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If the complex number $z$ satisfies the equation $\left\lvert z- {4\over z}\right\rvert = 2$ then the least and the greatest values of $|z|$ are ?

My try $\left\lvert z- {4\over z}\right\rvert = 2$

$\left| |z| - \left\lvert z- {4\over z}\right\rvert \right| \le 2$.

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  • $\begingroup$ See cut-the-knot.org/arithmetic/algebra/… $\endgroup$ Jan 19, 2016 at 11:39
  • $\begingroup$ How do you define $\mathrm{mod}(z)?$ $\endgroup$ Jan 19, 2016 at 11:40
  • $\begingroup$ mod(z) is defined as $\sqrt( x^2+y^2)$ if Z= x+iy $\endgroup$ Jan 19, 2016 at 11:41
  • $\begingroup$ @ labbhattacharjee i have seen it and know the inequalities but cannot understand how to proceed further $\endgroup$ Jan 19, 2016 at 11:44
  • $\begingroup$ Divide by $2$ and let $w = \frac{z}{2}$. The equation becomes $\lvert w - 1/w\rvert = 1$. Square it, you get (after a little rearranging) a quadratic equation in $x = \lvert w\rvert^2$. $\endgroup$ Jan 19, 2016 at 13:50

2 Answers 2

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Given $$\left|z-\frac{4}{z}\right| = 2$$ and here we have to find $\max$ and $\min$ of $|z|$

So $$|z| = \left|\left(z-\frac{4}{z}\right)+\frac{4}{z}\right|\leq \left|z-\frac{4}{z}\right|+\frac{4}{|z|}$$

Above we have used $\bf{\triangle \; Inequality}$

So $$|z|\leq 2+\frac{4}{|z|}\Rightarrow |z|^2-2|z|\leq 4$$

So $$\left(|z|-1\right)^2 \leq 5$$

Now after that You can solve it.

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    $\begingroup$ Hey. (excuse me for no mathjax. I will update this). If you consider the identity |z1(+-)z2| <= |z1| + |z2|, where z1=z and z2=4/z, then wont the inequality become 2<= |z|+|4/z|, which is totally the different thing? $\endgroup$ Dec 5, 2016 at 13:07
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If you know $\left|\frac zw\right|=\frac{|z|}{|w|}$, then you can solve your inequality for real $|z|$. Then see if real $z$ attains those max and min.
$$|r-4/r|\leq2\\-2r\leq r^2-4\leq2r\\5\leq(r+1)^2,(r-1)^2\leq5\\-1+\sqrt{5}\leq r\leq1+\sqrt{5}$$

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  • $\begingroup$ But how is this helpful?? $\endgroup$ Jan 19, 2016 at 13:39

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