Question #138183
Solve coshz=2 and separate real and Imaginary part?
1
Expert's answer
2020-10-17T14:50:12-0400

coshz=2ez+ez=4cosh z=2\Rightarrow e^z+e^{-z}=4 . Let ez=a.e^z=a. Then a+1a=4a24a+1=0.a+\frac{1}{a}=4\Rightarrow a^2-4a+1=0. Hence a=2±3.a= 2\pm\sqrt{3} . Hence ez=2±3.e^z= 2\pm\sqrt{3}. Now if z=x+iyex(cosy+isiny)=2±3.z=x+iy\Rightarrow e^x(cos y+i sin y)=2\pm\sqrt{3}. As it has no imaginary part, so y=nπ.y=n\pi. Since RHS >0,>0, excosy>0e^xcosy >0 . Hence n=2m.n=2m. ex=2±3x=ln(2±3)\therefore e^x=2\pm\sqrt{3}\Rightarrow x=ln(2\pm\sqrt{3}) \therefore z=ln(2±3)+i2mπ.z=ln(2\pm \sqrt{3})+i2m\pi.


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