Question

Find the square root of i.

Answer 1

Answer:

Here, we need to find √i

Here ,

i is a complex number.

i = √-1

We need to find √i

√i = √√-1

[ °.° ⁿ√√a = ²ⁿ√ a ]

√i = ⁴√-1

Concept:

Complex numbers : Complex numbers are the numbers which has a negative number under a square root

Here

√-1 = i

i² = (√-1)² = -1

i³ = i².i = - 1.i = - i

i⁴ = i².i² = - 1(-1) = 1

Answer 2

Heye

your answer ⏬

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$\sqrt{i} = ( - 1) \frac{1}{4} \\ let \: \sqrt{i} = a - + ib \\ i = (a + ib) \\ i = {a}^{2} + {i}^{2} + {b}^{2} + 2ab \\ i = {a}^{2} - {b}^{2} + 2ab \\ {a - {b}^{2} } = \: 2ab \: = 1 \\ {a}^{2} = {b}^{2} =a = \frac{1}{26} \\ ( \frac{ 1}{26}) = {6}^{2} \\ \frac{1}{4b} = {b}^{2} \\ \\ {b}^{4} = \frac{1}{4} \\ \\ b = \frac{1}{ \sqrt{2} } \: \: = a \: \: \: \: 1 \times \frac{1}{ \sqrt{3} } \\ \sqrt{i} = + ( \frac{1}{ \sqrt{2} } + \frac{1}{ \sqrt{2} } ) \\ \\ \sqrt{1 \frac{ + }{1} } ( \frac{ \sqrt{2} }{2} + i \frac{ \sqrt{2} }{2} \\ \\ \\ is \: your \: answer$

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mark me brainliest..........