What are the two solutions to this math problem?

Sqrt (55 + 48i)

a^2 + 2 i a b - b^2 = 55 + 48 i

ab = 24 

(a + b)(a - b) = 55 

(8 + 3)(8 - 3) = 55 or (64 - 9)

(8 + 3i)^2 = 55 + 48i

(-8 - 3i)^2 = 55 + 48i

(a + bi)^2 = a^2 – b^2 +2abi

ab = 24 = 6*4 = 8*3

a^2 – b^2 = 55 = 64 – 9

The roots were 8 + 3i and -8 - 3i

We have:

a + bi = √(55 + 48i)

If we square both sides and simplify the left side we get:

(a + bi)² = 55 + 48i

a² + 2abi + b²i² = 55 + 48i

i² = -1, so:

a² + 2abi - b² = 55 + 48i

Now we can group that into standard complex number form:

a² - b² + 2abi = 55 + 48i

(a² - b²) + (2ab)i = 55 + 48i

We can now compare the real portion of the two sides and make an equation and the complex portion of the two sides to make a second equation:

a² - b² = 55 and 2ab = 48

We now have a system of two equations and two unknowns that can be solved.

2ab = 48

ab = 24

b = 24/a

a² - b² = 55

a² - (24/a)² = 55

a² - 576/a² = 55

a⁴ - 576 = 55a²

a⁴ - 55a² - 576 = 0

If we substitute x = a²:

x² - 55x - 576 = 0

x = [ -b ± √(b² - 4ac)] / (2a)

x = [ -(-55) ± √((-55)² - 4(1)(-576))] / (2 * 1)

x = [ 55 ± √(3025 + 2304)] / 2

x = [ 55 ± √(5329)] / 2

x = (55 ± 73) / 2

x = -18/2 and 128/2

x = -9 and 64

If we substitute those back into a²:

x = -9 and x = 64

a² = -9 and a² = 64

a = ±3i and a = ±8

"a" must be real so we can throw out the first two solutions:

a = ±8

Now we can find the two "b" values:

b = 24/a

b = 24/(-8) and b = 24/8

b = -3 and b = 3

So your two solutions are:

-8 - 3i and 8 + 3i

√(55+48i) = +√(55+48i)  and -√(55+48i)   ............ANS