# $|e^a-e^b| \leq |a-b|$ for complex numbers with non-positive real parts

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!

• can you please mention in which Qualifying exam it has appeared,would be helpful for practice @dls Commented May 20, 2017 at 6:07
• @BAYMAX Idk if it's still useful to you, but it came in the TIFR PhD entrance test in 2017. Commented Mar 6, 2021 at 16:52

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

• Neat. But could anyone explain this idea further? Commented Dec 18, 2016 at 14:51
• This takes another approach to the mean-value theorem -- the integral approach. It works neatly with holomorphic functions (and the MV theorem itself isn't very elegant), but I still would've liked to see the connection drawn. Commented Aug 3, 2017 at 10:23

Prove and then use the following fact:

Let $D \subseteq \mathbb C$ be a convex region and let $f: D \to \mathbb C$ be holomorphic with $|f'|\le 1$ on $D$. Then for $a,b\in D$ we have

$$|f(b) - f(a)| \le |b-a|$$

• What I'd also like to see here is a reference to the mean value theorem, and why although it fails in general with complex functions the norm can still be bounded (similarly as with vector-valued real functions). Commented Aug 3, 2017 at 10:21

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.

A brute force approach, avoiding complex analysis, just to prove it can be done.

Let $$a=a_1+a_2i, b=b_1+b_2i.$$

Then \begin{align}\left|e^a-e^b\right|^2&=e^{2a_1}+e^{2b_1}-2\operatorname {Re} e^{a+\overline b}\\&=e^{2a_1}+e^{2b_1}-2e^{a_1+b_1}\cos(a_2-b_2)\\&=(e^{a_1}-e^{b_1})^2+2e^{a_1+b_1}(1-\cos(a_2-b_2)).\tag1\end{align}

And we need to show $$(1)$$ is $$< (a_1-b_1)^2+(a_2-b_2)^2$$ when $$a_1,b_1<0.$$ It turns out, we can show this term by term.

We get the real inequality $$\left|e^{a_1}-e^{b_1}\right|\leq |a_1-b_1|$$ for $$a_1,b_1<0,$$ from the mean value theorem.

So we finally need, since $$0

$$2(1-\cos \theta)\leq \theta^2.$$

But $$\frac{1-\cos\theta}2=\sin^{2}\frac{\theta}2\leq \frac{\theta^2}{4}$$ so $$2(1-\cos\theta)\leq \theta^2.$$

We get strict inequality in the first terms when $$a_1\neq b_1.$$ And if $$a_1=b_1,$$ we get strict inequality when $$1-\cos(a_2-b_2)\neq 0,$$ since, when non-zero, the term $$e^{a_1+b_1}$$ makes it strictly smaller.

But for strict equality with the left side zero, $$(a_2-b_2)^2=0.$$

So we get the strict inequality when $$a\neq b.$$