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 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.
First of all i would observe $$1 = e^{i2k\pi} \;\; k \in \mathbb{Z}$$ then i would write $(a + ib)^2 = \left( \rho e^{i\theta} \right)^2 = \rho^2 e^{i2\theta}$ so...

$$\rho^2 e^{i2\theta} = e^{i2k\pi}$$ which of course implies that $\rho = 1$, (since it is a positive real number) while 

$$ 2\theta = 2 k \pi \Rightarrow \theta = k \pi$$

So the solutions are of the form

$$a + ib = e^{ik\pi}$$ which implies for $k$ even 
$$a + ib = 1 \Rightarrow \left\{ \begin{array}{l} a = 1 \\ b = 0 \end{array} \right.$$

while for $k$ odd

$$a + ib = -1 \Rightarrow \left\{ \begin{array}{l} a = -1 \\ b = 0 \end{array} \right.$$