Range of $|z_1+z_2|$ for some $z_1,z_2\in\Bbb C$

Question

So if we have 2 complex numbers $z_1$ and $z_2$, then the following inequality holds :

$$||z_1|-|z_2||\leq|z_1+z_2|\leq|z_1|+|z_2|$$

We did a question in class that went "Find the range of $|z|$ if $\left|z-\dfrac4z\right|=2$"

The solution began with this : $$\left||z|-\dfrac4{|z|}\right|\leq\left|z-\dfrac4z\right|\leq|z|+\dfrac4{|z|}$$ and we proceeded with the assumption that the range of $\left|z-\dfrac4z\right|$ is $\left[\left||z|-\dfrac4{|z|}\right|,|z|+\dfrac4{|z|}\right]$, however this only works under the assumption that the angle between the 2 complex numbers $\Big(z$ and $\dfrac4z\Big)$ attains a value of $\pi$ as well as $0$, right?

So, shouldn't we first prove so, somehow?

If we were to take $z_1=z$ and $z_2=-z$, for example, in the original inequality, then the inequality leads us to : $0\leq|z-z|\leq2|z|$ which is true but the range of $|z-z|$ is not $\left[0,2|z|\right]$, because the angle between $z_1$ and $z_2$ is always $\pi$.

Thanks!

Answer 1

Given a complex number $z\not=0$, we can let $z_1=z$ and $z_2=-\dfrac4z$. Then $$z_1+z_2=z-\frac4{z}$$ $$|z_1|-|z_2|=|z|-\frac4{|z|}$$ $$|z_1|+|z_2|=|z|+\frac4{|z|}$$

Since $||z_1|-|z_2||\leq|z_1+z_2|\leq|z_1|+|z_2|$ hold for any two complex number $z_1$ and $z_2$, we have $$\left||z|-\frac4{|z|}\right|\le \left|z-\frac4z\right|\le \left||z|+\frac4{|z|}\right|$$

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If $\left|z-\frac4z\right|=2$, then $$\left||z|-\frac4{|z|}\right|\le 2\le \left||z|+\frac4{|z|}\right|$$

From $-2\le|z|-\frac4{|z|}\le2$, we get $\sqrt5-1\le|z|\le\sqrt5+1$
From $2\le|z|+\frac4{|z|}$, we get $0\le(|z|-1)^2+3$, which always holds.

Hence, $|z|\in[\sqrt5-1,\sqrt5+1]$.

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To complete the full story, we should show that for any $v\in [\sqrt5-1,\sqrt5+1]$, there is a complex number $z$ such that $\left|z-\frac4z\right|=2$ and $|z|=v$.

Consider $f(\theta)=\sqrt{4+e^{2i\theta}}+e^{i\theta}$, where $0\le\theta\le\pi$ and $\sqrt{\cdot}$ means the principle square root. Since $\frac4{f(\theta)}=\sqrt{4+e^{2i\theta}}-e^{i\theta}$, we know $\left|f(\theta)-\frac4{f(\theta)}\right|=\left|2e^{i\theta}\right|=2$.

Since $|f(0)|=\sqrt5+1$ and $|f(\pi)|=\sqrt5-1$ and the map $\theta\to|f(\theta)|$ from $[0,\pi]$ to $\Bbb R$ is a continuous real-valued function, for any $v\in[\sqrt5-1,\sqrt5+1]$, there is a $\theta\in [0,\pi]$ such that $|f(\theta)|=v$, by the intermediate value theorem.

Hence the range of $|z|$ is $[\sqrt5-1,\sqrt5+1]$.

Answer 2

This approach doesn't need the use of triangle inequalities: We have that $$ \left|z-\frac 4 z\right| = 2 \iff \left(z-\frac 4 z\right)\left(\overline z-\frac 4 {\overline z}\right)=4 \iff |z|^4+{16}-4|z|^2-8\Re(z^2)=0\,, $$ where $\Re (z^2)$ is the real part of $z^2$, so $\Re (z^2)=\lambda |z|^2$ for some $\lambda\in[-1,1]$, and the equation becomes $$ |z|^4-(4+8\lambda)|z|^2 + 16=0\,. $$ Note that for any solution $|z|$ of this equation (given $\lambda\in[-1,1]$) we can choose one value of $z$ with that module and $z^2=\lambda |z|^2$, thus satisfying the initial condition.

