The Complex Conjugate of a Complex Number

# The Complex Conjugate of a Complex Number

Recall that a complex number is in the form $z = a + bi$ where $a, b \in \mathbb{R}$, then the real part of $z$ is $\Re (z) = a$ and the imaginary part of $z$ is $\Im (z) = b$. Therefore we have that $z = \Re (z) + \Im (z) i$. We will now defined a special number call the complex conjugate of $z$ and look at its interesting properties. This will become useful when diving further into linear algebra.

 Definition: If $z = a + bi = \Re(z) + \Im(z) i$ is a complex number, then the Complex Conjugate of $z$ denoted $\bar{z} = a - bi = \Re(z) - \Im(z) i$.

For example, consider the complex number $z = 5 + 2i$. Then the complex conjugate of $z$ is $\bar{z} = 5 - 2i$. Another example is the complex number $z = i - 1$ whose complex conjugate is $\bar{z} = -1 - i$.

 Theorem 1: Let $z, z' \in \mathbb{C}$ be complex numbers $z = a + bi$ and $z' = a' + b'i$. Then: a) $z + \bar{z} = 2 \Re (z)$. b) $z - \bar{z} = 2 \Im (z) i$. c) $\bar{\bar{z}} = z$. d) $\overline{z + z'} = \bar{z} + \bar{z'}$ (Additivity Property). e) $\overline{zz'} = \bar{z} \bar{z'}$ (Multiplicativity Property).
• Proof a) We have that $z = a + bi$ and $\bar{z} = a - bi$. So $z + \bar{z} = (a + bi) + (a - bi) = (a + a) + (bi - bi) = 2a = 2 \Re (z)$.
• Proof b) We have that $z = a + bi$ and $\bar{z} = a - bi$. So $z - \bar{z} = (a + bi) - (a - bi) = (a - a) + (bi + bi) = 2bi = 2 \Im (z) i$.
• Proof c) If $\bar{z} = a - bi$ then $\bar{\bar{z}} = a - (-bi) = a + bi = z$.
• Proof d) We have that $z + z' = (a + bi) + (a' + b'i) = (a + a') + (b + b')i$, and so:
(1)
\begin{align} \quad \overline{z+z'} = (a + a') - (b + b')i = [a - bi] + [a' - b'i] = \bar{z} + \bar{z'} \end{align}
• Proof e) We have that $zz' = (a + bi)(a' + b'i) = aa' + ab'i + a'bi - bb' = (aa' - bb') + (ab' + a'b)i$, so $\overline{zz'} = (aa' - bb') - (ab' + a'b)i$. Now we also note that $(a - bi)(a' - b'i) = aa' - ab'i + a'bi - bb' = (aa' - bb') - (ab' - a'b)i$ Therefore $\overline{zz'} = \bar{z} \bar{z'}$. $\blacksquare$