Suppose f (z) is real-valued and differentiable for all imaginary points z. Show that f

Suppose f (z) is real-valued and differentiable for all imaginary points z. Show that f ´(z) is imaginary for at all imaginary points z.

The answer follows from the nature of analytic functions; specifically, from the Cauchy Riemann equations.

taking z = x + iy and
f(z) = u(x,y) + iv(x,y)
the Cauchy Riemann equations state that
∂u/∂x = ∂v/∂y
∂u/∂y = -∂v/∂x

Given the specifications of the problem, we know that f(z) is real valued, which means that v(0,y) = 0 for all y. Thus, at x=0, ∂v/∂y = ∂/∂y(0) = 0

Since f(z) is differentiable over the complex numbers (and therefore analytic), we know that it follows the Cauchy Riemann equations. Thus, we know that whenever x=0,
f'(z) = f'(0+iy) = ∂/∂x [u + iv] = ∂u/∂x + i * ∂v/∂x

However, by the C-R equations, ∂u/∂x = ∂v/∂y, and by the above argument, ∂v/∂y = 0. Thus,
f'(z) = 0 + i * ∂v/∂x
Since the real part of f'(z) is zero, f'(z) is purely imaginary.
QED