Range of modulus of Complex Number If $z\in \mathbb{C}$ and  $$ |z-1|+|z+3|\le 8$$ Find the Range of $$|z-4|$$
My Try: $$|2z-8|=|2z+2-10|\le |2z+2|+10=|(z-1)+(z+3)|+10\le|z-1|+|z+3|+10\le18$$ $\implies$
$$|z-4|\le9$$ I need Hint to find Minimum Value..
 A: You have already found the maximum, though as noted by 5xum, you need to show the upper bound is reached.  Both bounds in fact can be found using triangle inequalities in an almost mechanical way:
Upper Bound:
$$|z-4+3|+|z-4+7|\le 8 \implies |z-4|-3 + |z-4|-7 \le 8 \implies |z-4| \le 9$$
Now you need to note $z = -5$ does achieve this upper bound.
Lower Bound:
$$|z-4+3|+|z-4+7|\le 8 \implies 3-|z-4| + 7-|z-4| \le 8 \implies 1 \le |z-4|$$
Again note $z = 3$ achieves this bound.
A: An obvious lower bound here is of course $0$, since the absolute value is always $\geq 0$. The value $0$ is also acchieved at $z=4$.
As for the upper bound:
Actually, all you did was prove that $|z-4|\leq 9$, but that does not yet mean that the range actually includes $9$. It does only if you prove that there exists such a $z$ that $|z-4|=9$.
A: Some geometric ideas: 
The given inequality describes the points inside and on the ellipse with foci at $\;(1,0)\;,\;(-3,0)\;$ and major axis's length $\;8\;$ . (Can you find the usual rectangular description, i.e. in the real plane, of this ellipse?).
You're now asking about all the points around $\;(4,0)\;$ assuming they're within the above ellipse....so this is the same as asking what's the intersection of the above ellipse and the above circle (without fixed radius...yet).
Well, now do some geometry...:)
