$y_1=\max ∣ ∣z−ω∣−∣z−ω^2||$, where $|z|=2$ and $y_2 =\max ∣ ∣z−ω∣−∣z−ω^2∣∣$, where $|z|=\frac12$ and $ω$ and $ω^2$ are complex cube roots of unity 
If $y_1=\max ∣ ∣z−ω∣−∣z−ω^2||$, where $|z|=2$ and $y_2 =\max ∣ ∣z−ω∣−∣z−ω^2∣∣$, where $|z|=\frac12$ and $ω$ and $ω^2$  are complex cube roots of unity, then

*

*A) $y_1=\sqrt3$ ; $y_2=\sqrt3$

*B) $y_1\lt\sqrt3$ ; $y_2=\sqrt3$

*C) $y_1=\sqrt3$ ; $y_2\lt\sqrt3$

*D) $y_1\gt3$ ; $y_2\lt3$

Using $||a|-|b||\le|a-b|$,
$∣ ∣z−ω∣−∣z−ω^2||\le|w^2-w|$
So, $y_1=y_2=\sqrt3$. So, I pick option A). But the answer given is C).
Looks like it has something to do with $|z|$ but not able to make the connection.
 A: The application of reverse triangle inequality is correct. The problem comes when considering whether the upper bound can actually be attained. The reverse triangle inequality states that
$$ ||a| - |b|| \leq |a-b| \, \forall a,b \in \mathbb{C},$$ with equality when $ a = rb$ for some $r \in \mathbb{R} ^ {+}.$ (note the $+$).
In this case, equality holds when $z - \omega = r (z-\omega^2)$ for some $ r \in\mathbb{R}^{+}$, which geometrically means that $z, \omega, \omega^2$ are collinear, but with $\omega, \omega^2$ on the same side of $z$.
Since $\omega, \omega^2$ lie on the unit circle, in the $|z|=2$ case we can clearly reach equality. However when $|z|=\frac{1}{2}$, if $z, \omega, \omega^2$ are collinear, then $z$ necessarily lies between $\omega, \omega^2$ (just draw a circle of radius 1 and a circle of radius $\frac{1}{2}$ to convince yourself.)
Thus equality cannot be attained in this case, so the maximum value $y_2$ must be $<\sqrt{3}$.

(When $|z| = 2$, we attain maximum at $G$ and $H$. We can never attain maximum when $|z| = \frac{1}{2}$ since the blue dot is always "between" $\omega$ and $\omega^2$.)
A: The triangle inequality in the form of
$\;||z-\omega|-|z-\omega^2||\le|\omega-\omega^2|,\;$
or
$$\left|\left|z+\dfrac{1-i\sqrt3}2\right|-\left|z+\dfrac{1+i\sqrt3}2\right|\right|\le\sqrt3,$$
becomes the equality if $\;z=\dfrac{-1\pm i\sqrt{15}}2,\;|z|=2.\;$
Thus, $\;\mathbf{y_1 = \sqrt3.}\;$
If $\;|z|=\frac12,\;$ then $\;|z-\omega|,|z-\omega^2|\in[\frac12,\frac32].\;$
Thus, $\;\mathbf{y_2<\sqrt3.}\;$
Therefore, the answer is C).
Appendium.
Assuming of $\;|z|=2,\;z=u+iv\;$ leads to the systems
\begin{cases}
(2u+1)^2+4v^2+3-\sqrt{(2u+1)^2+(2v-\sqrt3)^2}\sqrt{(2u+1)^2+(2v+\sqrt3)^2}=6\\[4pt]
u^2+v^2=4,
\end{cases}
\begin{cases}
4u+14 = \sqrt{4u+20-4v\sqrt3}\sqrt{4u+20+4v\sqrt3}\\[4pt]
u^2+v^2=4,
\end{cases}
\begin{cases}
(2u+10)^2-(2u+7)^2 = 12v^2\\[4pt]
u^2+v^2=4,
\end{cases}
\begin{cases}
4u+17=4(4-u^2)\\[4pt]
u^2+v^2=4,
\end{cases}
with $\;z=\dfrac{-1\pm i\sqrt{15}}2,\;|z|=2.\;$
