# Evaluate the question √i

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√i

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Let, (a + ib)2 = 0 + i

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

⇒ a2 + (bi)2 + 2abi = 0 + i

Since i2 = -1

⇒ a2 - b2 + 2abi = 0 + i

Now, separating real and complex parts, we get

⇒ a2 - b2 = 0 …………..eq.1

⇒ 2ab = 1…….. eq.2

⇒ a = 1/2b

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

⇒ $(\frac{1}{2b})^2$ – b2 = 0

⇒ 1 – 4b4 = 0

⇒ 4b2 = 1

Simplify and get the value of b2 , we get,

⇒ b2 = -$\frac{1}{2}$ or b2$\frac{1}{2}$

As b is real no. so, b2 = 3

b = $\frac{1}{\sqrt2}$ or b = -$\frac{1}{\sqrt2}$

Therefore , a = $\frac{1}{\sqrt2}$ or a = -$\frac{1}{\sqrt2}$

Hence the square root of the complex no. is $\frac{1}{\sqrt2}$ + $\frac{1}{\sqrt2}$i and - $\frac{1}{\sqrt2}$ - $\frac{1}{\sqrt2}$i.