$|z-1/z|=2$ where $z$ is a complex number. How to find the greatest value of $|z|$ geometrically?

Question

$|z-\frac{1}{z}|=2$, where $z$ is a complex number.

How to find the greatest value of $|z|$ geometrically?

P.S: I know the algebraic method.

Tried to plot it on wolframalpha but it is treating $z$ as a real number and then plotting.

Answer 1

A first step. You ask for what the greatest $|z|$ is, satisfying $$ |z-1/z|=2. $$ But you also want a picture. The $z$ that satisfy the equation are represented by two circles in the complex plane. One way of seeing this is to square, let $z=x+iy$, and simplify to conclude that $$ \bigl((x-1)^2+y^2-2\bigr)\bigl((x+1)^2+y^2-2\bigr)=0. $$ I include a picture below. On the other hand, that does not provide a geometrical solution to your first problem, of finding the largest $|z|$.

On the other hand, drawing also a circle with center in $(0,0)$ with radius $1+\sqrt{2}$, it will be tangent to the other circles at $x=\pm(1+\sqrt{2})$, and the other two circles will be completely inside (which is clear since the new circle has larger radius), as the second image shows. Thus, the largest value of $|z|$ is obtained when $z$ is real, and is $\pm(1+\sqrt{2})$.

Answer 2

$|z|$ = $|(z+\frac{1}{z})- \frac{1}{z}| \le |(z+\frac{1}{z})|+ |\frac{1}{z}| = 2 + \frac{1}{|z|}$ which implies $|z| \le 1+\sqrt2.$ In no way it is $1+\sqrt5.$

Answer 3

I know you asked for a geometric approach, but this is much too long for a comment (or even several comments), and I wanted to clear up your confusion about my comment above. $\DeclareMathOperator{\dom}{dom}$ $\DeclareMathOperator{\re}{Re}$ $\DeclareMathOperator{\im}{Im}$

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First of all, an upper bound on a real-valued function $f$ is a real number $\alpha$ such that $\alpha\ge f(x)$ for every $x\in\dom f.$ If $\alpha>f(x)$ for all $x\in\dom f,$ we say that $\alpha$ is a strict upper bound. Given a real-valued function $f$ with a non-empty domain, if $f$ has an upper bound, then the set of all upper bounds of $f$ has a least element, which we call the least upper bound of $f$ (for obvious reasons).

For example, if we consider $f:\Bbb R\to\Bbb R$ given by $f(x)=\cos x,$ then the upper bounds of $f$ are the real numbers $\alpha\ge 1,$ the strict upper bounds of $f$ are the real numbers $\alpha>1,$ and $1$ is the least upper bound of $f.$

For another example, if we consider $g:\Bbb R\to\Bbb R$ given by $g(x)=\arctan x,$ then the upper bounds of $g$ are the real numbers $\alpha\ge\frac\pi2,$ every upper bound of $g$ is strict, and $\frac\pi2$ is the least upper bound of $g.$

Note that $f$ actually reaches the least upper bound--for example, when $x=0$--but $g$ never does. When a real-valued function with an upper bound actually reaches its least upper bound, we refer to the least upper bound as the maximum value of $f$.

In this particular case, we are considering the function $h:A\to\Bbb R$ given by $h(z)=|z|,$ where $$A=\left\{z\in\Bbb C:\left\lvert z-z^{-1}\right\rvert=2\right\}.$$ What my comment means, then, is that $1+\sqrt5$ is a strict upper bound of $h,$ while $1+\sqrt2$ is the least upper bound (and in fact the maximum) of $h.$

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But how can we see this? Well, the geometric approach is to note that the circle about $0$ of radius $1+\sqrt5$ doesn't intersect the set $A,$ which (as you've seen before) is:

Since you already have a geometric demonstration, here is an algebraic approach that can actually give you the maximum, and not just an upper bound.

