How can we show| sin(z| > 1

Show that, if z is a non real complex such that sin(z) is real, then|sin(z)| > 1.

Thank you.

Use the identity sin(x+iy)=sin(x)cosh(y)+icos(x)sinh(y) .

Since sin(z) = sin(x + iy) is real, cos(x)sinh(y) must be 0 (from the identity). Since z is non-real, y is not 0, so sinh(y) is not 0. So cos(x) = 0, which implies x = π/2 + πn for some integer n, and we must have:
|sin(z)| = |sin(π/2 + πn)cosh(y)| = |cosh(y)| > 1.