How can we show sin(z) > 1 or sin(z) < -1

Let z be a non real complex such that sin(z) is real. Show that sin(z) > 1 or sin(z) < -1.

Thank you.

sin(x + iy) = sin(x)cosh(y) + icos(x)sinh(y)

We are given that sin(z) is real, so if z = x + iy then
(1) sin(z) = sin(x) cosh(y)
and
(2) icos(x)sinh(y) = 0

From (2), either cos(x)=0 or sinh(y)=0.

If sinh(y) = 0 then y=0, so x + iy is real. We know that z is not real, so we know y is nonzero.

So cos(x)=0, which tells us x = (pi)/2 + n(pi) for some integer n. But then sin(x) = 1 or -1, giving us
(3) sin(z) = cosh(y) or -cosh(y)

But the range of cosh is (-infinity, -1] U [1, infinity), so (3) gives us:
sin(z) > 1 or sin(z) < -1
as required.

If sin(z) is real, then it cannot be greater than 1 or less than -1.
Remember: sin(x)= (e^(xi) -e^(-xi))/ (2i)