Show that $$|c|^2= \frac{4k^2}{k^2 +\gamma^2}$$ given (1)$$a+b=c$$ and (2)$$ik(a-b)=-\gamma c$$ This was given in a lecture without proof, so there's probably a very simple way of proving the equality. This is a purely algebraic problem.
Using (1) and (2) I can show that $$b=a(\frac{k-i\gamma}{k+i\gamma})$$ and similarly $$a=b(\frac{k+i\gamma}{k-i\gamma})$$ Which leads me to $$|b|^2=bb^*=a^*(\frac{k+i\gamma}{k-i\gamma})a(\frac{k-i\gamma}{k+i\gamma})=aa^*=|a|^2$$ Where $i=\sqrt-1$, $k$ and $\gamma$ are unknowns; $a$, $b$ and $c$ are real constants.
My attempt at showing the top relation was to take the complex conjugate of (2) for $c$ such that $$c=-\frac{ik(a-b)}{\gamma}$$ and $$c^*=\frac{ik(a-b)}{\gamma}$$ and then using the identity $$|c|^2=cc^*$$ and eliminating $a$ and $b$ but still unable to get the desired result.
Could someone please show me step by step how to obtain $$|c|^2= \frac{4k^2}{k^2 +\gamma^2}$$
Thank you.
Let's use the first equation to eliminate $b$. Then:
$$ik(2a - c) = -\gamma c$$
$$c = \frac{2ik}{ik - \gamma}a$$
Multiply by $-ik - \gamma$:
$$c = \frac{k - i\gamma}{k^2 + \gamma^2}2ka$$
Now we compute:
$$|c|^2 = \frac{k^2 + \gamma^2}{(k^2 + \gamma^2)^2} \cdot 4k^2|a|^2 = \frac{4k^2}{k^2 + \gamma^2}|a|^2$$
So I'm guessing $|a| = 1$, otherwise your identity is not true. (I've also assumed $k$ and $\gamma$ are real and not both zero, otherwise the expression $\frac{4k^2}{k^2 + \gamma^2}$ might not be negative or have an imaginary component which wouldn't make sense)