Answer to Question #138183 in Complex Analysis for Mubina

Question

Solve coshz=2 and separate real and Imaginary part?

Answer 1

$cosh z=2\Rightarrow e^z+e^{-z}=4$ . Let $e^z=a.$ Then $a+\frac{1}{a}=4\Rightarrow a^2-4a+1=0.$ Hence $a= 2\pm\sqrt{3} .$ Hence $e^z= 2\pm\sqrt{3}.$ Now if $z=x+iy\Rightarrow e^x(cos y+i sin y)=2\pm\sqrt{3}.$ As it has no imaginary part, so $y=n\pi.$ Since RHS $>0,$ $e^xcosy >0$ . Hence $n=2m.$ $\therefore e^x=2\pm\sqrt{3}\Rightarrow x=ln(2\pm\sqrt{3})$ $\therefore$ $z=ln(2\pm \sqrt{3})+i2m\pi.$