Complex no.

[man111]

If [tex]z_{1}[/tex] and [tex]z_{2}[/tex] are two distinct complex no. such that [tex]|z_{1} | = |z_{2}|[/tex]

and [tex]Re(z_{1})>0[/tex] and [tex]Im(z_{2})<0[/tex].

Then Calculate value of [tex]\frac{z_{1}+z_{2}}{z_{1}-z_{2}}=[/tex]

[mkmarinov]

Let [tex]z_1=e^{ia}, z_2=e^{ib}[/tex]
WLOG [tex]|z_1|=|z_2|=1[/tex]
From what is given it follows that [tex]cosa>0, sinb<0[/tex]

[tex]\frac{z_1+z_2}{z_1-z_2}=\frac{e^{ia}+e^{ib}}{e^{ia}-e^{ib}}=\frac{cosa+cosb+i(sina+sinb)}{cosa-cosb+i(sina-sinb)}=\\=\frac{\cos\frac{a+b}{2}\cos\frac{a-b}{2}+i\sin\frac{a+b}{2}\cos\frac{a-b}{2}}{-\sin\frac{a+b}{2}\sin\frac{a-b}{2}+i\sin\frac{a-b}{2}\cos\frac{a+b}{2}}=\frac{\cos\frac{a-b}{2}}{\sin\frac{a-b}{2}}.\frac{\cos\frac{a+b}{2}+i\sin\frac{a+b}{2}}{-\sin\frac{a+b}{2}+i\cos\frac{a+b}{2}}=\\=-i.cotg\frac{a-b}{2}.\frac{e^{i\frac{a+b}{2}}}{e^{\frac{a+b}{2}}}=-icotg\frac{a-b}{2}[/tex]

Couldn't simplify it any more.