# Demystifying Complex Conjugates

This is the third post in the Complex Numbers Primer.  In the previous post in this series, we have learned that complex numbers can be added and multiplied just like binomials.  We did not discuss subtraction since it follows from addition; that is, if $a + bi$ and $c + di$ are complex numbers, then

$a + bi - (c + di) = a - c + (b - d)i$.

Since we have completed the three fundamental operations on complex numbers, the next logical question would be how do we divide complex numbers? For example, how do we find the quotient of the expression

$\displaystyle\frac{5 + 2i}{3 - i}$?

Before we answer that question, recall that in simplifying rational expressions with radicals in the denominator, we multiply the numerator and the denominator of the expression by the ‘conjugate’ of the denominator.  For instance, to simplify

$\displaystyle\frac{5}{2 + \sqrt{3}}$.

we have to multiply the numerator and the denominator of the expression by $2 - \sqrt{3}$ which is the conjugate of $2 + \sqrt{3}$.  Performing the computation, we have

$\displaystyle \left (\frac{5}{2+\sqrt{3}}\right )\left (\frac{2-\sqrt{3}}{2-\sqrt{3}}\right ) = 10 - 5\sqrt{3}$.

Notice that this method can be applied in computing $\frac{5 + 2i}{3 - i}$ since

$\displaystyle\frac{5 + 2i}{3 - i} =\frac{5 + 2\sqrt{-1}}{3 - \sqrt{-1}}$.

Therefore, we multiply the expression above by $\displaystyle\frac{3 + i}{3+i}$.

So,

$\displaystyle\left (\frac{5+2i}{3-i} \right )\left (\frac{3+i}{3+i} \right )=\frac{15+11i+2i^2}{9-i^2}=\frac{13+11i}{10}$.

In complex numbers, we call  $3 + i$ the complex conjugate of $3-i$. In general, $a - bi$ is the complex conjugate of $a + bi$.

In performing the operations above, we note three observations:

1. The product of a complex number $a + bi$ and its complex conjugate $a - bi$ is always a real number since $(a+bi)(a-bi) = a^2-b^2i^2 = a^2 + b^2$.
2. The quotient of two complex numbers can be expressed as $a + bi$. For example, $\frac{5 + 2i}{3 - i} = \frac{13}{10} + \frac{11}{10}i$.
3. Since $a + bi = a + b\sqrt{-1}$, we can always simplify expressions containing denominators of the form $a + bi$, $b \neq 0$.

We have now completed the four fundamental operations of complex numbers: addition, subtraction, multiplication, and division. To summarize, if $a$, $b$, $c$, and $d$ are real numbers, for the complex number $z = a + bi$ and $w = c + di$ the following statements hold:

• $z + w = (a + c) + (b+d)i$.
• $zw = (ac - bd) + (ad + bd)i$.
• $\displaystyle\frac{z}{w} = \frac{ac+bd + (bc-ad)i}{c^2+d^2}$

The proof of the third bullet is left as an exercise.  In the next post, we will discuss the connection between complex numbers and geometry.