Let $z\in\mathbb{C}, z\ne0$, if $\left|z-\frac{1}{z}\right|=2$, then what is the maximum value of $|z| ?$(Solve preferably without using geometry)

Question

Let $z$ be a non-zero complex number such that $\left|z-\frac{1}{z}\right|=2$. What is the maximum value of $|z| ?$ (Solve preferably without using geometry.)

My attempt:

I tried to simplify things down by writing $\left|z-\frac{1}{z}\right|=2$ as $$\left(z-\frac 1z\right)\overline{\left(z-\frac 1z\right)}=\left(z-\frac 1z\right)\left(\overline z -\frac1{\overline z}\right)=4\implies |z|^2+\frac 1{|z|^2}+\frac{z}{\overline z}+\frac {\overline z}{z}=4.$$

But then, I am getting stuck at this step in the hope of proceeding with a breakthrough.

Then I tried using some standard inequalities:

$$|z|-\left|\frac 1z\right|\leq \left|z-\frac 1z\right|$$ and $$\left|z-\frac 1z\right|\leq |z|+\left|\frac 1z\right|$$

But this doesn't help much either.

Also, I tried writing $z=re^{i\theta}$ and neither does this make things easy. Till now, I am unable to find any major breakthrough. I am not interested in a geometric solution.

I think the link $|z-1/z|=2$ where $z$ is a complex number. How to find the greatest value of $|z|$ geometrically? asks for a geometrical solution specifically. But, I wanted a regular algebraic, approach to this problem. So, I am not sure if my post can be considered a duplicate or not as both of the posts, though focus on the same question, but the desired approach of the solution method (to the problem) , are different. (As, of now, I tried to make it specific in this post of mine to avoid further confusion)

Answer 1

Let $z=re^{i\theta}$, then you get

$$|r^2e^{2i\theta}-1|=2r$$

expand, we get

$$r^4-2r^2\cos2\theta+1=4r^2$$

Move terms,

$$\frac{1}{2}\left(r^2+\frac{1}{r^2}\right)=2+\cos2\theta$$

For LHS, the minimum value is $1$, when $r=1$, and we want $r$ to be as large as possible, apparently, $r_{max}=r_m>1$

For $r\in[1,\infty)$, the LHS is monotonic increasing, so the maximum value occures when the RHS reaches its maxima, so we get:

$$\frac{1}{2}\left(r_m^2+\frac{1}{r_m^2}\right)=2+1\Rightarrow r_m^2=3\pm2\sqrt2$$

Since $r_m>1$, we get

$$r_m=\sqrt{3+2\sqrt2}=\sqrt2+1$$

Answer 2

As an alternative, following your first idea without exponential form

$$\left(z-\frac 1z\right)\overline{\left(z-\frac 1z\right)}=\left(z-\frac 1z\right)\left(\overline z -\frac1{\overline z}\right)=4\iff |z|^2+\frac 1{|z|^2}\color{red}-\frac{z}{\overline z}\color{red}-\frac {\overline z}{z}=4$$

we have

$$\iff |z|^2+\frac 1{|z|^2}-2=\frac{z}{\overline z}+\frac {\overline z}{z}+2\iff \left(|z|-\frac1{|z|}\right)^2=\frac{z^2}{|z|^2}+\frac {\overline z^2}{|z|^2}+2$$

$$\iff \left(|z|-\frac1{|z|}\right)^2=\left(\frac{z}{|z|}+\frac {\overline z}{|z|}\right)^2\iff \left(|z|-\frac1{|z|}\right)^2=\frac{(z+\bar z)^2}{|z|^2}$$

with $(z+\bar z)=2|z|\cos \theta$ then last identity is equivalent to

$$\left(|z|-\frac1{|z|}\right)^2 =4\cos^2 \theta$$

and $|z|$ is maximum for $\cos^2 \theta =1$ that is

$$\left(|z|-\frac1{|z|}\right)^2 =4\implies |z|_{\text{max}}=\sqrt 2 +1$$

Answer 3

Well here's a generalized formula for your question;

$$\left|z ± \frac{b}{z}\right|=a$$ where a and b are constants;

Maximum and minimum values of $|Z|$ are

$$max = \frac{a+\sqrt{a^2+4|b|}}{2}$$

$$min = \frac{-a+\sqrt{a^2+4|b|}}{2}$$