of $$ \left| \frac{\chi_e - \chi_a}{\chi_e} \right| = \left| 1 - \frac{\chi_a}{\chi_e} \right| $$ to correctly deal with negative and nonzero exact values. See the brief discussion of "relative error" at [https://en.wikipedia.org/wiki/Approximation_error#Formal_definition].
Expanding to two terms, \begin{align*} \omega &> \omega_a &&: & \frac{|\chi_e - \chi_a|}{\chi_e} &\approx -\frac{\omega - \omega_a}{\omega_a} + \frac{4 (\omega - \omega_a)^3}{\omega_a \Delta\omega_a^2} \\ \omega &< \omega_a &&: & \frac{|\chi_e - \chi_a|}{\chi_e} &\approx \frac{\omega - \omega_a}{\omega_a} - \frac{4 (\omega - \omega_a)^3}{\omega_a \Delta\omega_a^2} \end{align*} We could solve this cubic, but determining which root is real requires knowing which of $\omega_a > \frac{10}{3 \sqrt{3}}\Delta\omega_a$, $\omega_a = \frac{10}{3 \sqrt{3}}\Delta\omega_a$, or $\omega_a \leq \frac{10}{3 \sqrt{3}}\Delta\omega_a$ holds. You didn't indicate what is "big" and what is "small", so we don't have the information to resolve this. If it happens that $\omega_a \gg \Delta \omega_a$, then the range is $\omega \in [\omega_a -r, \omega_a + r]$ where $$r = \frac{\sqrt[3]{\frac{5}{3}} \Delta\omega _a^2}{2^{2/3} \sqrt[3]{9 \Delta\omega _a^2 \omega _a+\sqrt{3} \sqrt{27 \Delta\omega _a^4 \omega _a^2-100 \Delta\omega _a^6}}}+\frac{\sqrt[3]{9 \Delta\omega _a^2 \omega _a+\sqrt{3} \sqrt{27 \Delta\omega _a^4 \omega _a^2-100 \Delta\omega _a^6}}}{2\ 3^{2/3} \sqrt[3]{10}} \text{.} $$ While that is certainly a thing which can be computed, I don't see that it provides much insight, so is probably not the desired answer.