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This problem was taken from the AMC 12 2015 contest B.

Rational numbers $a$ and $b$ are chosen at random among all rational numbers in the interval $[0,2)$ that can be written as fractions $\frac{n}{d}$ where $n$ and $d$ are integers with $1 \le d \le 5$. What is the probability that $$(\text{cos}(a\pi)+i\text{sin}(b\pi))^4$$

My approach: By De Moivre's Theorem, this could be rewritten as $4(\cos{4a\pi}+\sin{4b\pi})$. If this number is real, than the imaginary part must be $0$.

This occurs when $4b\pi = \pi{k}$ for any integer $k$. Or $4b = k$. Or $4\frac{p}{q}=k$ (p and q have GCF of $1$). Since $k$ is an integer, $q$ divides 4. So the denominators of the coveted fractions are factors of $4$.

These are the possible values of a or b: $0,1,\frac12,\frac32,\frac13,\frac23,\frac43,\frac53,\frac14,\frac34,\frac54,\frac74,\frac15,\frac25,\frac35,\frac45,\frac65,\frac75,\frac85,\frac95$ (Note that other possible values of n and d are either out of the interval or simplify to the above fractions.)

Out of these there are $8$ that do the job. I felt that no matter what $a$ is, it is only up to $b$ to determine if the number is real.

So I reached a probability of $\frac25$. The answer is $\frac{6}{25}$. Where did I go wrong? Thanks.

Is De Moivre's theorem even applicable at all?

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    De Moivre's theorem says that $(\cos{x}+i\sin{x})^n=\cos{nx}+i\sin{nx}$. Here, $a\pi \neq b\pi$. So, I think that's where the mistake is. – 2017-02-15

2 Answers 2

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Define $f(a,b) = (\cos a\pi + i \sin b\pi)^4$. If $\sin b\pi = 0$ or $\cos a\pi = 0$, then $f(a,b)$ is real for obvious reasons. If neither is zero, then $f(a,b)$ is real only if $$\cos^2 a \pi - \sin^2 b \pi = 0$$ which would imply $\cos a \pi = \pm \sin b \pi$, or equivalently, $$a \pm b - \frac{1}{2} \in \mathbb Z.$$ From here, enumeration is straightforward.

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You cannot use De Moivre's theorem unless $a = b$.