Q. 115.0( 1 Vote )

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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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