Maximum value of $|z|$ given $\lvert z-\frac 4z \rvert = 8$?

Question

The question is $$ \left|z-\frac 4z \right| = 8$$ Find the max value of $ |z|$

You know how the triangle inequality is: $$ \bigg| | z_1 | - | z_2| \bigg| \leqq | z_1 \pm z_2 | \leqq | z_1 | + | z_2 | $$

The solutions used only the left hand side inequality, and also ignoring the absolute values outside of $ | z_1 | - | z_2 | $, i.e. they solved

$ | z | - \left| \frac 4z \right| \leqq 8$ to obtain the answer $ | z |_{max} = 4 + 2 \sqrt{5} $

I am confused about this in two ways, firstly, the way they solved it aren't they assuming here that $| z | \geqq | \frac 4z |$ ? Also can you just ignore the right hand side inequality?

Answer 1

Your equation is of the form $|z^2-4|=8|z|$, which is a punctured ellipse parallel to the axes centered on the real axis. Then the farthest point away, i.e. the maximium value of $|z|$ is on the major ellipse at $z= 4+2\sqrt{5}$.

Answer 2

$|z|-|\frac 4 z| \leq |z-\frac 4 z| =8$ and solving this we get $|z| \leq 4 +2\sqrt 5$. This proves that any complex number such that $|z-\frac 4 z| =8$ necessarily satisfies $|z| \leq 4+2\sqrt 5$ (whether or not $|z| \geq |\frac 4 z|$). Now we have to see that the value $4+2\sqrt 5$ is actually attained. To see this just take $z=4+2\sqrt 5$. This number satisfies the given equation. Hence the maximum value is $4+2\sqrt 5$.

Answer 3

"aren't they assuming here that $|z|≥|\frac 4z|$"

No.

if $|z| < |\frac 4z|$ then $|z|-|\frac 4z| < 0 < 8$. That's fine.

And if $|z|\ge |\frac 4z|$ then $|z|-|\frac 4z| = ||z|-|\frac 4z||$. That's fine too.

In any event $|z|-|\frac 4z| \le ||z|-|\frac 4z||$. In fact $a\le |a|$ for all real numbers. Becase if $a < 0$ then $a < 0 < |a|$. And if $a \ge 0$ then $a=|a|$. So $a \le |a|$. Always.

"Also can you just ignore the right hand side inequality?"

Of course. We can ignore both sides and watch a movie instead if we want to.

But if I told you to mow the lawn and I gave you a lawn mower and a pair of scissors and you use the lawn mower and not the pair of scissors can I yell at you and tell you "No, you were supposed to use both"

We have $|z| -|\frac 4z| \le ||z| - |\frac 4z|| \le |z\pm \frac 4z| \le |z| +|\frac 4z|$.

That sentence contains NINE count them $9$ inequalities we can use.

1. $|z| -|\frac 4z|\le ||z| - |\frac 4z||$
2. $|z| -|\frac 4z|\le |z- \frac 4z|$
3. $|z| -|\frac 4z|\le |z+ \frac 4z|$
4. $|z| -|\frac 4z|\le |z| +|\frac 4z|$
5. $||z| - |\frac 4z|| \le |z- \frac 4z|$
6. $||z| - |\frac 4z|| \le |z+\frac 4z|$
7. $||z| - |\frac 4z||\le |z| +|\frac 4z|$
8. $|z -\frac 4z| \le |z| +|\frac 4z|$
9. $|z+\frac 4z| \le |z| + |\frac 4z|$.

Do we have to use all of them? No. We just need to use the ones that help us.

And they figured $|z| -|\frac 4z| \le |z -\frac 4z| = 8$ was the one that helps us.

=======

Okay..... let's go through this in detail.

$||z| -|\frac 4z|| \le |z-\frac 4z| = 8 \le |z| +|\frac z4|$.

Case 1: $|z| < |\frac 4z|$.

That means $|z|^2 < 4$ and $|z| < 2$. We're looking for the maximum value of $|z|$ so whatever result we get here will not be useful if we later find that $|z|$ might be $\ge 2$ is some other cases.

But lets go on...

$-8 \le |z| - |\frac 4z|<0$

$0\le |z|^2 + 8|z| - 4< 8|z|$

$20 \le (|z|^2 + 8|x| + 16< 8|z| + 20$

$20 \le (|z|+4)^2 < 8|z|+20$

Now $|z|<2$ so $2\sqrt 5 \le 4+|z| < \sqrt{8|z|+20}$

$2\sqrt 5 - 4 \le |z| < \sqrt{8|z|+20}-4$

But we also have $|z| < 2$ so $2\sqrt 5-4\le |z| < \min(2,\sqrt{8|z|+20}-4)$.

As $\sqrt{8|z|+20}-4 \ge \sqrt{8(2\sqrt 5-4)+20}-4=2\sqrt{4\sqrt 5-3}> 2$ we hav $2\sqrt 5-4\le |z| < 2$.

