Deep root

Algebra Level 5

The real solution of the equation below can be expressed as $x=z+z^{*}$ , where $z$ is a complex number, with $|z|=1$ . If $\arg{z}=\frac{a}{b}\pi$ , where $a$ and $b$ are co-prime, find $b-a$ . $x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$

The answer is 7.

2 solutions

Alan Enrique Ontiveros Salazar
Aug 6, 2014

The identity $\cos \dfrac{\theta}{2}=\sqrt{\dfrac{1+\cos \theta}{2}}$ will be very useful here. If we multiply both sides of that identity by $2$ , we get: $2\cos \dfrac{\theta}{2}=\sqrt{2+2\cos \theta}$ . Also, the identity $\sin \dfrac{\theta}{2}=\sqrt{\dfrac{1-\cos \theta}{2}}$ will help us. That suggest us to let $x=2\cos \theta$ :

$2\cos \theta=\sqrt{2+\sqrt{2-\sqrt{2+2\cos \theta}}}$

The "deep roots" will be deleted by a chain reaction like this, using the identities:

$2\cos \theta=\sqrt{2+\sqrt{2-2\cos \dfrac{\theta}{2}}}$

$2\cos \theta=\sqrt{2+2\sin \dfrac{\theta}{4}}$

$2\cos \theta=\sqrt{2+2\cos \left(\dfrac{\pi}{2}-\dfrac{\theta}{4}\right)}$

$2\cos \theta=2 \cos \left(\dfrac{\pi}{4}-\dfrac{\theta}{8}\right)$

Since we are looking only for the solutions in the first quadrant:

$\theta=\dfrac{\pi}{4}-\dfrac{\theta}{8} \Longrightarrow \dfrac{9\theta}{8}=\dfrac{\pi}{4} \Longrightarrow \theta = \dfrac{2\pi}{9}$

So, $x=2\cos\left(\dfrac{2\pi}{9}\right)$ .

Finally, if we let $z=R \text{ cis } \theta$ , clearly $\arg z=\theta$ , and we know that $z+z^*=2 R \cos \theta$ .

Equating it with $x$ we have:

$x=z+z^* \Longrightarrow 2\cos\left(\dfrac{2\pi}{9}\right)=2R\cos\theta \Longrightarrow \theta=\dfrac{2\pi}{9} \Longrightarrow \arg z=\dfrac{2\pi}{9}$

Hence, $a=2$ , $b=9$ and $b-a=\boxed{7}$ .

Great! The only suggestion I give to you is that there's a smoother way of you noticing that you can make a trigonometric substitution. First, you test the conditions that there's $x$ such it's root of the equation. These conditions are:

1) $x \geq 0$ , since $x$ is a radical.

2) $2+x \geq 0 \Rightarrow -2 \leq x$ , which doesn't help us too much, since the first condition already says that.

3) $2-\sqrt{2+x}\geq 0 \Rightarrow 2 \geq \sqrt{2+x} \Rightarrow x \leq 2$

From 1 and 3 , we got: $0 \leq x \leq2$ , therefore we can make a trigonometric substitution, such that:

$x=2\cos{\theta}, \theta \in \left[0, \frac{\pi}{2} \right]$ and then you don't need to worry about modulus, as long as to this domain of $\theta$ (first quadrant), all the values of $\sin{\theta}$ and $\cos{\theta}$ are positive.

Dieuler Oliveira - 6 years, 10 months ago

I'm a bit confused by the solution. In the step $2\cos(\frac{2\pi}{9})=2R\cos\theta$ , how do we get that $\theta=\frac{2\pi}{9}$ ? We have two variables $R$ and $\theta$ , so if we had, for example, $R=\cos\frac{2\pi}{9}$ and $\theta=0$ , couldn't those also be solutions to the equation? For every value of $\theta$ , there's a value for $R$ that also works.

Raymond Lin - 6 years, 10 months ago

Actually you are right. There's infinite integer solutions for this problem. I hadn't payed attention on that, when I set this part of the problem. At first I thought of fixing the modulus of $z$ as 1, but I thought it was unnecessary, without giving enough attention for that. I've already fixed the problem. Thank you!

Dieuler Oliveira - 6 years, 10 months ago

Incredible Mind
Jan 18, 2015

i suggest

x=cos(8u)

then we need not mess with fractions inside cosines

Deep root

The answer is 7.

2 solutions

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