Derivative of complex functions Prove that if $f'(z) = 1$ everywhere in a domain $D$ then $f(z) = z + c$
for some complex constant $c.$
And $z$ is a complex no.
 A: Hint. Try showing that $g(z) = f(z) - z$ is holomorphic and constant. (I'm guessing your definition of a domain requires it to be connected.)
A: $D$ is a domain and connected.  Assuming each point can be connected by a piecewise smooth curve $\gamma$ to every other point in the domain, and since $f^\prime$ is holomorphic and has a primitive $f$:
Without loss of generality, assume the origin is in the domain and $f(0)=z_0.$
$$\int_\gamma f^\prime(z) dz = \int_0^z f^\prime(\zeta) \,d\zeta=f(z)-f(0).$$
But $$\int_0^z d\zeta = z,$$
So $$f(z) =z + z_0.$$
A: The chain rule is true of complex-valued functions of a real variable. If $s(t)$ moves smoothly within the domain as $r\in\mathbb R$ changes, then $\dfrac d {dt} \big(f(s(t)) -s(t)\big) = f'(s(t))s'(t)-s'(t).$ If $f'(s(t))=1$ for all values of $t,$ then this derivative is $0.$ A theorem from real variables says that if the derivative of a real-valued function of a real variable is $0$ on an interval, then the function is constant on the interval. That that is also true of complex-valued functions of a real variable follows just by considering the real and imaginary parts.
