How can I find maximum and minimum modulus of a complex number?

Question

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?

Do you want find the max and the min by algebraic way or it doesn't matter? — haqnatural, Jun 08 '17 at 09:35
Nice question. It led me to find out about the Cartesian oval that I hadn't seen before. — Paul Aljabar, Jun 08 '17 at 11:12
thank you Paul for the Keyword. I was so confusing this with Apollonius circle. The right keyword is always some kind of satisfied discover. — Curiosity, Jun 08 '17 at 12:10

score 3 · Answer 1 · answered Jun 08 '17 at 11:16

3

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

answered Jun 08 '17 at 11:16

CY Aries

23,393

How a bout general problem? Find max of |z| knowing that $m |z-z_1|+n|z+z_1|=k$. – minhthien_2016 Jun 08 '17 at 15:55
The general problem seems to be quite difficult. – CY Aries Jun 09 '17 at 09:04

score 2 · Answer 2 · answered Jun 08 '17 at 12:08

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.

score 1 · Answer 3 · answered Jun 10 '17 at 07:56

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 $\dfrac{22}{25}$.

How can I find maximum and minimum modulus of a complex number?

3 Answers3

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