Complex Equation Formula

Question

Can someone show me how the following two expressions are equivalent:

$$\Gamma = \frac{i X - R_c}{i X + R_c} = -e^{-i 2 \mathrm{tan}^{-1} (\frac{X}{R_c})}$$

I'm working through a calculation and I am not sure how this step is done.

Answer 1

Hint

Start with $$\Gamma = \frac{i X - R_c}{i X + R_c}=\frac{i X - R_c}{i X + R_c}\times\frac{i X - R_c}{i X - R_c}=\frac{X^2-R_c^2}{X^2+R_c^2}+i\frac{2 R_c X}{X^2+R_c^2}$$ and work with Euler formula and a few trigonometric identities.

Answer 2

you may be able to get some mileage from an argument along the following lines. at least while you seek a more purely algebraic solution

call the two expressions $L(X)$ and $R(X)$ as functions of $X$ $$ L(X) = \frac{i X - R_c}{i X + R_c} \\ R(X) = -e^{-i 2 \mathrm{tan}^{-1} (\frac{X}{R_c})} $$

then $$ L' = \frac{2iR_c}{(iX+R_c)^2} $$ and $$ R' = \Gamma\frac{d}{dX}\left(-2i\tan^{-1} \frac{X}{R_c} \right) = -2i\frac{\Gamma}{R_c} \frac1{1+\left(\frac{X}{R_c}\right)^2} \\ = \frac{-2iR_c\Gamma}{X^2+R_c^2} $$ substituting $\Gamma = -\frac{R_c-i X }{{R_c+i X }}$ shows that $L'=R'$

moreover, the expressions $L(X)$ and $R(X)$ give the same result, $i$, when $X=R_c$