x² – x + (1 + i) = 0

Given x² – x + (1 + i) = 0

Recall that the roots of quadratic equation ax² + bx + c = 0, where a ≠ 0, are given by

Here, a = 1, b = –1 and c = (1 + i)

By substituting –1 = i² in the above equation, we get

We can write 3 + 4i = 4 – 1 + 4i

⇒ 3 + 4i = 4 + i² + 4i [∵ i² = –1]

⇒ 3 + 4i = 2² + i² + 2(2)(i)

⇒ 3 + 4i = (2 + i)² [∵ (a + b)² = a² + b² + 2ab]

By using the result 3 + 4i = (2 + i)², we get

[∵ i² = –1]

∴ x = i or 1 – i

Thus, the roots of the given equation are i and 1 – i.