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If $z$ is a complex no. such that $\displaystyle \left|z+\frac{2}{z}\right| = 2\,$ Then find max. and min. value of $\left|z\right|$.

$\bf{My\; Try:}$ Given $\displaystyle \left|z+\frac{2}{z}\right| = 2\Rightarrow \left|z+\frac{2}{z}\right|^2 = 2^2=4$.

So $\displaystyle \left(z+\frac{2}{z}\right)\cdot \left(\bar{z}+\frac{2}{\bar{z}}\right) = 4\Rightarrow \left|z\right|^2+2\left(\frac{z}{\bar{z}}+\frac{\bar{z}}{z}\right)+\frac{1}{|z|^2} = 4$.

Now how can I find the max. and min. values of $|z|$?

Help me please.

Thanks

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$$|z|-\frac{2}{|z|}\le \left|z+\frac{2}{z}\right|=2 \implies\\ |z|^2-2|z|-2\le0\implies\\ 1-\sqrt{3}<0\le|z|\le1+\sqrt{3}$$

Maximum value is $1+\sqrt{3}$ provided we show that equality holds.

Recall that equality holds when $z$, $-\frac{2}{z}$ have same argument. Hence $\arg{z}=-\frac{\pi}{2}$.

Similarly using $\frac{2}{|z|}-|z|\le \left|z+\frac{2}{z}\right|=2$ we can show that the minimum is $\frac{2}{1+\sqrt{3}}=\sqrt{3}-1$

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Let $z=\color{red}{\sqrt{2}}re^{i\theta}$, then $\left|z+\frac{2}{z}\right|=2$ reduces to $$r^2+\frac{1}{r^2}=2-2\cos{2\theta} \tag{1}$$

Observe that $(1)$(considering it as a quadratic in $r^2$) will have two real roots of the form $\alpha$ and $\frac{1}{\alpha}$ only if $\cos{2\theta} \le 0$. Thus if the maximum value is $\alpha$, the minimum would be $\frac{1}{\alpha}$.

Without loss of generality assume maximum(of $r^2$) is $\ge 1$, notice that $r^2+\frac{1}{r^2}$ is an increasing function when $r\ge 1$, hence maximum is attained when R.H.S is maximum, i.e. when $\cos{2\theta}=-1$. Solving for $r$, get $r=\sqrt{2+\sqrt{3}}\quad,\quad\frac{1}{\sqrt{2+\sqrt{3}}}$

Maximum value of $|z|=\sqrt{2}\cdot \sqrt{2+\sqrt{3}}=1+\sqrt{3}$ and minimum value of $|z|=\frac{2}{1+\sqrt{3}}=\sqrt{3}-1$.

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  • $\begingroup$ I think your formula (1) should have $4-2 \cos(2 \theta)$? $\endgroup$ – copper.hat Jul 2 '14 at 7:05
  • $\begingroup$ @copper.hat. I was trying to prove for $|z+\frac{\color{red}{2}}{z}|=2$. and took $z=\color{red}{\sqrt{2}}{re^{i\theta}}$ $\endgroup$ – jdoicj Jul 2 '14 at 7:06
  • $\begingroup$ Crap, I goofed. Thanks! $\endgroup$ – copper.hat Jul 2 '14 at 7:07
  • $\begingroup$ @copper.hat yours is nothing compared to mistakes I commit! $\endgroup$ – jdoicj Jul 2 '14 at 7:09
  • $\begingroup$ Well, I have made an uncountable number of mistakes already. As an aside, I think you need to adjust your minimum value to $ { \sqrt{2} \over \sqrt{2 + \sqrt{3}}} = \sqrt{2} \sqrt{2 - \sqrt{3}}$. $\endgroup$ – copper.hat Jul 2 '14 at 7:24
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Here is one approach:

Let $z=r e^{i\theta}$, then we want to compute the $\min,\max$ of $I=\{ r \mid \, |r e^{i2\theta} + {2 \over r}| = 2 \}$. We see that $r \in I$ iff $|r \cos ( 2 \theta)+ {2 \over r} +i r \sin ( 2 \theta)|^2 = r^2 +{4 \over r^2} + 4 \cos(2 \theta)= 4$ for some $\theta$.

If we let $f(r) = 1-{1 \over 4} (r^2 +{4 \over r^2} )$, we see that $I = f^{-1} [-1,1]$.

The function $f$ is strictly concave, $\lim_{r \downarrow 0} f(r) = \lim_{r \to \infty} f(r) = -\infty$, $f$ has a maximum value of $0$ at $r=\sqrt{2}$. Hence $I = [r_0,r_1]$ where $r_k$ solve $f(r_k) = -1$.

Solving $f(r) = -1$ yields the equation $r^2 +{4 \over r^2} = 8$ which has positive solutions $r = \sqrt{2} \sqrt{2 \pm \sqrt{3}}$.

Hence the minimum value of $|z|$ is $\sqrt{2} \sqrt{2 - \sqrt{3}}$ and the maximum value if $\sqrt{2} \sqrt{2 + \sqrt{3}}$.

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Let $z=re^{i\theta}$, then your last step can be written as $$r^2+2\cos (2\theta)+\frac{1}{r^2}=4.$$ From this we can get \begin{align*} \left(r+\frac{1}{r}\right)^2 & = 6-2\cos (2\theta) \end{align*} Thus $2 \leq \left(r+\frac{1}{r}\right) \leq 2\sqrt{2}.$ Now try to find the range of $r$.

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