Let a be a positive real number and let $M_a =\{z \in C^*: |z+\frac{1}{z}|=a\}$ Find the minimum and maximum value of $|z|$ when z$\in M_a$

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    $\begingroup$ What is $\mathbb{C}^*$? $\endgroup$ – copper.hat Feb 22 '13 at 5:52
  • $\begingroup$ $C*$ = non zero complex number $\endgroup$ – Sachin Sharmaa Feb 22 '13 at 6:19

Let $z=r(\cos \theta+i\sin \theta)$

$z^{-1}=r^{-1}(\cos \theta-i\sin \theta)$

We have ,

$|z+\frac{1}{z}|=|(r+1/r)\cos \theta +(r-1/r)i\sin\theta|$

So we have,

$\displaystyle|(r+1/r)\cos \theta +(r-1/r)i\sin\theta|=\sqrt{(r^2+1/r^2)+2\cos 2\theta}$

Thus we have ,

$(r^2+1/r^2)+2\cos 2\theta=a^2$

Let $r^2=x$,

$\Rightarrow x^2+1=x(a^2-2\cos2\theta)$

$\Rightarrow x^2-x(a^2-2\cos2\theta)+1=0$

Solving the quadratic we have,

$\displaystyle x=\frac{(a^2-2\cos 2\theta)+\sqrt{(a^2-2\cos 2\theta)^2-4}}{2}$ and $\displaystyle x=\frac{(a^2-2\cos 2\theta)-\sqrt{(a^2-2\cos 2\theta)^2-4}}{2}$

Clearly x attains its maximum value at $2\theta=0$ using the first expression,

The value is $\displaystyle \frac{(a^2+2)+\sqrt{(a^2+2)^2-4}}{2}$

So the maximum value of $r=\sqrt{\displaystyle \frac{(a^2+2)+\sqrt{(a^2+2)^2-4}}{2}}$

This on simplification turns out to be, $\displaystyle r=\frac{a+\sqrt{a^2+4}}{2}$

Minimum value can also be obtained in that way.


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