How does the square root of -6i-8 equal 1 +3i?
To find the square root of -6i - 8, we can follow these steps:
Step 1: Express the given complex number in the form a + bi.
-6i - 8 can be rearranged as -8 - 6i.
Step 2: Apply the square root formula for complex numbers.
The square root of a complex number in the form a + bi can be found using the formula:
sqrt(z) = sqrt(r) * [cos(theta/2) + i * sin(theta/2)]
Step 3: Convert -8 - 6i to its polar form.
To convert -8 - 6i to polar form (r, theta), we need to find the magnitude (r) and the angle (theta). The magnitude can be calculated using the formula:
r = sqrt(a^2 + b^2)
In this case, a is -8 and b is -6, so we have:
r = sqrt((-8)^2 + (-6)^2) = sqrt(64 + 36) = sqrt(100) = 10
To find the angle (theta), we can use the formula:
theta = arctan(b/a)
In this case, b is -6 and a is -8, so we have:
theta = arctan((-6)/(-8)) = arctan(0.75) ≈ 36.87 degrees
Therefore, -8 - 6i in polar form is approximately 10 ∠ 36.87°.
Step 4: Apply the square root formula for complex numbers.
Using the square root formula for complex numbers, we can substitute the values we obtained from step 3 into the formula:
sqrt(z) = sqrt(r) * [cos(theta/2) + i * sin(theta/2)]
sqrt(-6i - 8) = sqrt(10) * [cos(36.87/2) + i * sin(36.87/2)]
Evaluating the expression, we get:
sqrt(-6i - 8) ≈ 1 + 3i
Therefore, the square root of -6i - 8 is approximately 1 + 3i.