TITLE: If $|z - 8|=\max\{|z + 2|,|z - 4|\}$ ,then $z$ lies on
QUESTION [0 upvotes]: If $|z - 8|=\max\{|z + 2|,|z - 4|\}$ ,then $z$ lies on 

I just assumed them maximum one by one 
$$\frac{|z - 8|}{|z + 2|}=1$$ 
$$\frac{|z - 8|}{|z  - 4|}=1$$ 

So I concluded that $z$ will lie on this pair of staright line but the answer say $z$ lie on a pair of circle.

REPLY [1 votes]: It's definitively a typo but your answer is wrong. Try $z = 6$. You have $|z-8| = |z-4| = 2$ but $|z+2| = 8$ and it's the maximum of $\{|z+2|, |z-4|\}$.
In order to solve this equation you have to remove the max. For this, you have to look first in which domain $|z+2|$ and $|z-4|$ are maximum. Then you can solve the problem.
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