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2015-12-11T01:06:09+05:30

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Suppose that a+bi is a square root of 5 + 12i. 
Then, (a+bi)^2 = (a^2 - b^2) + (2ab)i = 5 + 12i. 
Equate real and imaginary parts: 
a^2 - b^2 = 5 
2ab = 12 ==> b = 6/a. 

So, a^2 - (6/a)^2 = 5 
==> a^2 - 36/a^2 = 5 
==> a^4 -5a^2 - 36 = 0. 
==> (a^2 -9)(a^2 + 4) = 0. 
Since a must be real, a = 3 or -3. 
This gives b = 2 or -2, respectively. 

Thus, we have two square roots: 3+2i or -3-2i.
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