$|e^a-e^b| \leq |a-b|$

Came across this problem on an old qualifying exam: Let $a$ and $b$ be complex numbers whose real parts are negative or 0. Prove the inequality $|e^a-e^b| \leq |a-b|$.

If $f(z)=e^z$ and $z=x+iy$, then $|f'(z)|=e^x\leq 1$ given that $x \leq 0$. I played around with the limit definition of the derivative, but wasn't able to get anywhere. Not sure what else to try; a hint would be very helpful!

-

Consider integrating $f'(z) dz$ along the line segment from $a$ to $b$

-

An interesting related article A norm inequality for Hermitian operators by Ritsuo Nakamoto

The American Mathematical Monthly; Mar 2003; 110, 3;

-

If you put $\exp(z)$ in here and turn what you see about how two complex numbers $z$, $w$ in the left half plane $\Re(z)\le 0$ suffer under the map $\exp$ into a statement, that statement would be exactly the wanted inequality.

-