Now all that work we did we the $\sqrt{8|z|+20}-4$ was wasted because we knew that $|z| < 2$. And if discover that it is possible for $|z|\ge 2$ then all our work on case 1) will have been wasted.

Case 2:

$|z|\ge|\frac 4z|$ and so $|z| \ge 2$

$0 \le |z|-|\frac 4z| \le 8$

$-8|z| \le |z|^2 -8|z| -4 \le 0$

$-8|z|+20 \le |z|^2 - 8|z| +16 \le 20$

$\sqrt{\max(0,20-8|z|)}\le |z|-4\le 2\sqrt 5$

$4+ \sqrt{\max(0,20-8|z|)} \le |z| \le 4 +2\sqrt 5$.

Now we don't care about the LHS side indicates as we are looking for the maximum value of $|z|$ so doing that work was wasted. Also as it is possible that $|z | < |\frac 4z|$ there wasn't any real reason to insiste $|z|$ MUST be $\ge |\frac 4z|$ so there was never any reason to do the LHS side at all in the first place.

And because can have $|z| \ge 2$ there was no reason to do case 1 at all and all that work was wasted.

And if $z= 4 + 2\sqrt 5$ then $z -\frac 4z=(4+2\sqrt 5) + \frac 4{4+2\sqrt 5}=$

$4+2 \sqrt 5 - \frac {4(1-\frac 12\sqrt 5)}{4(1+\frac 12\sqrt 5)(1-\frac 12\sqrt 5)}=$

$(4 + 2\sqrt 5) -\frac {4-2\sqrt 5}{4(1-\frac 54)}=$

$(4+2\sqrt 5) + (4 - 2\sqrt 5) = 8$.

Now let's go on and continue to so wasted work.

Let's deal with $|z -\frac z4| =8 \le |z| +|\frac z4|$

So

$0\le |z|^2 -8|z| + 4$ and $12 \le (|z|-4)^2$

$2\sqrt 3 \le ||z| - 4|$

Case 3: $|z| < 4$ then

$2\sqrt 3 \le 4 - |z|$ and $|z|\le 4-2\sqrt 3$.

This was wasted work because we had already determined that $|z| = 4 + 2\sqrt 5$ was possible.

Case 4: $|z| \ge 4$ then

$2\sqrt 3 \le |z|-4$ and $|z| \ge 4+2\sqrt 3$.

Okay.... that's fine but we also proved that $|z| \le 4+2\sqrt 5$ so this result doesn't help us at all.

so all this was wasted.

====

.....

And that actually matters is that

1. $z = 4 + 2\sqrt 5$ is possible so $|z|=4+2\sqrt 5$ is possible.

2. $-\infty < |z| -|\frac 4z| \le ||z|-|\frac 4z|| < |z+\frac 4z| = 8$.

So $|z| \le 4+2\sqrt 5$.

All other conditions about $|z| \ge 2\sqrt 5-4$ or if $|z| \ge 4$ then $|z| \ge 4+2\sqrt 3$ don't matter. They are true. But they dont matter.

Answer 4

Redo:

We have three potential inequalities

$|z| - |\frac z4| \le ||z| - |\frac z4|| \le |z -\frac z 4|=8$ or

1. $|z| -|\frac z4| \le 8$.

$-|z| + |frac z4| \le ||z| - |\frac z4|| \le |z -\frac z 4|=8$ or

2. $|\frac z4| -|z| \le 8$.

$8=|z-\frac z4|\le |z| + |\frac z4|$ or

3. $|z| +|\frac z4| \ge 8$.

All three of these equations are true for any possible value of $z$ where $|z +\frac 4z| =8$.

The first yields $|z|\le 4 + 2\sqrt 5$ AND $|z|\ge 2\sqrt 5-4$ so $2\sqrt 5-4\le |z|\le 4+2\sqrt 5$. That is always true.

The second yields $|z|\ge 2\sqrt 5-4$ OR that $|z|\le -4 -\sqrt 5$ but that is impossible. So $|z|\ge 2\sqrt 5 -4$. That is always true.

The third yields $|z| \ge 2\sqrt 3$ OR $|\le 4-2\sqrt 3$.

Putting these results together and noting that $2\sqrt 5-4 \le 4-2\sqrt 3 \iff \sqrt 5 + \sqrt 3 < 4 \iff 5+2\sqrt {15}+3 < 16\iff 8+2\sqrt {15}< 16=8+2\sqrt {16}$ which it is we get.

$2\sqrt 5 - 4\le |z| \le 4-2\sqrt 3$ or $4+2\sqrt 3 \le |z| \le 4+ 2\sqrt 5$

So $|z|_{max} = 4+2\sqrt 5$.

As we were only asked for the maximum value and we get that from inequality 1), inequality 1) is the only one we need to consider.