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If z and w are distinct complex numbers such that $|z| =|w| = r$, prove that

$|\frac{1}{2}(z + w)| < r$.

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Your title is a little bit wrong. You can't order complex numbers so they can't have inequality. However it is clear in your question you are talking about the modulus. Have you tried forming an argument with the complex triangle inequality? – user42538 Feb 12 '13 at 0:46
What do you mean that you cannot order complex numbers so they cant have inequality? I see nothing wrong with the wording of this question. – CogitoErgoCogitoSum Feb 12 '13 at 0:50
It's in the title. The question is fine. – user42538 Feb 12 '13 at 0:52
Just for the reference, this is the strict convexity of $\mathbb{C}$ with the Euclidean norm. All inner product spaces are strictly convex: – 1015 Feb 12 '13 at 0:54
yes, I've tried the triangle inequality, unfortunately it doesnt have a strictly <. – Jerome Turner Feb 12 '13 at 0:57
up vote 0 down vote accepted

Hint: Prove that $\left|\frac{1}{2}(z+w)\right| \le r$ via the triangle inequality. When does equality in the triangle inequality hold? Show that this case is not possible if $z$ and $w$ are distinct with $|z|=|w|$.

This approach actually generalizes to show that any inner product space is strictly convex as noted by @julien in the comments.

If you look through the standard proof of the triangle inequality for complex numbers (and more generally inner product spaces), you will see that equality holds iff:

$$ \Re (z \overline{w}) = z\overline{w} = |z|\cdot |w| $$

It can be shown with a simple calculation that this means $z$ and $w$ are linearly dependent, $z = \lambda w$, and $\lambda \ge 0$.

But $|z| = |w|$ here, which forces $\lambda = 1$. This means that $z$ and $w$ cannot be distinct when equality holds.

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I like your idea, I'm confused how to proceed from $|z + w|^2 = |z|^2 + |w|^2 + 2|z||w|$ to get the desired formula. – Jerome Turner Feb 12 '13 at 1:09
@user61916 I added an explanation. Hope this helps. – Ayman Hourieh Feb 12 '13 at 10:49

Hint: Use the geometric interpretations of complex numbers. It is given that $z$ and $w$ are located on a circle of radius $r>0$ (if $r=0$ the claim is not true) with centre at the origin. Use the parallelogram law for addition of complex numbers to locate $1/2\cdot (z+w)$ to see it lies inside said circle, proving the desired inequality.

Now you can try to find a purely algebraic proof. Hint for that: compute $[(1/2)\cdot (z+w)]^2$.

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Let $$z=re^{j\theta_1}$$ and $$w=re^{j\theta_2}$$

$$ \mid \frac{z+w}{2}\mid=\frac{r}{2}|\left(Cos(\theta_1)+Cos(\theta_2)\right)+j\left(Sin(\theta_1)+Sin(\theta_2)\right)|$$ $\implies$

$$ \mid \frac{z+w}{2}\mid=\frac{r}{2}\sqrt{2+2Cos(\theta_1-\theta_2)}= r |Cos(\frac{\theta_1-\theta_2}{2})| \lt r$$

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