Yahoo Answers: Answers and Comments for What’s the square root of i? [Mathematics]
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From Anonymous
enCA
Thu, 02 Apr 2020 12:28:54 +0000
3
Yahoo Answers: Answers and Comments for What’s the square root of i? [Mathematics]
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From Anonymous: Let √i = x + iy (i.e. assume √i is a complex n...
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Thu, 02 Apr 2020 13:03:36 +0000
Let √i = x + iy (i.e. assume √i is a complex number and see if you get a sensible answer).
Squaring:
i = x² + 2ixy  y² (equation 1)
Equating real parts of equation 1:
x²  y² = 0
x = y
Equating imaginary parts of equation 1:
2xy = 1
Since x = y
2x² = 1
x = 1/√2 = √2/2
y =√2/2
√i = √2/2 + i√2/2
. . = (√2/2)(1 + i)
Check: If you square (√2/2)(1 + i) you will find you get i.

From Pinkgreen: i=e^(i pi/2)
=>
sqr(i)=e^(i pi/4)
=>
sqr...
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Fri, 03 Apr 2020 02:40:21 +0000
i=e^(i pi/2)
=>
sqr(i)=e^(i pi/4)
=>
sqr(i)=sqr(2)(1+i)/2

From Jim: Complex numbers are 'closed' for all o...
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Sat, 04 Apr 2020 19:06:59 +0000
Complex numbers are 'closed' for all operations.
√i = (1 + i)/√(2), or ½ (1 + i) √(2)

From Krishnamurthy: (1)^(1/4)
= 0.707106781... + 0.707106781... i
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Sat, 04 Apr 2020 18:59:59 +0000
(1)^(1/4)
= 0.707106781... + 0.707106781... i

From sepia: The square root of i:
0.7071067811865475244008...
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Thu, 02 Apr 2020 20:42:46 +0000
The square root of i:
0.70710678118654752440084436210484903928483593768847403658... +
0.70710678118654752440084436210484903928483593768847403658... i

From la console: Recall:
Z = a + ib ← this is a complex number...
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Thu, 02 Apr 2020 14:14:33 +0000
Recall:
Z = a + ib ← this is a complex number
M = √(a² + b²) ← this is its modulus
tan(α) = b/a → then you can deduce α ← this is the argument
Your case
Z = i → you can see that: a = 0 and you can see that: b = 1
m = √(0 + 1²) = 1 ← this is its modulus of Z
tan(α) = b/a → then you can deduce α = π/2
Then you must find a complex number z, such as: z² = Z
The modulus of z is: m = √M = √1 = 1
The argument of z is: β = α/2 = (π/2)/4 = π/4
So you can deduce that the first root of Z is:
z₁ = m.[cos(β) + i.sin(β)] → and to obtain the second, you add an angle of: (2π/2) → i.e.: π
z₂ = m.[cos(β + π) + i.sin(β + π)]
We've seen that: m = 1
z₁ = cos(β) + i.sin(β)
z₂ = cos(β + π) + i.sin(β + π)
We've seen that: β = π/4
z₁ = cos(π/4) + i.sin(π/4)
z₂ = cos[(π/4) + π] + i.sin[(π/4) + π] → recall: cos(x + π) =  cos(x)
z₂ =  cos(π/4) + i.sin[(π/4) + π] → recall: sin(x + π) =  sin(x)
z₂ =  cos(π/4)  i.sin(π/4)
Resume:
z₁ = cos(π/4) + i.sin(π/4)
z₂ =  cos(π/4)  i.sin(π/4)
You know that: cos(π/4) = sin(π/4) = (√2)/2
z₁ = (√2)/2 + i.(√2)/2
z₂ =  (√2)/2  i.(√2)/2
Simplification
z₁ = [(√2)/2].(1 + i)
z₂ =  [(√2)/2].(1 + i)
…and you can see, of course that: z₂ =  z₁

From hayharbr: (1 + i) / (√2)
That's what my TI84...
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Thu, 02 Apr 2020 12:39:37 +0000
(1 + i) / (√2)
That's what my TI84 calculator says anyway

From robert2020: If you mean the number 'one.' Then 1 ...
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Thu, 02 Apr 2020 12:35:40 +0000
If you mean the number 'one.' Then 1 has no square root other than itself.
1 ×1=1. Then one devided by itself is 1
This is why it's. Called an irrational number.