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!
Best Answer
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 $[\sqrt5-1,\sqrt5+1]$.