On $f' = f$ in $\mathbb{C}$ via Cauchy-Riemann equations and on the associated Cauchy problem

2

What is a straightforward way (using only the Cauchy-Riemann equations) to find the solutions of $f' = f$ in $\mathbb{C}$?

In particular, how do you prove that the complex esponential is the unique solution of the Cauchy problem with initial condition $f(0)=1$?

asked 2017-01-06

user id: user

0

Consider $g(z)=\exp(-z)f(z)$ to get $g'(z)=0$ which almost trivially leads to a constant as solution. – 2017-01-06
0

Or use the validity of the Cauchy-Riemann equation for the equivalent condition that a power series expansion for $f$ exists and compare coefficients in the differential equation. – 2017-01-06

0 Answers 0