Addition and Multiplication of Complex Numbers Examples 1

# Addition and Multiplication of Complex Numbers Examples 1

Recall from the Addition and Multiplication of Complex Numbers page that if $z = a + bi, w = c + di \in \mathbb{C}$ then addition between $z$ and $w$ is defined by:

(1)
\begin{align} \quad z + w = (a + bi) + (c + di) = (a + c) + (b + d)i \end{align}

Also, multiplication between $z$ and $w$ is defined by:

(2)
\begin{align} \quad z \cdot w = (a + bi) \cdot (c + di) = (ac - bd) + (ad + bc)i \end{align}

We will now look at some example problems regarding these operations on $\mathbb{C}$.

## Example 1

Prove that if $z, w \in \mathbb{C}$ then $\mathrm{Re} (z + w) = \mathrm{Re} (z) + \mathrm{Re} (w)$.

Let $z = a + bi$ and $w = c + di$. Then $\mathrm{Re} (z) = a$ and $\mathrm{Re} (w) = c$. So:

(3)
\begin{align} \quad \mathrm{Re} (z + w) = \mathrm{Re} [(a + c) + (b + d)i] = a + c = \mathrm{Re} (z) + \mathrm{Re}(w) \end{align}

## Example 2

Prove that if $z, w \in \mathbb{C}$ then $\mathrm{Re} (z \cdot w) \neq \mathrm{Re} (z) \cdot \mathrm{Re} (w)$ in general by counter example.

Let $z = w = 1 + i$. Then $\mathrm{Re} (z) = 1$ and $\mathrm{Re} (w) = 1$. Note that:

(4)
\begin{align} \quad z \cdot w = (1 + i) \cdot (1 + i) = 1 + 2i + i^2 = 2i \end{align}

Therefore we have that:

(5)
\begin{align} \quad \mathrm{Re} (z \cdot w) = 0 \neq 1 = 1 \cdot 1 = \mathrm{Re} (z) \cdot \mathrm{Re} (w) \end{align}

## Example 3

Prove that if $z \in \mathbb{C}$ then $\mathrm{Re} (iz) = - \mathrm{Im} (z)$ and $\mathrm{Im} (iz) = \mathrm{Re} (z)$.

Let $z = a + bi \in \mathbb{C}$. Then $\mathrm{Re} (z) = a$ and $\mathrm{Im} (z) = b$. Also:

(6)
\begin{align} \quad iz = i(a + bi) = ai + bi^2 = -b + ai \end{align}

Thus $\mathrm{Re} (iz) = -b = -\mathrm{Im} (z)$, and $\mathrm{Im}(z) = \mathrm{Re}(z)$.

## Example 4

Prove that if $\lambda \in \mathbb{R}$ and $z \in \mathbb{C}$ then $\mathrm{Re} (\lambda z) = \lambda \mathrm{Re} (z)$ and that $\mathrm{Im} (\lambda z) = \lambda \mathrm{Im} (z)$

Let $z = a + bi \in \mathbb{C}$. Then $\mathrm{Re} (z) = a$, $\mathrm{Im} (z) = b$ and:

(7)
\begin{align} \quad \lambda z = \lambda ( a + bi) = \lambda a + \lambda b i \end{align}

So $\mathrm{Re} (\lambda z) = \lambda a = \lambda \mathrm{Re} (z)$, and $\mathrm{Im} (\lambda z) = \lambda b = \lambda \mathrm{Im}(z)$.