Can I have your help with an algebra/complex numbers related question?
Question :
What is the least possible integer for n, such that [(1+i)(1-i)]^n = 1.
Note that i is the squre root of -1.
First I multiplied both numerator & denominated by (1+i) to get,
[(1+i)/(1-i)]^n = [ (1 + 2i - 1)/2]^n = (i)^n
==> i^n = 1
I know a real valued multiple of 4 would do, but how can we find the least possible integer valued n which suits This, How do we present it?
Since you know that n must be a multiple of 4,
Just pick n=4
It's clear that
i^1 = i
i^2 = -1
i^4 = (i^2)^2 = (-1)^2 = 1
Of course, i^0 is also 1, but maybe you wanted a positive integer...
So the question is asking for the least possible integer for n. In this case-4 , -8 etc. Would work too right?
How do we could pick the smallest one?
To find the least possible integer value for n that satisfies the equation i^n = 1, we can use Euler's formula. Euler's formula states that e^(ix) = cos(x) + isin(x), where e is the base of the natural logarithm. Applying this formula to i^n = 1, we have e^(i*pi/2 * n) = 1.
To find the least possible integer value for n, we need to find the smallest positive value for n that satisfies the equation. In this case, n must be a multiple of 4 because the period of the complex exponential function is 4. This means that any multiple of 4 will give us a solution.
So, the least possible integer value for n that satisfies the equation is n = 4.