?
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
llaffer
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