Friday, February 28, 2014

What is the square root of -24 + 70i?

Let the square root of -24 + 70i be a +
ib.


(a + ib)^2 = -24 +
70i


=> a^2 - b^2 + 2abi = -24 +
70i


equating the real and complex
parts


a^2 - b^2 = -24


2ab =
70


=> ab = 35


=>
b = 35/a


substitute in a^2 - b^2 =
-24


=> a^2 - (35/a)^2 =
-24


=> a^4 - 35^2 =
-24a^2


=> a^4 + 24a^2 - 35^2 =
0


let x = a^2


=> x^2 +
24x - 35^2 = 0


=> x^2 + 49x - 25x - 1225 =
0


=> x(x + 49) - 25(x + 49) =
0


=> (x - 25)(x + 49) =
0


=> x = 25 and x =
-49


but x = a^2


a^2 = 25
=> a = 5 and -5


a^2 = -49 gives complex values of a
but a is a real number, so we ignore this root.


a = 5 , b =
35/5 = 7


a = -5 , b = 35/-5 =
-7


The required square root is 5 + 7i and -5
- 7i

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