Change of expression of complex number

Question

Given that $\dfrac{5+z}{5-z} = e^{i\beta}$ where $z$ is a complex number, how do I show that $z$ can be expressed as $5i\cdot \tan\left(\dfrac \beta2\right)$?

Answer 1

$\frac {5 + z}{5-z} = e^{i\beta}$

$5 + z = 5e^{i\beta} - ze^{i\beta}$

$z + ze^{i\beta} = z(1 + e^{i\beta}) = 5(e^{i\beta}-1)$

$z = 5\frac {e^{i\beta}-1}{1 + e^{i\beta}}$

$z = 5(\frac {\cos \beta +i\sin \beta - 1}{1 + \cos \beta + i\sin \beta})$

Okay... we know we want to get something in terms of $\frac \beta 2$ so ...

$z = 5(\frac {\cos 2(\frac\beta 2) +i\sin 2(\frac\beta 2)- 1}{1 + \cos 2(\frac\beta 2) + i\sin 2(\frac\beta 2)})$

$z = 5\frac {(\cos^2 \frac \beta 2 - \sin^2 \frac \beta 2) + i(2 \sin \frac \beta 2 \cos \frac \beta 2) -1}{1 + (\cos^2 \frac \beta 2 - \sin^2 \frac \beta 2) + i(2 \sin \frac \beta 2 \cos \frac \beta 2)}$

$z = 5\frac { - \sin^2 \frac \beta 2 +2i \sin \frac \beta 2 \cos \frac \beta 2+(\cos^2 \frac \beta 2 -1)}{(1 - \sin^2 \frac \beta 2) + \cos^2 \frac \beta 2 +2i \sin \frac \beta 2 \cos \frac \beta 2}=5\frac { - 2\sin^2 \frac \beta 2 +2i \sin \frac \beta 2 \cos \frac \beta 2}{2\cos^2 \frac \beta 2+2i \sin \frac \beta 2 \cos \frac \beta 2}$

$= 5\frac {\sin \frac \beta 2 (- \sin \frac \beta 2 +i \cos \frac \beta 2)}{\cos \frac \beta 2(\cos \frac \beta 2+i \sin \frac \beta 2)}$

$= 5\frac {\sin \frac \beta 2 *i(i \sin \frac \beta 2 + \cos \frac \beta 2)}{\cos \frac \beta 2(\cos \frac \beta 2+i \sin \frac \beta 2)}$

$= 5i\frac {\sin \frac \beta 2 *i(\cos \frac \beta 2+ i \sin \frac \beta 2 )}{\cos \frac \beta 2(\cos \frac \beta 2+i \sin \frac \beta 2)}=5i\tan{\frac \beta 2}$