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

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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    By no mean this is true...2012-04-24
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    I think that being in the complex plane if this inequality can be fulfilled2012-04-24
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    Oh... Sorry I read a little bit too fast, there is a condition with the half-plane. But that's funny, nobody seems to see it, two wrong answers already ;)2012-04-24
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    http://math.stackexchange.com/questions/54507/ea-eb-leq-a-b2012-04-24
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    The old question only asks for a non-strict inequality, but a minimally careful bound 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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    Mean value theorem does not hold for complex-valued functions.2012-04-24
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    @dls — I think I gave a wrong name, what is the english name for the theorem I used ?2012-04-24
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    the mean value theorem holds for complex functions2012-04-24
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    The mean value theorem for derivatives does not hold for complex-valued functions. The standard example is $f(x)=e^{ix}$ on the interval $[0,2\pi]$. Then $f(2\pi)-f(0)=0$ but $2\pi|f'(x)|=2\pi$ for every $x$.2012-04-25
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    @dls — But still, the $|f(x)-f(y)| \leq \max_{z\in[x,y]} |\nabla f(z)|$, by integrating. What is the name of this inequality ?2012-04-25
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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