show that the equation tan(z)= -i has no solutions?

Answer 1

Hint: Use that tan^2(z) + 1 = sec^2(z)

And then show that sec(z) cannot be zero.

Answer 2

Tangent is based upon real, measurable dimensions. The tan(90) is infinity because the adjacent side of that "triangle" is 0. It doesn't get any freakier than that: There may be negative numbers if direction, etc. is important. But, there is nothing imaginary anywhere.

Answer 3

tan z=( e^iz -e^-iz)/(e^iz+e^-iz)*(-i) =-i
calling eîz= u
(u-1/u)=u+1/u so 2/u=0 impossible