Mock AIME I 2012 Problems/Problem 8

Problem

Suppose that the complex number $z$ satisfies $\left|z\right| = \left|z^2+1\right|$ . If $K$ is the maximum possible value of $\left|z\right|$ , $K^4$ can be expressed in the form $\dfrac{r+\sqrt{s}}{t}$ . Find $r+s+t$ .

Solution

We begin by dividing both sides by $|z|\ne 0$ to obtain $1 = \frac{\left|z^2+1\right|}{\left|z\right|} = \left|\frac{z^2+1}{z}\right| = \left|z+\frac{1}{z}\right|$ . Now, consider that we may write $z = re^{i\theta}=r\left(\cos(\theta)+i\sin(\theta)\right)$ with $r$ a positive real number so that $|z| = r$ and $\theta$ some real number. Then, $\left|z+\frac{1}{z}\right| = \left|re^{i\theta} + \frac{1}{r}\cdot e^{-i\theta}\right|= \left|r\left(\cos(\theta)+i\sin(\theta)\right)+\frac{1}{r}\left(\cos(-\theta)+i\sin(-\theta)\right)\right|$ $=\left|r\left(\cos(\theta)+i\sin(\theta)\right)+\frac{1}{r}\left(\cos(\theta)-i\sin(\theta)\right)\right|=\left|\left(r+\frac{1}{r}\right) \cos(\theta) +\left(r-\frac{1}{r}\right)i\sin(\theta)\right|$ $= \sqrt{\left(r+\frac{1}{r}\right)^2\cos^2(\theta)+\left(r-\frac{1}{r}\right)^2\sin^2(\theta)}=\sqrt{\left(r^2+\frac{1}{r^2}\right)\left(\cos^2(\theta)+\sin^2(\theta)\right)+2\left(\cos^2(\theta)-\sin^2(\theta)\right)}$ $= \sqrt{r^2+\frac{1}{r^2} + 2\cos(2\theta)}.$ Since we need $\left|z+\frac{1}{z}\right| = 1$ , we must have $r^2+\frac{1}{r^2} + 2\cos(2\theta) = 1$ , or equivalently, $r^4 + \left(2\cos(2\theta) - 1\right)r^2 + 1 =0$ . By the quadratic equation, this has roots $r^2 = \frac{1-2\cos(2\theta)\pm \sqrt{ \left(2\cos(2\theta) - 1\right)^2-4}}{2},$ and to maximize $r$ , we take the larger root $r^2 = \frac{1-2\cos(2\theta)+\sqrt{ \left(2\cos(2\theta) - 1\right)^2-4}}{2}$ which is clearly maximized when $2\cos(2\theta)$ is minimized. Since $-1\le \cos(2\theta)\le 1$ , the maximum value of $r$ will occur where $2\cos(2\theta) = -2$ , so the maximum value of $r$ occurs where $r^2 = \frac{3 + \sqrt{5}}{2}$ and finally we find that the maximum value of $|z|=r>0$ is $r= \sqrt{\frac{3+\sqrt{5}}{2}} = \frac{\sqrt{6+2\sqrt{5}}}{2} = \frac{1+\sqrt{5}}{2},\text{ the golden ratio.}$ Taking the fourth power, the desired answer is $\dfrac{7+\sqrt{45}}{2} \implies r+s+t=\boxed{054}$ .

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Mock AIME I 2012 Problems/Problem 8

Problem

Solution