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√(16 - 30i)

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Let, (a + ib)2 = 16 -30i

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

⇒ a2 + (bi)2 + 2abi = 16 - 30i

Since i2 = -1

⇒ a2 - b2 + 2abi = 16 - 30i

Now, separating real and complex parts, we get

⇒ a2 - b2 = 16…………..eq.1

⇒ 2ab = - 30…….. eq.2

⇒ a = -15/b

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

⇒ \((-\frac{15}{b})^2\) – b2 = 16

⇒ 225 – b4 = 16b2

⇒ b4 +16b2 - 225= 0

Simplify and get the value of b2  we get,

b2 = -25 or b2 = 9

As b is real no. so, b2 = 9

b = 3 or b = -3

Therefore, a = - 5 or a = 5

Hence the square root of the complex no. is - 5 + 3i and 5 - 3i.

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