Now write the equation as $$ \big(|z|^2-(2+4\lambda)\big)^2 = (2+4\lambda)^2-16\,. $$ We need the last term to be nonnegative, so there are solutions for exactly $\lambda\in[1/2,1]$. Let $t=2+4\lambda\in[4,6]$, and then the solutions take one of these two forms: $$ |z|^2 = t-\sqrt{t^2-16} \quad {\rm or} \quad |z|^2 = t+\sqrt{t^2-16}\,. $$ The first expression in $t$ is decreasing from its value in 4 ($=4$) to the one in 6 ($=6-\sqrt{20}$), and the second one is increasing from 4 to its value in 6 ($=6+\sqrt{20}$). Hence the range of $|z|$ is $$ \big[\,\sqrt{6-\sqrt{20}},\sqrt{6+\sqrt{20}}\,\big]\,, $$ which happens to be the same as $[\,\sqrt 5 - 1,\sqrt 5 +1\,]$.

Answer 3

(Some days before i started an answer, could not finish it, but since it may be of interest, i've closed the open point, submit follows.)

In order to have a better version of the exercise, i will work with $w=z/2$. Then dividing by two in the given relation for $z$ leads to the equivalence: $$ \left| \frac z2-\frac 2z \right|=1\qquad \Leftrightarrow \qquad \left| w-\frac 1w \right| =1\ . $$ Let us find the range of $|w|$ for all $w$ satisfying the above relation. We will denote by $r=r(w)\in\Bbb R_{>0}$ the modulus of $w$, $r=|w|$. Because of its symmetry w.r.t. $w\to 1/w$ it is enough to find all $$r\in [1,\infty)$$ (that can be reached through a complex $w\ne 0$). So we restrict below the search to such $r$ values. We have two steps, first try to economically find an upper set for the range of $r$, then show that each $r$ in this range can be obtained through a $w$.

For the first step, use the mentioned inequality, so $$ r - \frac 1r = \left| \ r - \frac 1r\ \right| = \left| \ |w| - \frac 1{|w|}\ \right| \le \left| \ w - \frac 1{w}\ \right| =1\ . $$ The obtained inequality (for left most and right most parts above) gives an upper set for the needed range: $$ r\in[1, \varphi]\ ,\qquad\text{ where }\varphi=\frac 12(1+\sqrt 5) $$ is the golden number. Its inverse is $\displaystyle \frac 1\varphi = \frac 12(-1+\sqrt 5)=\varphi-1$.

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Let us show now that each such value can be obtained. Fix $r$ in the given range. Write $w=ru$, $u=\cos t+i\sin t$, $t$ real, being (searched) on the unit circle. The wanted relation gives $$ 1 =\left|ru-\frac 1{ru}\right|^2 =\left|ru^2-\frac 1r\right|^2 =\left(r\cos 2t-\frac 1r\right)^2 + r^2\sin^2 2t =r^2\color{blue}{-2\cos 2t} +r^{-2}\ . $$ The function $r\to r^2+r^{-2}$ is strictly increasing for $r\in[1,\infty)$, takes the minimal value $2$ in $r=1$, and in $r=\varphi$ the value is $$ \varphi^2+ \varphi^{-2} =\frac 14(\ (\sqrt 5+1)^2+(\sqrt 5-1)^2\ )=3 \ . $$
So for $r\ge 1$ the expression $1-(r^2+r^{-2})$ takes values between $1-2=-1$ and $1-3=-2$, and for each value in $[-1,-2]$ we can arrange with an appropriate $t$ that $\color{blue}{-2\cos 2t}$ matches such a value.

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Since $w$ covers exactly the range $[\varphi^{-1},\varphi]$, the corresponding range for $z=2w$ is $[2\varphi^{-1},2\varphi]= [\sqrt 5+1,\ \sqrt 5-1]$.