Answer :

Let, (a + ib)2 = 3 + 4i


Now using, (a + b)2 = a2 + b2 + 2ab


a2 + (bi)2 + 2abi = 3 + 4i


Since i2 = -1


a2 - b2 + 2abi = 3 + 4i


now, separating real and complex parts, we get


a2 - b2 = 3 …………..eq.1


2ab =4…….. eq.2


a =


Now, using the value of a in eq.1, we get


– b2 = 3


12 – b4 = 3b2


b4 + 3b2 - 28= 0


Simplify and get the value of b2, we get,


b2 = -7 or b2 = 4


as b is real no. so, b2 = 4


b= 2 or b=


Therefore , a= or a= -


Hence the square root of the complex no. is + 2i and - -2i.


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