What is the answer to a^2=i ?

The answer to a^2=i is a= 1/(sqrt(2))+1/(sqrt(2))i,a=-1/(sqrt(2))-1/(sqrt(2))i

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a^2=i

a^{2} = i

a = \frac{1}{2} + \frac{1}{2} i, a = - \frac{1}{2} - \frac{1}{2} i

Solution steps

a^{2} = i

Substitute

a = x + y i

(x + y i)^{2} = i

Expand

(x + y i)^{2} : (x^{2} - y^{2}) + 2 i x y

(x^{2} - y^{2}) + 2 i x y = i

Rewrite

i

in standard complex form:

0 + i

(x^{2} - y^{2}) + 2 i x y = 0 + i

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[x^{2} - y^{2} = 0 2 x y = 1]

[x^{2} - y^{2} = 0 2 x y = 1] : (x = \frac{1}{2}, x = - \frac{1}{2}, y = \frac{1}{2} y = - \frac{1}{2})

Substitute back

a = x + y i

a = \frac{1}{2} + \frac{1}{2} i, a = - \frac{1}{2} - \frac{1}{2} i

i x^{2} + (1 - 2 i) x + 3 i - 1 = 0

x - 2 i = 2 + y i

z - \frac{1}{z} = 4 + 4 i

x^{2} + 2 i x - 1 - i = 0

z^{2} - (5 + 4 i) z - (3 - 5 i) = 0

What is the answer to a^2=i ?
The answer to a^2=i is a= 1/(sqrt(2))+1/(sqrt(2))i,a=-1/(sqrt(2))-1/(sqrt(2))i