# Minimum and Maximum value of |z|

This is a question that I came across today:

If $|z-(2/z)|=1$...(1) find the maximum and minimum value of |z|, where z represents a complex number.

This is my attempt at a solution:

Using the triangle inequality, we can write:

$||z|-|2/z||≤|z+2/z|≤|z|+|2/z|$

Let $|z|=r$ which implies that $|r-2/r|≤1≤r+2/r$ (From (1))

Since $|z-2/z| = 1$, we have \begin{aligned} 1 = |z- 2/z|^2 &= (z - 2/z)(\overline{z} - 2/\overline{z}) \\ &= |z|^2 - 2z/\overline{z} - 2\overline{z}/z + 4/|z|^2 \end{aligned} Write $z = re^{i\theta}$. Then $\overline{z} = re^{-i\theta}$. Substituting into the above, we obtain \begin{aligned} 1 &= r^2 - 2e^{i2\theta} - 2^{-i2\theta} + 4/r^2 \\ &= r^2 - 4\cos(2\theta) + 4/r^2 \\ \end{aligned} We can rearrange this to get $$r^2 + 4/r^2 \leq 4\cos(2\theta) + 1 \leq 5$$ or equivalently, $$r^4 -5r^2 + 4 \leq 0$$ Factor the left hand side to obtain $$(r-1)(r+1)(r-2)(r+2) \leq 0$$ As $r$ must be positive, this means that $r+1$ and $r+2$ are positive. Therefore, $r-1$ and $r-2$ must have opposite signs (or one of them is zero), which forces $1 \leq r \leq 2$.

You can check that $z=1$ and $z=2$ are solutions to the original equation, so $1$ and $2$ are the minimum and maximum values of $|z|$.

Here is a picture showing the set of solutions. From this, we might be tempted to speculate that we are looking at two circles of radius $1/2$, centered at $z=3/2$ and $z=-3/2$. But this is not the case. To see this, consider the point $z = 3/2 + i(1/2)$. This point is on the circle of radius $1/2$ centered at $z=3/2$. However, it does not satisfy the given equation: \begin{aligned} |z-2/z| &= |3/2 + i(1/2) - 2/(3/2 + i(1/2))| \\ &= |3/2 + i(1/2) - 6/5 + i(2/5)| \\ &= |3/10 + i(9/10)| \\ &= \sqrt{9/10} \neq 1 \end{aligned} I don't think it's an ellipse, either, since these are of the form $|z-a| + |z-b| = c$. If there is a way to transform $|z-2/z| = 1$ into that form, it's not obvious to me (also, our figure has not one but two "ellipses"). Using Complex Inequalities, $$||z|-|w||\le|z+w|\le|z|+|w|$$

$w=-\dfrac2z$ and writing $|z|=r$

$$\left|r-\dfrac2r\right|\le\left|z-\dfrac2z\right|\le r+\dfrac2r$$

$$\implies\left|r-\dfrac2r\right|\le1\le r+\dfrac2r$$

Now as $r>0,$ $$\dfrac{r+\dfrac2r}2\ge\sqrt{r\cdot\dfrac2r}=\sqrt2\iff r+\dfrac2r\ge2\sqrt2>1$$

So, we need $$\left|r-\dfrac2r\right|\le1\iff-1\le r-\dfrac2r\le1$$

$$r-\dfrac2r\le1\iff r^2-r-2\le0$$

Now we know $(x-a)(x-b)\le0$ with $a\le b;a\le x\le b$

Here $-1\le r\le2$ But $r>0\implies0<r\le2$

Can check for $-1\le r-\dfrac2r$ to find $r\ge1$

So, the final range $$1\le r\le2$$

Using this equality we can find both minimum and maximum. $$\left|z-\frac2z \right|\ge \left||z|-\frac2{|z|}\right|$$ $$\left||z|-\frac2{|z|}\right| \le 1$$ $$-1 \le |z|-\frac2{|z|}\le 1$$ $$|z|-\frac2{|z|}\ge -1$$ $$\left(|z| +\frac12\right)^2\ge \frac94$$ $$|z| \ge 1$$ $$|z|-\frac2{|z|}\le 1$$ $$\left(|z| - \frac12\right)^2\le \frac94$$ $$|z| \le 2$$ $$\implies 1\le |z| \le 2$$

• How did you arrive at that very last inequality by solving the one directly above it? – user361896 Aug 23 '16 at 10:30
• @KaumudiHarikumar Using last inequality,we found the minimum and maximum was founded above . – Aakash Kumar Aug 23 '16 at 14:59
• My question is how? – user361896 Aug 23 '16 at 23:48