Problem:
Rational numbers a and b are chosen at random among all rational numbers in the interval [0,2) that can be written as fractions dn where n and d are integers with 1≤d≤5. What is the probability that
(cos(aπ)+isin(bπ))4
is a real number?
Answer Choices:
A. 503
B. 254
C. 20041
D. 256
E. 5013
Solution:
There are 20 possible values for each of a and b, namely those in the set
S={0,1,21,23,31,32,34,35,41,43,45,47,51,52,53,54,56,57,58,59}
If x and y are real numbers, then (x+iy)2=x2−y2+i(2xy) is real if and only if xy=0, that is, x=0 or y=0. Therefore (x+iy)4 is real if and only if x2−y2=0 or xy=0, that is, x=0,y=0, or x=±y. Thus ((cos(aπ)+isin(bπ))4 is a real number if and only if cos(aπ)=0,sin(bπ)=0, or cos(aπ)=±sin(bπ). If cos(aπ)=0 and a∈S, then a=21 or a=23 and b has no restrictions, so there are 40 pairs (a,b) that satisfy the condition. If sin(bπ)=0 and b∈S, then b=0 or b=1 and a has no restrictions, so there are 40 pairs (a,b) that satisfy the condition, but there are 4 pairs that have been counted already, namely (21,0),(21,1),(23,0), and (23,1). Thus the total so far is 40+40−4=76.
Note that cos(aπ)=sin(bπ) implies that cos(aπ)=cos(π(21−b)) and thus a≡21−b(mod2) or a≡−21+b(mod2). If the denominator of b∈S is 3 or 5 , then the denominator of a in simplified form would be 6 or 10 , and so a∈/S. If b=21 or b=23, then there is a unique solution to either of the two congruences, namely a=0 and a=1, respectively. For every b∈{41,43,45,47}, there is exactly one solution a∈S to each of the previous congruences. None of the solutions are equal to each other because if 21−b≡−21+b(mod2), then 2b≡1 (mod2); that is, b=21 or b=23. Similarly, cos(aπ)=−sin(bπ)=sin(−bπ) implies that cos(aπ)=cos(π(21+b)) and thus a≡21+b(mod2) or a≡−21−b (mod2). If the denominator of b∈S is 3 or 5 , then the denominator of a would be 6 or 10 , and so a∈/S. If b=21 or b=23, then there is a unique solution to either of the two congruences, namely a=1 and a=0, respectively. For every b∈{41,43,45,47}, there is exactly one solution a∈S to each of the previous congruences, and, as before, none of these solutions are equal to each other. Thus there are a total of 2+8+2+8=20 pairs (a,b)∈S2 such that cos(aπ)=±sin(bπ). The requested probability is 40076+20=40096=256.
Note: By de Moivre's Theorem the fourth power of the complex number x+iy is real if and only if it lies on one of the four lines x=0,y=0,x=y, or x=−y. Then the counting of (a,b) pairs proceeds as above.
The problems on this page are the property of the MAA's American Mathematics Competitions