Problem:
Let w=23β+iβ and z=2β1+i3ββ, where i=β1β. Find the number of ordered pairs (r,s) of positive integers not exceeding 100 that satisfy the equation iβ wr=zs.
Solution:
We start by writing the given complex numbers in polar form. Observe that w=23β+iβ=eiβ 6Οβ, z=2β1+i3ββ=eiβ 32Οβ,
and i=eiβ 2Οβ.
Substituting into the equation iβ wr=zs, we get: ei(2Οβ+rβ 6Οβ)=eiβ 32Οsβ
Multiplying both sides by 6 gives: 3+rβ‘4s(mod12), or rβ‘4sβ3(mod12).
We seek the number of ordered pairs (r,s) with 1β€r,sβ€100 satisfying this condition.
Let r=4sβ3+12k for some integer k. Rewriting, let t=s+3k, so r=4tβ3. We require 1β€rβ€100, so 4β€4tβ3β€100β4β€tβ€25.
Thus, tβ4,5,β¦,25, which gives 22 values for t. For each such t, the corresponding r=4tβ3 is fixed, and s must satisfy sβ‘t(mod3) and 1β€sβ€100.
Now we break into cases by t(mod3):
When tβ‘0(mod3): there are 8 such t, and sβ‘0(mod3) gives 33 valid s.
When tβ‘1(mod3): there are 9 such t, and sβ‘1(mod3) gives 34 valid s.
When tβ‘2(mod3): there are 5 such t, and sβ‘2(mod3) gives 33 valid s.
So the total number of valid (r,s) pairs is: 8β 33+9β 34+5β 33=264+306+165=834β.