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Let be given the complex $z$ such that $$\left |z + \dfrac{1}{z}\right |=3.$$ Find maximum and minimum modul of complex $z$.
I tried. Put $z=x+i y$. From $$\left |z + \dfrac{1}{z}\right |=3.$$ We have $z^2 + 3z|$ = 3|z|.$ Threrefore
$$1 + 2 x^2 + x^4 - 2 y^2 + 2 x^2 y^2 + y^4 - 9 (x^2 + y^2)=0.$$ From here, I cann't find max and min of the $\sqrt{x^2 + y^2}.$
Hint. By using samjoe's comment $z\bar{z}=|z|^2$, we have that the given equation is equivalent to $$|z|^2+2\cos(\theta)+\frac{1}{|z|^2}=\left |z + \dfrac{1}{z}\right |^2=3^2=9, \;\;|z|^2+\frac{1}{|z|^2}=9-2\cos(\theta)$$ where $z=|z|(\cos(\theta)+i\sin(\theta))$. Note that $$7=9-2\cos(0)\leq|z|^2+\frac{1}{|z|^2}\leq 9-2\cos(\pi)= 9+2=11.$$ Now try to show that the maximum of $|z|$ is attained when $$|z|^2+\frac{1}{|z|^2}=11,$$ that is for $|z|=(11+3\sqrt{13})/2$.
What about the minimum?
