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 $d$.
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

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