Saturday, February 21, 2015

What is the complex number that has the square of 5+12i ?

Let the number that has a square of 5 + 12i be a +
bi


(a + bi)^2 = 5 + 12i


use (a
+ b)^2 = a^2 + b^2 + 2*a*b


=> a^2 + b^2*i^2 +
2*a*b*i = 5 + 12i


i^2 =
-1


=> a^2 - b^2 + 2*a*b*i = 5 +
12i


equate the real and complex
coefficients


=> a^2 - b^2 = 5 and ab =
6


a = 6/b


substitute in a^2 -
b^2 = 5


=> 36/b^2 - b^2 =
5


=> 36 - b^4 =
5b^2


=> b^4 + 5b^2 - 36 =
0


=> b^4 + 9b^2 - 4b^2 - 36 =
0


=> b^2( b^2 + 9) - 4(b^2 + 9) =
0


=> (b^2 - 4)(b^2 + 9) =
0


=> b^2 = 4 and b^2 =
-9


b is a real number, so we eliminate b^2 =
-9


b^2 = 4


=> b = 2 and
b = -2


a = 3 and a =
-3


The required number can be 3 + 2i and -3
-2i

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