[Cambridge/2004] The use of calculator is not allowed in this question.

1. Expand the squared expression

$(3 - i)^2 = 3^2 - 2(3)(i) + i^2$

2. Substitute $i^2 = -1$ and simplify into $a + ib$ form

$= 9 - 6i - 1 \\ = 8 - 6i$

1. Compute the result of part (i) and use it to rewrite the right-hand side of the given equation.

From part (i), we have:
$(3 - i)^2 = 9 - 6i + i^2 = 8 - 6i$

The given equation is $(z + i)^2 = -8 + 6i$.
Notice that the right-hand side is the exact negative of the result from part (i):
$(z + i)^2 = -(8 - 6i)$
$(z + i)^2 = -(3 - i)^2$

2. Incorporate the negative sign into the square to express the right-hand side as a perfect square.

Using the fact that $-1 = i^2$:
$(z + i)^2 = i^2(3 - i)^2$
$(z + i)^2 = [i(3 - i)]^2$
$(z + i)^2 = (3i - i^2)^2$
$(z + i)^2 = (1 + 3i)^2$

3. Take the square root of both sides to solve for the two possible values of $z$.

Taking the square root of both sides gives two possibilities:
$z + i = \pm(1 + 3i)$

Case 1: Positive root
$z + i = 1 + 3i \implies z = 1 + 2i$

Case 2: Negative root
$z + i = -(1 + 3i) = -1 - 3i \implies z = -1 - 4i$