View source for Mock AIME I 2012 Problems/Problem 8

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

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

<math>z=re^{i\theta}</math>.

Note that

<math>|z|^2 = r^2</math>.

To compute <math>|z^2+1|^2</math>, connect <math>z^2</math> to the origin and construct an altitude to the x-axis. Extend this line one unit above <math>z^2</math> and connect that point to the origin to create <math>z^2+1</math>.

Now, we use the Law of Cosines on the triangle with vertices at the origin, <math>z^2</math>, and <math>z^2+1</math>, giving

<math>|z^2+1|^2 = r^4 + 1 - 2r^2 \cos(2\theta+90) = r^4 + 2r^2 \sin(2\theta)+1</math>

(or you could use the Pythagorean Theorem). Either way, we proceed from here as we did with Solution 1 to get <math>r + s + t = \boxed{054}</math>.

-Solution by '''thecmd999'''

==See Also==
*[[Mock AIME I 2012 Problems/Problem 9| Next Problem]]
*[[Mock AIME I 2012 Problems/Problem 7| Previous Problem]]
*[[Mock AIME I 2012 Problems]]