I have this problem. Let be given complex number $z$ such that
$$|z+1|+ 4 |z-1|=25.$$
Find the greastest and the least of the modulus of $z$.
I tried with minimum.
Put $A(-1,0)$, $B(1,0)$ and $M(x,y)$ present of $z$.
We have $O(0,0)$ is the midpoint of the segment $AB$. Therefore
$$OM^2 = \dfrac{AM^2 + BM^2}{2}-\dfrac{AB^2}{4}.$$
Another way
$$25=AM+4BM \leqslant \sqrt{(1^2 + 4^2)(AM^2 + BM^2)},$$
Therefore
$$AM^2 + BM^2 \geqslant \dfrac{625}{17}.$$
$$OM^2 \geqslant \dfrac{625}{17} -1 = \dfrac{591}{17}.$$
Thus, minimum of $z$ is $\sqrt{\dfrac{591}{17}}$.
This answer is not true with Mathematica. Mathematica give $\dfrac{22}{5}$.
Where is wrong in my solution and how can I find the maximum?
Best Answer
For maximum $|z|$, we have
\begin{align} |5z|&=|(z+1)+4(z-1)+3|\\ &\le|z+1|+4|z-1|+|3|\\ &\le25+3\\ |z|&\le \frac{28}{5} \end{align}
with the equality holds if and only if $\displaystyle z=\frac{28}{5}$.