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

Now using, (a + b)2 = a2 + b2 + 2ab a2 + (bi)2 + 2abi = 3 + 4 i

Since i2 = -1 a2 - b2 + 2abi = 3 + 4 i

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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Evaluate:

(i) (iii) .

RS Aggarwal - Mathematics