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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?

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  • $\begingroup$ Your solution isn't wrong, it's less strong inequality. $\endgroup$ Jun 8, 2017 at 9:29
  • $\begingroup$ Do you want find the max and the min by algebraic way or it doesn't matter? $\endgroup$
    – haqnatural
    Jun 8, 2017 at 9:35
  • $\begingroup$ $AM+4BM=25 \neq 5$ $\endgroup$
    – stity
    Jun 8, 2017 at 9:40
  • $\begingroup$ @haqnatural any method. $\endgroup$ Jun 8, 2017 at 9:46
  • $\begingroup$ Then,draw the graphic of equation then you will see $\endgroup$
    – haqnatural
    Jun 8, 2017 at 9:50

3 Answers 3

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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 $\displaystyle z=\frac{28}{5}$.

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  • $\begingroup$ How a bout general problem? Find max of |z| knowing that $m |z-z_1|+n|z+z_1|=k$. $\endgroup$ Jun 8, 2017 at 15:55
  • $\begingroup$ The general problem seems to be quite difficult. $\endgroup$
    – CY Aries
    Jun 9, 2017 at 9:04
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This can be seen as following

"Given $x,y \in \mathbb{R}$ such that $$x \ge 0 , y \ge 0, |x-y| \ge 2, x+4y=25$$ Find the the range of the function $f(x,y)=\frac{x^2+y^2}{2}-1$"

Here there is several ways. You can substitute $x=25-4y$ and consider $f$ as quadratic function of $y$ and exam it on a interval (which one you need to find out). Or you can take the Jacobian, find the vanish point of the Jacobian inside the domain of definition, then compare with the value on the boundary.

Geometrically, the trajectory of $M$ is a circle but I can't recall how to prove. That would be another way to solve the problem with more geometric flavor.

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Put $A(-1,0)$, $B(1,0)$, $M(x,y)$ is the point present for $z$. Note that $O(0,0)$ is midpoint of the segment $AB$ and $AB=2$. We have $AM+4BM=25$. We need to find the least and the greastest of the segment $OM$. Put $a=AM$, $b=BM$ $(a,\, b >0)$, we have $a+4b=25$ or $a=25-4b$.

Because of $|MA-MB|\leqslant AB$ or $$|a-b|\leqslant 2 \Leftrightarrow |25-4b-b| \leqslant 2 \Leftrightarrow \dfrac{23}{5}\leq b\leq \dfrac{27}{5}.$$ Anotherway, $$OM^2 = \dfrac{AM^2 + BM^2}{2}-\dfrac{AB^2}{4}=\dfrac{(25-4b)^2+b^2}{2}-1=\frac{17 b^2}{2}-100 b+\frac{623}{2}.$$ Consider the function $f(b)=\dfrac{17 b^2}{2}-100 b+\dfrac{623}{2}$ here $\dfrac{23}{5}\leq b\leq \dfrac{27}{5}$, we get the maximum of $f(b)$ is $\dfrac{784}{25}$ and the minimum of $f(b)$ is $\dfrac{484}{25}$. Then, the maximum of $|z|$ is $\dfrac{28}{5}$ and the minimum of $|z|$ is $\dfrac{22}{25}$.

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