I am asked to prove the following: (Note: z = x + iy) |cos(z)|2 = cos2x + sinh2y --------------- So I started the following way: |cos(z)|2 = |cos(x+iy)|2 = |cos(x)cosh(y) - i(sin(x)sinh(y))|2 = cos2(x)cosh2(y) + sin2(x)sinh2(y) [after having square root squared removed] once I got here I was stuck. I am just not seeing how we can get this to equal cos2x + sinh2y Is there some silly trig identity I don't know? Or did I make a mistake? Any ideas? Thanks!