# Let, (a + ib)2 = 1 - i

Now using, (a + b)2 = a2 + b2 + 2ab a2 + (bi)2 + 2abi = 1 – i

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

Now, separating real and complex parts, we get a2 - b2 = 1…………..eq.1 2ab = -1…….. eq.2 a = Now, using the value of a in eq.1, we get – b2 = 1 1 – 4b4 = 4b2 4b4 + 4b2 -1= 0

Simplify and get the value of b2, we get, b2 = As b is real no. so, b2 = b2 = b= or b= - Therefore , a= - or a= Hence the square root of the complex no. is + i and  i.

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Evaluate:

(i) (iii) .

RS Aggarwal - Mathematics