Prove that $|e^{a}-e^{b}|<|a-b|$ Possible Duplicate:
  $|e^a-e^b| \leq |a-b|$  




Could someone help me through this problem? 
Let a, b be two complex numbers in the left half-plane. Prove that $|e^{a}-e^{b}|&lt;|a-b|$
By mean value theorem,
$$ |e^a - e^b| \leqslant |a-b|\max_{x\in [a,b]} e^x $$
But $a$ and $b$ have a negative real part, and then all $x$ in $[a,b]$ also have a negative real part. Hence the $\max$ is less than one. And thus
$$ |e^a - e^b| &lt|a-b|. $$
By mean value theorem,
$$ |e^a - e^b| \leqslant |a-b|\max_{x\in [a,b]} e^x $$
But $a$ and $b$ have a negative real part, and then all $x$ in $[a,b]$ also have a negative real part. Hence the $\max$ is less than one. And thus
$$ |e^a - e^b| &lt|a-b|. $$