# SImplifying help in solve two simultaneous equations

I have $\lambda \in \mathbb{C}$ being a complex number satisfying the equation $f(\lambda) = 0$ where

$$f(\lambda) = \lambda + r \frac{a + b e^{-\lambda T}}{a + b}.$$

My question says:

Show that if parameter $T$ is continuosly increased from $0$, the instabililty of the equilibirum is first ovserved when the following two conditions are simultaneously satisfied:

$$\cos(\omega T) = -\frac{a}{b}$$ $$\sin(\omega T) = (1 + a/b) \frac{\omega}{r}$$

where $\omega$ is the imaginary part of $\lambda$. Thus, or otherwise, find the minimal value of the explicit time delay $T$ for which the equilibirum will become unstable, as a function of $r,a$ and $b$.

So I basically need to solve these two simultaneous equations for $T$. From the first one, we get that

$$\cos(\omega T) = -\frac{a}{b} \implies \cos^2 (\omega T) = \frac{a^2}{b^2} \implies \sin(\omega T) = \pm \sqrt{1 - \frac{a^2}{b^2}}.$$

Subbing this into the second one gives me

$$\pm \sqrt{1 - \frac{a^2}{b^2}} = (1 + a/b) \frac{\omega}{r}$$

I then get $\omega$ to be

$$\pm\frac{r\sqrt{(b+a)(b-a)}}{b+a} = \omega$$

but the answers say it should be

$$\pm r \sqrt{\frac{b - a}{b + a}}$$

How do they get this?

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$\frac{\sqrt{x}}{x} = \frac{1}{\sqrt{x}}$. – copper.hat May 1 '13 at 20:08
@copper.hat Thanks – Kaish May 1 '13 at 20:11
You're welcome. Your effort is appreciated. – copper.hat May 1 '13 at 20:11