Derivative of Real Valued Complex Function Is Zero

Theorem

Let f:U→C be a differentiable function for U⊆C.


If f(z)∈R for all z∈U then f′(z)=0 for all z∈U and hence f is a constant function.

Proof

Write f(x+iy)=u(x,y)+iv(x,y) and notice that because f(x+iy)∈R, we have that v(x,y)=0 and subsequently

∂v∂x=∂v∂y=0.

By the Cauchy-Riemann equations we also then have that

∂u∂x=∂u∂y=0

and subsequently that f′=0.

f must then be the constant function by zero derivative implies constant complex function.