Subsection 1.2.1 Absolute value
ΒΆ
VIDEO Recall that \(\vert \cdot
\vert: \mathbb C \rightarrow \mathbb R \) is the function that returns the absolute value of the input. In other words, if \(\alpha = \alpha_r + \alpha_c i \text{,}\) where \(\alpha_r \) and \(\alpha_c \) are the real and imaginary parts of \(\alpha \text{,}\) respectively, then
\begin{equation*}
\vert \alpha \vert = \sqrt{ \alpha_r^2 + \alpha_c^2 }.
\end{equation*}
The absolute value (magnitude) of a complex number can also be thought of as the (Euclidean) distance from the point in the complex plane to the origin of that plane, as illustrated below for the number \(3 + 2i \text{.}\)
Alternatively, we can compute the absolute value as
\begin{equation*}
\begin{array}{l}
\vert \alpha \vert \\
~~~=~~~ \\
\sqrt{ \alpha_r^2
+ \alpha_c^2 } \\
~~~=~~~ \\
\sqrt{ \alpha_r^2 - \alpha_c \alpha_r i + \alpha_r \alpha_c i
+ \alpha_c^2 } \\
~~~=~~~ \\
\sqrt{ ( \alpha_r - \alpha_c i ) ( \alpha_r + \alpha_c i
)} \\
~~~=~~~ \\
\sqrt{ \overline \alpha \alpha }~~ ,
\end{array}
\end{equation*}
where \(\overline \alpha \) denotes the complex conjugate of \(\alpha \text{:}\)
\begin{equation*}
\overline \alpha = \overline{\alpha_r + \alpha_c i} = \alpha_r
- \alpha_c i\text{.}
\end{equation*}
The absolute value function has the following properties:
\(\alpha \neq 0 \Rightarrow \vert \alpha \vert > 0 \) (\(\vert
\cdot \vert \) is positive definite),
\(\vert \alpha \beta \vert = \vert \alpha \vert \vert \beta
\vert \) (\(\vert \cdot \vert \) is homogeneous), and
\(\vert \alpha + \beta \vert \leq \vert \alpha \vert + \vert \beta \vert \) (\(\vert \cdot \vert \) obeys the triangle inequality).
Norms are functions from a domain to the real numbers that are positive definite, homogeneous, and obey the triangle inequality. This makes the absolute value function an example of a norm.
The below exercises help refresh your fluency with complex arithmetic.
Homework 1.2.1.1 .
\(( 1 + i )( 2 - i ) = \)
\(( 2 - i )( 1 + i ) = \)
\(\overline{( 1 - i )}( 2 - i ) = \)
\(\overline{ \overline{( 1 - i )} ( 2 - i )} = \)
\(\overline{( 2 - i )} ( 1 - i ) = \)
\((1-i) \overline{( 2 - i )} = \)
Solution
\(( 1 + i )( 2 - i ) = 2 + 2i - i - i^2 = 2 + i +
1 = 3 + i \)
\(( 2 - i )( 1 + i ) = 2 -i + 2i - i^2 = 2 + i +
1 = 3 + i \)
\(\overline{( 1 - i )}( 2 - i ) = ( 1 + i )( 2 -
i ) = 2 -i + 2i - i^2 = 3 + i \)
\(\overline{ \overline{( 1 - i )} ( 2 - i )} =
\overline{ {( 1 + i )} ( 2 - i )} =
\overline{ 2 - i + 2 i - i^2 } =
\overline{ 2 + i +1} =
\overline{ 3 + i } = 3 - i\)
\(\overline{( 2 - i )} ( 1 - i ) =
( 2 + i ) ( 1 - i ) = 2 -2 i + i - i^2 = 2 - i + 1 =
3 - i \)
\((1-i) \overline{( 2 - i )} =
( 1 - i ) ( 2 + i ) = 2 + i - 2 i - i^2 = 2 - i + 1
= 3 - i \)
Homework 1.2.1.2 .
Let \(\alpha, \beta \in \mathbb C \text{.}\)
ALWAYS/SOMETIMES/NEVER: \(\alpha \beta =
\beta \alpha \text{.}\)
ALWAYS/SOMETIMES/NEVER: \({\overline \alpha \beta} = \overline
\beta \alpha \text{.}\)
Hint Let \(\alpha = \alpha_r + \alpha_c i \) and \(\beta = \beta_r + \beta_c i \text{,}\) where \(\alpha_r , \alpha_c,
\beta_r, \beta_c \in \mathbb R \text{.}\)
Answer
ALWAYS: \(\alpha \beta =
\beta \alpha \text{.}\)
SOMETIMES: \({\overline \alpha \beta} = \overline
\beta \alpha \text{.}\)
Solution
ALWAYS: \(\alpha \beta =
\beta \alpha \text{.}\)
Proof:
\begin{equation*}
\begin{array}{l}
\alpha \beta \\
~~~~=~~~~~ \lt \mbox{substitute} \gt \\
(\alpha_r + \alpha_c i) (\beta_r + \beta_c i) \\
~~~~=~~~~~ \lt \mbox{multiply out} \gt \\
\alpha_r \beta_r + \alpha_r \beta_c i +
\alpha_c \beta_r i - \alpha_c \beta_c \\
~~~~=~~~~~ \lt \mbox{commutativity of real multiplication} \gt \\
\beta_r \alpha_r + \beta_r \alpha_c i +
\beta_c \alpha_r i - \beta_c \alpha_c \\
~~~~=~~~~~ \lt \mbox{factor} \gt \\
(\beta_r + \beta_c i) (\alpha_r + \alpha_c i) \\
~~~~=~~~~~ \lt \mbox{substitute} \gt \\
\beta \alpha.
