A Complex Variable ODE

Question

suppose $f$ is a holomorphic function on some domain $D$ satisfying $f'(z)=af(z) $ for some >constant a. show that $f(z)=Ce^{az}$, for some constant $C$

Answer 1

Since you showed no work, I am only going to get you started. Oberve, $f^\prime(z) = a f(z) \iff a = \frac{ f^\prime(z)}{f(z)}$. Integrate both sides. Reduce.

Answer 2

Multiplying the equation by $\mathrm{e}^{-az}$ (i.e., by the integrating factor) we obtain: $$ 0=\mathrm{e}^{-az}\big(f'(z)-af(z)\big)=\big(\mathrm{e}^{-az}f(z)\big)'. $$ Thereofore $\mathrm{e}^{-az}f(z)$ is a constant function, i.e., $$ \mathrm{e}^{-az}f(z)=c, $$ for some $c\in\mathbb C$ or $$ f(z)=c\,\mathrm{e}^{-az}. $$