First, observe that $$A=\left\{z\in\Bbb C:\left\lvert z-z^{-1}\right\rvert^2=4\right\}.$$ Second, observe that for all $w\in\Bbb C,$ we have $$|w|^2=w\overline{w},$$ so by properties of the modulus function, we have $$\begin{align}\left\lvert z-z^{-1}\right\rvert^2 &= \left(z-z^{-1}\right)\overline{\left(z-z^{-1}\right)}\\ &= \left(z-z^{-1}\right)\left(\overline{z}-\overline{z^{-1}}\right)\\ &= \left(z-z^{-1}\right)\left(\overline{z}-\overline{z}^{-1}\right)\\ &= z\overline z-z\overline z^{-1}-\overline z z^{-1}+z^{-1}\overline z^{-1}\\ &= z\overline z-z\overline z^{-1}-\overline{\overline z z^{-1}}+\left(z\overline z\right)^{-1}\\ &= \lvert z\rvert^2-2\re\left(z\overline z^{-1}\right)+\left(|z|^2\right)^{-1}\end{align}$$ for all non-zero $z\in\Bbb C.$ For $z\in A,$ this becomes $$4=\lvert z\rvert^2-2\re\left(z\overline z^{-1}\right)+\left(|z|^2\right)^{-1}.\tag{$\diamondsuit$}$$ Equivalently, $$0=\lvert z\rvert^2-4-2\re\left(z\overline z^{-1}\right)+\left(|z|^2\right)^{-1},$$ so multiplying by $|z|^2$ yields $$0=\lvert z\rvert^4-\left(4+2\re\left(z\overline z^{-1}\right)\right)|z|^2+1$$ for all $z\in A.$ Making the substitution $u=|z|^2$ gives us $$0=u^2-\left(4+2\re\left(z\overline z^{-1}\right)\right)u+1\tag{$\star$}$$ for all $z\in A,$ and since the modulus function is non-negative, then maximizing $|z|$ on $A$ is equivalent to maximizing $u$ on $A.$ Applying the quadratic formula to $(\star)$ yields $$\begin{align}u &= \frac12\left(4+2\re\left(z\overline z^{-1}\right)\pm\sqrt{\left(4+2\re\left(z\overline z^{-1}\right)\right)^2-4}\right)\\ &= 2+\re\left(z\overline z^{-1}\right)\pm\sqrt{\left(2+\re\left(z\overline z^{-1}\right)\right)^2-1}\end{align}$$ for all $z\in A,$ and since we wish to maximize the value of $u$ on $A$ then this is equivalent to maximizing the function $$z\mapsto 2+\re\left(z\overline z^{-1}\right)+\sqrt{\left(2+\re\left(z\overline z^{-1}\right)\right)^2-1}$$ on $A,$ which is equivalent to maximizing $$w(z)=\re\left(z\overline z^{-1}\right)$$ on $A.$

Now, observe that $$z\overline{z}^{-1} = z\left(\frac{|z|^2}{z}\right)^{-1} = \left(\frac{z}{|z|}\right)^2 = \frac1{|z|^2}(\re z+i\im z)^2 = \frac{(\re z)^2-(\im z)^2+2i(\re z)(\im z)}{|z|^2},$$ so since $|z|^2=(\re z)^2+(\im z)^2,$ then $$w(z)=\frac{(\re z)^2-(\im z)^2}{(\re z)^2+(\im z)^2},\tag{$\heartsuit$}$$ so it's enough to maximize $\frac{x^2-y^2}{x^2+y^2},$ where $x=\re z$ and $y=\im z$. Observe that for a fixed, non-zero $x$ value, increasing $|y|$ yields a greater denominator and a smaller numerator, so a smaller quotient! Similarly, for a fixed, non-zero $x$ value, decreasing $|y|$ yields a smaller quotient. Hence, for a fixed, non-zero $x$ value, $\frac{x^2-y^2}{x^2+y^2}$ is maximized when $y=0,$ in which case $\frac{x^2-y^2}{x^2+y^2}=1.$ On the other hand, for $x=0$ and $y\ne0,$ we have$\frac{x^2-y^2}{x^2+y^2}=-1,$ and so $\frac{x^2-y^2}{x^2+y^2}$ is maximized exactly when $x\ne 0$ and $y=0.$

Translating this back into terms of $(\heartsuit),$ we have that $w(z)$ is maximized when $\re z\ne 0$ and $\im z=0.$ In other words, we obtain the desired maximum when $z$ is real, or equivalently $\overline z=z.$ By $(\diamondsuit),$ $z$ maximizes $w(z)$ on $A$ if and only if $z$ is real and $$4=\lvert z\rvert^2-2\re\left(zz^{-1}\right)+\left(|z|^2\right)^{-1}=\lvert z\rvert^2-2+\left(|z|^2\right)^{-1}=\lvert z\rvert^2-2+\left(|z|^{-1}\right)^2,$$ meaning $z$ is real and $$4=\left(z-z^{-1}\right)^2.$$ Hence, either $z-z^{-1}=2$ or $z-z^{-1}=-2$ (with $z$ real) will provide the desired maximum value of $w(z)$ on $A,$ and so provide the desired maximum value of $|z|$ on $A.$ The only possible solutions $z,$ then, are $\pm(1+\sqrt2)$ and $\pm(\sqrt2-1).$ The first two provide the maximum value: $1+\sqrt2.$

My, wasn't that a pain? The geometric solution looks even better now, doesn't it?

Answer 4

I hope this is what you're looking for: $$|z-\frac{1}{z}|=2$$ If we multiply by $|z|$: $$|z^2-1^2|=2|z|$$ This equation resembles the form of an hyperbola $$x^2-y^2=c^2$$