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: 
Your post says $i^2=1$, surely you mean $i^2=-1$ the complex unit, no?

Well assuming $i^2=-1$, firstly
$$(c+id)^{2}=(c+id)(c+id)$$

After multiplying we find
$$c^{2}+2icd+(id)^{2}=1$$
Since $i^{2}=-1$
$$c^{2}-d^{2}+2icd=1$$
Since $1$ is real and there is no imaginary part on the r.h.s we can split to two equations
\begin{align}
c^{2}-d^{2} &= 1 \\
cd &= 0
\end{align}
Two equations in two unknowns, can you take it form here?

Best,
Bacon.

**EDIT**
In response to your comment below (I don't have sufficient reputation to make a comment yet) I would say that in order for the product of two quantities to be identically zero, at *least one* needs to be zero. Therefore using this fact and the difference of two squares observation above it, you have several cases to consider before you find the values of $c$ and $d$.
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.