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