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 dnβ where n and d are integers with 1β€dβ€5. What is the probability that
(cos(aΟ)+isin(bΟ))4
is a real number?
Answer Choices:
A. 503β
B. 254β
C. 20041β
D. 256β
E. 5013β
Solution:
There are 20 possible values for each of a and b, namely those in the set
S={0,1,21β,23β,31β,32β,34β,35β,41β,43β,45β,47β,51β,52β,53β,54β,56β,57β,58β,59β}
If x and y are real numbers, then (x+iy)2=x2βy2+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 x2βy2=0 or xy=0, that is, x=0,y=0, or x=Β±y. Thus ((cos(aΟ)+isin(bΟ))4 is a real number if and only if cos(aΟ)=0,sin(bΟ)=0, or cos(aΟ)=Β±sin(bΟ). If cos(aΟ)=0 and aβS, then a=21β or a=23β and b has no restrictions, so there are 40 pairs (a,b) that satisfy the condition. If sin(bΟ)=0 and bβS, 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 (21β,0),(21β,1),(23β,0), and (23β,1). Thus the total so far is 40+40β4=76.
Note that cos(aΟ)=sin(bΟ) implies that cos(aΟ)=cos(Ο(21ββb)) and thus aβ‘21ββb(mod2) or aβ‘β21β+b(mod2). If the denominator of bβS is 3 or 5 , then the denominator of a in simplified form would be 6 or 10 , and so aβ/S. If b=21β or b=23β, then there is a unique solution to either of the two congruences, namely a=0 and a=1, respectively. For every bβ{41β,43β,45β,47β}, there is exactly one solution aβS to each of the previous congruences. None of the solutions are equal to each other because if 21ββbβ‘β21β+b(mod2), then 2bβ‘1 (mod2); that is, b=21β or b=23β. Similarly, cos(aΟ)=βsin(bΟ)=sin(βbΟ) implies that cos(aΟ)=cos(Ο(21β+b)) and thus aβ‘21β+b(mod2) or aβ‘β21ββb (mod2). If the denominator of bβS is 3 or 5 , then the denominator of a would be 6 or 10 , and so aβ/S. If b=21β or b=23β, then there is a unique solution to either of the two congruences, namely a=1 and a=0, respectively. For every bβ{41β,43β,45β,47β}, there is exactly one solution aβS 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)βS2 such that cos(aΟ)=Β±sin(bΟ). The requested probability is 40076+20β=40096β=256ββ.
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=βy. 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