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

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