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

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 $c^2=1 \Rightarrow c=\pm 1$.
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.