<p>sinZ=cosh3</p> Solution sin z = cosh 3 e^(iz) - e^(-iz) = 2i cosh(3) ==> e^(2

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sin z = cosh 3 e^(iz) - e^(-iz) = 2i cosh(3) ==> e^(2iz) - 2icosh(3) e^(iz) -1 =0. This is of quadratic type. So e^(iz) = (1/2) ( 2i cosh(3) + sqrt( 4 - 4 cosh^2(3))) or e^(iz) = (1/2) ( 2i cosh(3) - sqrt( 4 - 4 cosh^2(3))) We can use that 1 - cosh^2(3) = - sinh^2(3). So e^(iz) = i (cosh(3) + sinh(3)) or e^(iz) = i (cosh(3) - sinh(3)) e^(iz) = i e^3, or e^(iz) = i e^(-3) Taking logs gives iz = 3 + pi/2 i or iz = -3 +pi/2i. You can solve for z and add multiples of 2pi to generalize the solutions. I'll finish the other in a few minutes. ***Update*** The statement tan(z) = i is equivalent to itan(z) = -1. itan(z) = (e^(iz) - e^(-iz))/(e^(iz) + e^(-iz)) If this were equal to -1, we can multiply through by the denominator to get the equation e(iz) - e^(-iz) = - (e(iz) + e^(-iz)) = -e(iz) - e^(-iz) This gives e^(iz) = - e^(iz) which would require e^(iz) = 0. But e^(iz) is never zero. Hence the original equation has no solution.

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