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?

  • $\begingroup$ Your solution isn't wrong, it's less strong inequality. $\endgroup$ – Jaideep Khare Jun 8 '17 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 '17 at 9:35
  • $\begingroup$ $AM+4BM=25 \neq 5$ $\endgroup$ – stity Jun 8 '17 at 9:40
  • $\begingroup$ @haqnatural any method. $\endgroup$ – minhthien_2016 Jun 8 '17 at 9:46
  • $\begingroup$ Then,draw the graphic of equation then you will see $\endgroup$ – haqnatural Jun 8 '17 at 9:50

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

  • $\begingroup$ How a bout general problem? Find max of |z| knowing that $m |z-z_1|+n|z+z_1|=k$. $\endgroup$ – minhthien_2016 Jun 8 '17 at 15:55
  • $\begingroup$ The general problem seems to be quite difficult. $\endgroup$ – CY Aries Jun 9 '17 at 9:04

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.


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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