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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 $|e^{a}-e^{b}|<|a-b|$

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    The ol$d$ question only asks for a non-strict inequality, but a minimally careful boun$d$ on the integral suggested by GEdrgar there will show the stricy inequality as asked here.2012-04-24

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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| <|a-b|.

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    @Lierre I'm not aware of a name. In the one-variable real case, I just know it as a property of the integral.2012-04-25