# What’s the square root of i?

### 8 Answers

- JimLv 72 months ago
Complex numbers are 'closed' for all operations.

√i = (1 + i)/√(2), or ½ (1 + i) √(2)

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- KrishnamurthyLv 72 months ago
(-1)^(1/4)

= 0.707106781... + 0.707106781... i

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- PinkgreenLv 72 months ago
i=e^(i pi/2)

=>

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

=>

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

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- sepiaLv 72 months ago
The square root of i:

0.70710678118654752440084436210484903928483593768847403658... +

0.70710678118654752440084436210484903928483593768847403658... i

- Keith ALv 62 months agoReport
(Slightly pointless!)

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- la consoleLv 72 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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- Anonymous2 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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- robert2020Lv 62 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.

You have the wrong definition of irrational.

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