An equation with real coefficients has $1+i$ and $i$ among its roots. What is the lowest possible degree of the equation? 1 2 3 4

Step 1 ：Since the equation has real coefficients, the complex roots must come in conjugate pairs. We have two complex roots given: \(1+i\) and \(i\). The conjugate of \(1+i\) is \(1-i\), and the conjugate of \(i\) is \(-i\). So, we need to find the lowest degree polynomial that has these four roots.

Step 2 ：The polynomial with these roots is \((x - (1+i))(x - (1-i))(x - i)(x + i)\), which expands to \(x^4 - 2x^3 + 3x^2 - 2x + 2\).

Step 3 ：\boxed{4} is the lowest possible degree of the equation.