Question: I need to solve this complex equation:
$$
(c + id)^2 = 1
$$

where $i^2=-1$.

What am I supposed to calculate here? Just $c$ and $d$?

Answer: 
By far, the simplest solution is simply to take the square root of both sides:

$c+di = \pm 1$

which allows us to immediately equate real and imaginary parts to get $c = \pm 1, d = 0$.

Much simpler than squaring the LHS.
By the binomial formulas we know that 
$$
0=z^2-1=(z-1)(z+1)
$$
has the only solutions $z=\pm 1$. This is true in the real numbers and stays true in the field of complex numbers.