Range of $|z_1+z_2|$ for some $z_1,z_2\in\Bbb C$ 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!
 A: 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
$$\left||z|-\frac4{|z|}\right|\le \left|z-\frac4z\right|\le \left||z|+\frac4{|z|}\right|$$

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

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 $[\sqrt5-1,\sqrt5+1]$.
A: 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 $[\,\sqrt 5 - 1,\sqrt 5 +1\,]$.
A: (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
$\displaystyle \frac 1\varphi = \frac 12(-1+\sqrt 5)=\varphi-1$.

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.

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