Answer :

Let, (a + ib)2 = -11 - 60i


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


a2 + (bi)2 + 2abi = -11 - 60i


Since i2 = -1


a2 - b2 + 2abi = -11 - 60i


now, separating real and complex parts, we get


a2 - b2 = -11…………..eq.1


2ab = -60…….. eq.2


a =


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


– b2 = -11


900 – b4 = -11b2


b4- 11b2 - 900= 0


Simplify and get the value of b2, we get,


b2 = 36 or b2 = -25


as b is real no. so, b2 = 36


b= 6 or b= -6


Therefore , a= -5 or a= 5


Hence the square root of the complex no. is -5 + 6i and 5 – 6i.


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