There are three very common operations that are new to complex numbers.
Conjugates
We start by introducing the complex conjugate.
Given a complex number z = x + yi, we define its conjugate to be
\overline{z} = x - yi
The conjugate of a number is very simple, and just flips the sign of the complex numbers imaginary part.
Let z,w \in \mathbb{C}. Then we have the following
(a) \overline{z + w} = \overline{z} + \overline{w}
(b) \overline{z - w} = \overline{z} - \overline{w}
(c) \overline{z \cdot w} = \overline{z} \cdot \overline{w}
(d) \overline{\left(\frac{z}{w}\right)} = \frac{\overline{z}}{\overline{w}}
(e) \overline{\overline{z}} = z
Parts
We define the following unary functions
The Real part of a complex number is given by the function \text{Re} \colon \mathbb{C} \to \mathbb{R} such that
\text{Re}(z) = \frac{z +\overline{z}}{2}
The Imaginary part of a complex number is given by the function \text{Im} \colon \mathbb{C} \to \mathbb{R} such that
\text{Im}(z) = \frac{z - \overline{z}}{2i}.
Let z = 2 + 3i. Then the conjugate is \overline{z} = 2 - 3i. This gives us for our real and imaginary parts
\text{Re}(z) = \frac{z +\overline{z}}{2} = \frac{(2+3i) + (2 - 3i)}{2} = \frac{4 + 0i}{2} = 2
\text{Im}(z) = \frac{z - \overline{z}}{2i} = \frac{(2+3i) - (2 - 3i)}{2} = \frac{0 + 6i}{2} = 3
We have that a complex number is equivalent to it’s real and imaginary parts.
Let z \in \mathbb{C}. Then z = \text{Re}(z) + \text{Im}(z) \cdot i.
This could be proved directly, but instead I opt for some simple one line lemmas.
Let z = x + yi. Then \text{Re}(z) = x
We know \overline{z} = x - yi. Using our real part function we obtain
\text{Re}(z) = \frac{z +\overline{z}}{2} = \frac{2x}{2} = x.
Let z = x + yi. Then \text{Im}(z) = y
We know \overline{z} = x - yi. Using our imaginary part function we obtain
\text{Im}(z) = \frac{z - \overline{z}}{2} = \frac{2yi}{2i} = y.
Then our proof is a pretty simple one-liner
Let z = x + yi. Then from our previous lemmas we have \text{Re}(z) = x and \text{Im}(z) = y. Substitution gives us
z = x + yi = \text{Re}(z) + \text{Im}(z) \cdot i.
Our next theorem states that if two complex numbers have the same real and imaginary parts, they are the same number.
Let z,w \in \mathbb{C}. z = w if and only if \text{Re}(z) = \text{Re}(w) and \text{Im}(z) = \text{Im}(w).
(\Leftarrow) Let z = (x,y) and w = (u,v). If \text{Re}(z) = \text{Re}(w) then we have x = u. Similarly, if \text{Im}(z) = \text{Im}(w) then we have y = v. This gives us our equivalence
z = (x,y) = (u,v) = w.
(\Rightarrow) Let z = w. Then (x,y) = (u,v). We know that ordered pairs are only equivalent if x = u and y = v. Since \text{Re}(z) = x and \text{Re(w) = u}, we have \text{Re}(z) = \text{Re}(w). A similar argument applies to the imaginary part.