I'm going to assume that you mean to say that i = √(-1) is a root of the auxiliary equation. That is, if the Cauchy-Euler DE is
x ²y'' + axy' + by = 0
then the auxiliary equation obtained by substituting y = xᵐ is
x ² (m (m - 1) xᵐ ⁻ ²) + ax (m xᵐ ⁻ ¹) + bxᵐ = 0
which reduces to
m (m - 1) + am + b = 0
or
m ² + (a - 1) m + b = 0
By the fundamental theorem of algebra, we can write the quadratic in terms of its roots r₁ and r₂,
(m - r₁) (m - r₂) = 0
Given that one root is the imaginary unit i, and the coefficients of the aux. equation are real, it follows that the other root is -i, because complex roots must occur with their conjugates. So we have as our aux. equation,
(m - i ) (m + i ) = 0
or
m ² + 1 = 0
Then a - 1 = 0 and b = 1, so that the given root and general solution correspond to the DE,
x ²y'' + xy' + y = 0