Problem
Let A = {(1967 + 1686isinθ)/(7 – 3icosθ): θ ∈ R}. If A contains exactly one positive integer n, then find the value of n.
Solution
To begin, we’ll equate the imaginary part of the given expression to 0. This will give us some values of θ.
Then, from these set of values, we’ll find the values of θ for which the real part of the expression is a positive integer.
Finally, we’ll find that positive integer n.
First, let’s try to ‘clean up’ the given expression. If we factorize both the numbers in the denominator, we get 1967 = 281 x 7 and 1686 = 281 x 6.
So, the expression simplifies to 281(7 + 6isinθ)/(7 – 3icosθ) – much easier to manipulate.
To proceed further, let’s multiply and divide it by the conjugate of the denominator, i.e. (7 + 3icosθ).
I’ll skip the calculations. What we’ll finally get is 281[(49 – 9sin2θ) + 21i(2sinθ + cosθ)]/(49 + 9cos2θ).
Now, if we equate the imaginary part to 0, we’ll get 2sinθ + cosθ = 0, which gives tanθ = -1/2.
The real part of the expression equals 281(49 – 9sin2θ)/(49 + 9cos2θ).
Using tanθ = -1/2, we’ll get sin2θ = -4/5 and cos2θ = 4/5.
Finally, we’ll substitute these values in the previous expression to get 281(49 – 9 x (-4/5))/(49 + 9 x 4/5).
The expression in the brackets are the same, and get cancelled out. What we’re left with is 281, a positive integer.
So, the value of n is 281.
Comments
The problem was fairly easy, except for the scary numbers. The level could’ve been amped up by having multiple possible values for the real part, from which a positive integer had to be chosen.