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 $d$.
$$(c+id)^2=1 \Rightarrow c^2+2cdi+i^2d^2=1 \Rightarrow c^2+2cdi-d^2=1 \Rightarrow (c^2-d^2)+2cdi=1+0i$$ 

The real part of the LHS must be equal to the real part of the RHS and the imaginary part of the LHS must equal to the imaginary part of the RHS. 

So, you have to solve the following system : $$c^2-d^2=1 \\ 2cd=0$$ 

---

From the first equation we get $c^2=1+d^2$. 

Substituting this in $2cd=0 \Rightarrow cd=0 \Rightarrow c^2d^2=0$ we get $$(1+d^2)d^2=0 \\ \Rightarrow 1+d^2=0 \text{ or } d^2=0 \\ \Rightarrow d^2=-1 \ \ \text{ or } \ \ d=0 \\ \Rightarrow d=\pm i \ \ \text{ or } \ \ d=0$$ 

For $d=\pm i$, substituting at $c^2=1+d^2$ we get $c^2=0 \Rightarrow c=0$. 

For $d=0$, substistuting at we get $c^2=1 \Rightarrow c=\pm 1$.