Complex Number Exercise

I'll post a long solution. Maybe there is an easier one:

$\large \displaystyle S = |z| + |2z-w|$

Let $\ z = re^{i \phi}$ and $w = e^{i \theta}$

$\large \displaystyle S = r + \sqrt{(2r \cos(\phi) - \cos(\theta) )^2 + (2r \sin(\phi) - \sin(\theta))^2 }$

$\large \displaystyle S = r + \sqrt{ 4r^2 + 1 - 4r( \cos(\phi)\cos(\theta) + \sin(\phi)\sin(\theta) ) }$

$\large \displaystyle S = r + \sqrt{ 4r^2 + 1 - 4r \cos(\theta- \phi) }$

The cosine will be alternating between $-1$ and $1$ . Since we want the minimum value of $S$ , we want it to be $1$ . Thus:

$\large \displaystyle S = r + \sqrt{ 4r^2 + 1 - 4r }$

$\large \displaystyle S = r + |2r-1|$

Since $|2r-1| \geq 0$ , it's easy to see that $S$ will be minimum when we have $|2r-1| = 0$ , or $r = \frac12$ . In this situation:

$\color{#3D99F6} \boxed{ \large \displaystyle m = S_{min} = \frac12 }$

$|w| = 1 \implies w = 1 \text{ OR } w = -1$

We have two cases here, each one of the cases will again have two subcases with all values of z $\in \{ 0, \frac{1}{2}, \frac{-1}{2} \}$

For $z = 0$ , we have the value of $4m = 4$ . But for $z = \frac{1}{2}$ or $z = \frac{-1}{2}$ , the value of $4m = 2$