Solve coshz=2 and separate real and Imaginary part?
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Expert's answer
2020-10-17T14:50:12-0400
coshz=2⇒ez+e−z=4 . Let ez=a. Then a+a1=4⇒a2−4a+1=0. Hence a=2±3. Hence ez=2±3. Now if z=x+iy⇒ex(cosy+isiny)=2±3. As it has no imaginary part, so y=nπ. Since RHS >0,excosy>0 . Hence n=2m.∴ex=2±3⇒x=ln(2±3)∴z=ln(2±3)+i2mπ.
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