Hey, how do I solve this? A bit of an explanation too please
To solve this problem all you need to know is that the question is asking you to find the smallest number n such that the expression is equal to some a*i where a is a real number. In my explanation I solved it by using the exponential form.
So to the solution:
You can turn the inner part (2-2sqrt(3)) into the form re^(itheta)
r = sqrt(a^2 + b^2)
a = 2 and b = -2sqrt(3) in this case so r = 4
factor out the 4 and you get 4*(1/2 - (sqrt(3)/2))
1/2 = cos(theta) since e^(i*theta) = cos(theta) + isin(theta)
therefore theta = pi/3
So our number is 4*e^(i pi/3)
Raise that to the nth power and you get 4*e^(i n pi/3)
This has to be equal to some imaginary number a*i where a is some real number
You can then divide both sides by 4 and take the natural log on both sides to get i*n pi/3 = ln(ia/4) = ln(i) + ln(a/4) = i pi/2 + ln(a/4)
Then multiply both sides by 3/(i*pi) to get n = 3/2 + something times the natural log term.
Since the minimum value the natural log term can have is 0, the minimum value of n = 3/2
Let me know if anything was unclear in my explanation of the solution.