Show that cos²Z + sin²Z = 1 in complex variables?

Question

It requires the use of trigonometry in proving this.

Answer 1

QUESTION: Show that cos²Z + sin²Z = 1 in complex variables?

Let z = x + iy

cos(z) = cos(x + iy) = cos(x).cos(iy) – sin(x).sin(iy)
cos²(z) = cos²(x).cos²(iy) – 2.cos(x).cos(iy).sin(x).sin(iy) + sin²(x).sin²(iy)

sin(z) = sin(x + iy) = sin(x).cos(iy) + cos(x).sin(iy)
sin²(z) = sin²(x).cos²(iy) +2.sin(x).cos(iy).cos(x).sin(iy) + cos²(x).sin²(iy)

Adding together the expressions for cos²(z) and sin²(z),

cos²(z) + sin²(z) = cos²(x).cos²(iy) + sin²(x).sin²(iy)+ sin²(x).cos²(iy) + cos²(x).sin²(iy)
cos²(z) + sin²(z) = (cos²(x) + sin²(x)).cos²(iy) + (sin²(x) + cos²(x)).sin²(iy)
cos²(z) + sin²(z) = cos²(iy) + sin²(iy)
cos²(z) + sin²(z) = 1