Complex number inequality. If z and w are distinct complex numbers such that $|z| =|w| = r$, prove that  
$|\frac{1}{2}(z + w)| < r$.
 A: 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 $[(1/2)\cdot (z+w)]^2$.
A: 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.
A: Let $$z=re^{j\theta_1}$$ and
$$w=re^{j\theta_2}$$
$$ \mid \frac{z+w}{2}\mid=\frac{r}{2}|\left(Cos(\theta_1)+Cos(\theta_2)\right)+j\left(Sin(\theta_1)+Sin(\theta_2)\right)|$$ $\implies$
$$ \mid \frac{z+w}{2}\mid=\frac{r}{2}\sqrt{2+2Cos(\theta_1-\theta_2)}= r |Cos(\frac{\theta_1-\theta_2}{2})| \lt r$$
