If the area of the triangle whose vertices are the complex numbers z, iz and z+iz is 50 square units, then |z| equals

Take the sides from the vertex $z+iz$: the vectors to the other vertices are $(z+iz)-z=iz$ and $(z+iz)-iz=z$. Multiplication by $i$ rotates by $90^\circ$, so these two sides are perpendicular, and each has length $|z|$. The triangle is therefore right-angled and isosceles with legs $|z|$, and its area is $\frac12|z|^2$. Setting $\frac12|z|^2=50$ gives $|z|^2=100$, so $|z|=10$. Taking the area as $|z|^2$ would wrongly give $5\sqrt2$.