How to simplify a given expression?

Question

Suppose $$ y = 10 \log \left|1 + \frac{a}{2b} \left[ \frac{1+\phi+(\phi-1) e^{-2i \theta_{2}}}{1+\phi-(\phi-1)e^{-2i\theta_{2}}} \right] \right|^{2} $$ where $$ \phi = \frac{c}{a} \left( \frac{e^{2i \theta_{1}}-1}{e^{2i \theta_{1}}+1} \right) .$$ Here $a,b,c,\theta_{1}$ and $\theta_{2}$ are fixed constants. We need to prove that $$ y = 10 \log \left[ 1+ \left( \frac{a}{2b} \frac{\tan{\theta_{2}+\frac{c}{a} \tan{\theta_{1}}}}{1-\frac{c}{a} \tan{\theta_{2}} \tan{\theta_{1}}} \right)^{2}\right]. $$ My idea: I try to rationalize $y$ and $\phi$ separately and then divided whole expression by $cosine$ function, but not able to remove imaginary parts, please help me to simplify this.

Answer 1

Note $$|1+z|^2=(1+z)(1+\overline{z})=1+(z+\overline{z})+|z|^2$$ If $z$ is purely imaginary, then $z=-\overline{z}$ and the above reduces to $1+|z|^2$. I claim that's the situation we have here, so I'll reduce the problem to showing that, defining $m:=c/a$ and $t_i := \tan\theta_i$, $$\frac{1+\phi+(\phi-1)\exp(-2i\theta_2)}{1+\phi-(\phi-1)\exp(-2i\theta_2)}= i\;\frac{t_2+mt_1}{1-mt_1t_2} \tag1$$ (where I'm assuming all constants are real). First, we observe that $$\phi = imt_1 \qquad \exp(-2i\theta_2) = \frac{i+t_2}{i-t_2} \tag{2}$$ so that we can write the left-hand side of $(1)$ as $$\frac{(1+imt_1)(i-t_2)+(imt_1-1)(i+t_2)}{(1+imt_1)(i-t_2)-(imt_1-1)(i+t_2)} \tag3$$ which readily simplifies to the right-hand side of $(1)$. $\square$