Complex Multiplication

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Multiplication of complex numbers looks weird.

Complex Multiplication

\[ (a + b \, i) \cdot (c + d \, i) = (ac-bd) + (ad+bc) \, i \]

It is defined this way to maintain the distributive property, which we want to keep from the real numbers. The distributive property is vital to our algebra.

$\blacktriangleright $ Division requires a little discussion.

With real numbers, we have stories that give meaning to division.

$10 \div 5$ has two meanings. This expresion might represent how many groups are needed to split a group of $10$ marbles into groups of size $5$. That would be $2$ groups. It could also represent how many marbles would be in each group, if $10$ marbles were split up into $5$ equal groups. That would be $2$ marbles per group.

But, even real numbers stretch this story. $10 \div \sqrt {2}$ ? What does it mean to split a pile of marbles into $\sqrt {2}$ piles?

This marble story is also going to break with complex numbers. Fortunately, the real numbers have a different view of division, which we can bring with us over to complex numbers.

Just from a strictly algebaic viewpoint, $10 \div 5$ represents the number that you multiply $5$ by, to get $10$. In this way, we can think about division in terms of multiplication. This thought leads to reciprocals and fractions. Luckily, every real number, except $0$ has a reciprocal.

Note: $10 \div 0$ has no meaning in real numbers, because there is no real number that you can mulitply by $0$ and get $10$.

What about complex numbers? Do they have reciprocals? Does each complex number have a partner complex number such that their product equals $1$?

$\blacktriangleright $ Can we solve $1 = (2 + 3 \, i) \cdot (a + b \, i) $?

Multiplicative Inverse

Solve $1 = (2 + 3 \, i) \cdot (a + b \, i) $?

\[ 1 = (2 + 3 \, i) \cdot (a + b \, i) = (2a - 3b) + (2b+3a) \, i \]

We need $1 = \answer {2a - 3b}$ and $0 = \answer {2b+3a}$.

The second equation tells us that $\frac {-3a}{2} = b$. Substituting this into the first equation tells us that $1 = 2a - 3 \cdot \frac {-3a}{2}$. Now we can solve for $a$.

\begin{align*} 1 & = 2a - 3 \left ( \frac {-3a}{2} \right ) \\ & = 2a + \answer {\frac {9a}{2}} \\ & = \frac {13a}{2} \\ \frac {2}{13} & = a \end{align*}

With that, we can get a value for $b$.

\[ b = \frac {-3a}{2} = \frac {2}{13} \cdot \frac {-3}{2} = -\frac {3}{13} \]

Let’s check:

\[ (2 + 3 \, i) \cdot \left (\frac {2}{13} - \frac {3}{13} \, i \right ) = \left ( 2 \cdot \frac {2}{13} - 3 \cdot \left (- \frac {3}{13}\right )\right ) + \left ( 3 \cdot \frac {2}{13} - 2 \cdot \frac {3}{13} \right ) \, i = \frac {4+9}{13} + \frac {6-6}{13} \, i = 1 \]

Does this process work for all nonzero complex numbers?

$\blacktriangleright $ We know it works for $r + 0 \, i$ with $r \ne 0$, because these are real numbers.

\[ (r + 0 \, i) \cdot \left (\frac {1}{r} + 0 \, i \right ) = 1 \]

$\blacktriangleright $ Similarly, it works for $0 + r \, i$ with $r \ne 0$.

\[ (0 + r \, i) \cdot \left (0 - \frac {1}{r} \, i \right ) = 1 \]

$\blacktriangleright $ Now for complex numbers where neither component is $0$

Given real numbers $A \ne 0$ and $B \ne 0$, solve the following for $x$ and $y$.

\[ 1 = (A + B \, i) \cdot (x + y \, i) \]

\begin{align*} 1 & = (A + B \, i) \cdot (x + y \, i) \\ & = (Ax-By) + (Bx+Ay) \, i \end{align*}

For this to work, we need $\answer {A x - B y} = 1$ and $\answer {B x + A y} = 0$.

The second equation tells us that $x = \answer {-\frac {A y}{B}}$, which is ok, since $B \ne 0$ in this case.

Substituting that into the first equation, for $x$, gives us

\[ A \cdot \left (-\frac {A y}{B}\right ) - B y = 1 \]

\[ -\frac {A^2 y}{B} - B y = 1 \]

\[ -\frac {A^2 y}{B} - \frac {B^2}{B} y = 1 \]

\[ -\frac {A^2 + B^2}{B} y = 1 \]

\[ y = \frac {-B}{A^2 + B^2} \]

This is ok, since $A^2 + B^2 \ne 0$, because neither $A$ nor $B$ equals $0$.

This gives us $x = -\frac {A y}{B} = -\frac {A}{B} \cdot \frac {-B}{A^2 + B^2} = \frac {A}{A^2 + B^2} $.

\[ (A + B \, i) \cdot \left ( \frac {A}{A^2 + B^2} - \frac {B}{A^2 + B^2} \, i \right ) = 1 \]

$\blacktriangleright $ Every nonzero complex number has a reciprocal. We will use this to define division as multiplication by the reciprocal.

Complex Conjugate

The modulus of a complex number is its distance from the origin or $0$. The symbol for the modulus of a complex number, $z$, is $|z|$.

If $z = a + b \, i$, then

\[ |z| = |a + b \, i| = \sqrt {a^2 + b^2} \]

This is helpful, because a sum of two squares factors over the Complex numbers.

\[ (a + b \, i) \cdot (a - b \, i) = a^2 + b^2 \]

$\blacktriangleright $ $a - b \, i$ is called the complex conjugate of $a + b \, i$ and it will help us with division, by converting complex denominators to real numbers.

\[ \frac {1}{a + b \, i} = \frac {1}{a + b \, i} \cdot 1 = \frac {1}{a + b \, i} \cdot \frac {a - b \, i}{a - b \, i} = \frac {a - b \, i}{(a + b \, i)(a - b \, i)} = \frac {a - b \, i}{a^2 + b^2} \]

\[ \frac {1}{a + b \, i} = \frac {a}{a^2 + b^2} - \frac {b}{a^2 + b^2} \, i \]

We can use the complex conjugate of the denominator to get us back to standard form.

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more examples can be found by following this link
More Examples of Complex Arithmetic

2025-05-17 22:22:25