[root2(cos2pi/11+isin2pi/11)]^4
To simplify the expression [√2(cos(2π/11) + isin(2π/11))]^4, we can use De Moivre's theorem, which states that for any complex number z = r(cosθ + isinθ), the nth power of z is given by:
z^n = r^n(cos(nθ) + isin(nθ))
Let's apply this formula to simplify the given expression:
First, we have z = √2(cos(2π/11) + isin(2π/11)).
Next, we need to find z^4 using De Moivre's theorem. Therefore, we calculate z raised to the power of 4:
z^4 = (√2)^4 [cos(4(2π/11)) + isin(4(2π/11))]
Simplifying further, we get:
z^4 = 2 [cos(8π/11) + isin(8π/11)]
Hence, the simplified expression is 2(cos(8π/11) + isin(8π/11)).