Answer:
The square roots of -36+36√3i are 3+3i and -3-3i.
Explanation:
We can first factor out a common factor of 36 from -36+36√3i to get:
-36+36√3i = 36(-1+√3i)
To find the square roots of this expression, we can use the fact that the square roots of -1 are ±i. Thus, the square roots of -1+√3i are of the form:
±(a+bi)
where a and b are real numbers that we need to find. We can square this expression and set it equal to -1+√3i:
(a+bi)^2 = -1+√3i
Expanding the left side using FOIL, we get:
a^2 + 2abi - b^2 = -1+√3i
Equating the real and imaginary parts, we get two equations:
a^2 - b^2 = -1 (Equation 1)
2ab = √3 (Equation 2)
From Equation 2, we can solve for b:
b = √3/(2a)
Substituting this into Equation 1, we get:
a^2 - (√3/2a)^2 = -1
Simplifying, we get:
a^4 - 3 = 0
This quadratic equation has two real roots, which are:
a = ±√(3)/2
Substituting these values into the equation for b, we get:
b = ±1/2
Therefore, the two square roots of -36+36√3i are:
(√3/2 + i/2)√36 = 3 + 3i
-(√3/2 + i/2)√36 = -3 - 3i
Hence, the square roots of -36+36√3i are 3+3i and -3-3i.