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I want to solve for $\gamma$ in the following equation:

$$ \frac{1-e^{-2\cdot 10^{-3}\gamma}}{1+e^{-2\cdot 10^{-3}\gamma}}=1.54e^{i0.24\pi} $$

Attempt: \begin{align} 1-e^{-2\cdot10^{-3}\gamma}&=1.54e^{i0.24\pi}(1+e^{-2\cdot 10^{-3}\gamma}) \\ &=1.54e^{i0.24\pi}+1.54e^{i0.24\pi}e^{-2\cdot 10^{-3}\gamma} \end{align} \begin{align} 1-e^{-2\cdot10^{-3}\gamma}-1.54e^{i0.24\pi}e^{-2\cdot 10^{-3}\gamma} =1.54e^{i0.24\pi} \end{align} \begin{align} 1-e^{-2\cdot 10^{-3}\gamma} (-1-1.54e^{i0.24\pi}) =1.54e^{i0.24\pi} \end{align} \begin{align} -e^{-2\cdot 10^{-3}\gamma} (-1-1.54e^{i0.24\pi}) =1.54e^{i0.24\pi}-1 \end{align} \begin{align} e^{-2\cdot 10^{-3}\gamma} (-1-1.54e^{i0.24\pi}) =-1.54e^{i0.24\pi}+1 \end{align} \begin{align} e^{-2\cdot 10^{-3}\gamma} = \frac{-1.54e^{i0.24\pi}+1}{-1-1.54e^{i0.24\pi}} \end{align} Stuck with this fraction, how can I simplify it? \begin{align} -2\cdot 10^{-3}\gamma = \ln \bigg (\frac{-1.54e^{i0.24\pi}+1}{-1-1.54e^{i0.24\pi}}\bigg ) \end{align} \begin{align} \gamma =\frac{1}{-2\cdot 10^{-3}} \ln \bigg (\frac{-1.54e^{i0.24\pi}+1}{-1-1.54e^{i0.24\pi}}\bigg ) \end{align}

Correct answer is approximately: $$ 2\cdot 10^{-3}\gamma=\ln (2.24 e^{i0.68\pi}) $$ $$ \gamma= 0.0403+i0.108=0.115 e^{i0.386\pi} $$ Thanks!

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You can use $e^{i\theta }=\cos \theta + i\sin \theta $, and then multiply and divide by the conjugate of denominator to simplify further.


By the way, it would have been easier to solve, if you had introduced another variables, I.e.

$$e^{-2 \times 10^{-3} \gamma}=t ; \quad 1.54 e^{i 0.24\pi} =k$$

Now rewrite the equation as $$\frac{1-t}{1+t}=k \implies^{\text{(C & D)}} \; t =\frac{1-k}{k+1}$$

And now substitue in terms of $\gamma$ back into obtained equation.

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