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Anonymous asked in Science & MathematicsMathematics · 2 months ago

What’s the square root of i?

8 Answers

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  • Jim
    Lv 7
    2 months ago

    Complex numbers are 'closed' for all operations.

    √i = (1 + i)/√(2), or ½ (1 + i) √(2)

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  • 2 months ago

    (-1)^(1/4)

    = 0.707106781... + 0.707106781... i

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  • 2 months ago

    i=e^(i pi/2)

    =>

    sqr(i)=e^(i pi/4)

    =>

    sqr(i)=sqr(2)(1+i)/2

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  • sepia
    Lv 7
    2 months ago

    The square root of i:

    0.70710678118654752440084436210484903928483593768847403658... +

    0.70710678118654752440084436210484903928483593768847403658... i

    • Keith A
      Lv 6
      2 months agoReport

      (Slightly pointless!)

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  • 2 months ago

    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

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  • Anonymous
    2 months ago

    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.

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  • 2 months ago

    (1 + i) / (√2)      

    That's what my TI-84 calculator says anyway

    • Philip
      Lv 6
      2 months agoReport

      But it doesn't put (+/-) before your answer.

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  • 2 months ago

    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.

    • Puzzling
      Lv 7
      2 months agoReport

      You have the wrong definition of irrational. 

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