Is d/dz(Im(f(z))) always equal to Im(f'(z)) for an analytic function f(z)?

[bmanbs2]

So does d/dz(Im(f(z))) = Im(f'(z))?

For context, if Im(f'(z)) > 0, is Im(f(z)) always increasing and one-to-one?

f(z) is analytic

 

[Hurkyl]

You can compute what it should be from the definitions, can't you? Are you sure I am f(z) is even complex differentiable?

 

[bmanbs2]

All I know is that f(z) is analytic.

 

[HallsofIvy]

You didn't say that before! But, given that f is analytic, f and its real and imaginary parts are certainly differentiable- now, let f(z)= g(z)+ ih(z) and simply write out the two sides of your equation. What is Im(df/dz)? What is d(Im(f(z))/dz?

 

[Hurkyl]

But, given that f is analytic, f and its real and imaginary parts are certainly differentiable

... as real-valued functions of two real variables (the real and imaginary parts of z), but not necessarily as a function of one complex variable.


edit: or are you thinking about the differential-geometry-like treatment of differentiation where, e.g., df splits into a linear combination of dz and d\bar{z}, and such? (this is directed both at HallsofIvy and the original poster)