What is the answer to i(x)=x^2+1 ?

The answer to i(x)=x^2+1 is x=-(-1+sqrt(5))/2 i,x=(1+sqrt(5))/2 i

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i(x)=x^2+1

i (x) = x^{2} + 1

x = - \frac{- 1 + 5}{2} i, x = \frac{1 + 5}{2} i

Solution steps

i (x) = x^{2} + 1

Substitute

x = a + bi

i (a + bi) = (a + bi)^{2} + 1

Expand

- b + ia = (a^{2} - b^{2} + 1) + 2 iab

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

[- b = a^{2} - b^{2} + 1 a = 2 ab]

[- b = a^{2} - b^{2} + 1 a = 2 ab] : (a = 0, a = 0, b = - \frac{- 1 + 5}{2} b = \frac{1 + 5}{2})

Substitute back

x = a + bi

x = - \frac{- 1 + 5}{2} i, x = \frac{1 + 5}{2} i

z^{2} = 4 (- 1 - i 3)

i (t) = 4 t^{2} - 48 t + 154

i (t) = 8 t^{2} - 4 t A

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

∣ z + i ∣ = 3 + 0 i

What is the answer to i(x)=x^2+1 ?
The answer to i(x)=x^2+1 is x=-(-1+sqrt(5))/2 i,x=(1+sqrt(5))/2 i