# 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 +...

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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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