\end{array}
\end{equation*}
SOMETIMES: \({\overline \alpha \beta} = \overline
\beta \alpha \text{.}\)
An example where it is true: \(\alpha = \beta =
0 \text{.}\)
An example where it is false: \(\alpha = 1 \) and \(\beta = i \text{.}\) Then \(\overline \alpha \beta = 1 \times i = i \) and \(\overline \beta \alpha = -i \times 1 = -i
\text{.}\)
Homework 1.2.1.3 .
Let \(\alpha, \beta \in \mathbb C \text{.}\)
ALWAYS/SOMETIMES/NEVER: \(\overline{\overline \alpha \beta} = \overline
\beta \alpha \text{.}\)
Hint Let \(\alpha = \alpha_r + \alpha_c i \) and \(\beta = \beta_r + \beta_c i \text{,}\) where \(\alpha_r , \alpha_c,
\beta_r, \beta_c \in \mathbb R \text{.}\)
Answer Solution 1
\begin{equation*}
\begin{array}{l}
\overline{ \overline{ \alpha } \beta } \\
~~~~ = ~~~ \lt \alpha = \alpha_r + \alpha_c i ;
\beta = \beta_r + \beta_c i
\gt \\
\overline{ \overline{ (\alpha_r + \alpha_c i ) }
(\beta_r + \beta_c i) } \\
~~~~ = ~~~ \lt \mbox{ conjugate } \alpha \gt \\
\overline{ (\alpha_r - \alpha_c i )
(\beta_r + \beta_c i) } \\
~~~~ = ~~~ \lt \mbox{ multiply out } \gt \\
\overline{ (\alpha_r \beta_r - \alpha_c \beta_r i +
\alpha_r \beta_c i + \alpha_c \beta_c ) }
\\
~~~~ = ~~~ \lt \mbox{ conjugate } \gt \\
\alpha_r \beta_r + \alpha_c \beta_r i -
\alpha_r \beta_c i + \alpha_c \beta_c \\
~~~~ = ~~~ \lt \mbox{ rearrange } \gt \\
\beta_r \alpha_r + \beta_r \alpha_c i -
\beta_c \alpha_r i + \beta_c \alpha_c \\
~~~~ = ~~~ \lt \mbox{ factor } \gt \\
( \beta_r - \beta_c i ) ( \alpha_r + \alpha_c i ) \\
~~~~ = ~~~ \lt \mbox{ definition of conjugation } \gt \\
\overline{( \beta_r + \beta_c i) } ( \alpha_r +
\alpha_c i ) \\
~~~~ = ~~~ \lt \alpha = \alpha_r + \alpha_c i ;
\beta = \beta_r + \beta_c i
\gt \\
\overline{\beta} \alpha
\end{array}
\end{equation*}
Solution 2
Proofs in mathematical textbooks seem to always be wonderfully smooth arguments that lead from the left-hand side of an equivalence to the right-hand side. In practice, you may want to start on the left-hand side, and apply a few rules:
\begin{equation*}
\begin{array}{l}
\overline{ \overline{ \alpha } \beta } \\
~~~~ = ~~~ \lt \alpha = \alpha_r + \alpha_c i ;
\beta = \beta_r + \beta_c i
\gt \\
\overline{ \overline{ (\alpha_r + \alpha_c i ) }
(\beta_r + \beta_c i) } \\
~~~~ = ~~~ \lt \mbox{ conjugate } \alpha \gt \\
\overline{ (\alpha_r - \alpha_c i )
(\beta_r + \beta_c i) } \\
~~~~ = ~~~ \lt \mbox{ multiply out } \gt \\
\overline{ (\alpha_r \beta_r - \alpha_c \beta_r i +
\alpha_r \beta_c i + \alpha_c \beta_c ) } \\
~~~~ = ~~~ \lt \mbox{ conjugate } \gt \\
\alpha_r \beta_r + \alpha_c \beta_r i -
\alpha_r \beta_c i + \alpha_c \beta_c
\end{array}
\end{equation*}
and then move on to the right-hand side, applying a few rules:
\begin{equation*}
\begin{array}{l}
\overline{ \beta } \alpha \\
~~~~ = ~~~ \lt \alpha = \alpha_r + \alpha_c i ;
\beta = \beta_r + \beta_c i
\gt \\
\overline{(\beta_r + \beta_c i) }
(\alpha_r + \alpha_c i )
\\
~~~~ = ~~~ \lt \mbox{ conjugate } \beta \gt \\
(\beta_r - \beta_c i)
(\alpha_r + \alpha_c i ) \\
~~~~ = ~~~ \lt \mbox{ multiply out } \gt \\
\beta_r \alpha_r + \beta_r \alpha_c i -
\beta_c \alpha_r i + \beta_c \alpha_c.
\end{array}
\end{equation*}
At that point, you recognize that
\begin{equation*}
\alpha_r \beta_r + \alpha_c \beta_r i -
\alpha_r \beta_c i + \alpha_c \beta_c \\
~~~= ~~~~ \\
\beta_r \alpha_r + \beta_r \alpha_c i -
\beta_c \alpha_r i + \beta_c \alpha_c
\end{equation*}
since the second is a rearrangement of the terms of the first. Optionally, you then go back and presents these insights as a smooth argument that leads from the expression on the left-hand side to the one on the right-hand side:
\begin{equation*}
\begin{array}{l}
\overline{ \overline{ \alpha } \beta } \\
~~~~ = ~~~ \lt \alpha = \alpha_r + \alpha_c i ;
\beta = \beta_r + \beta_c i
\gt \\
\overline{ \overline{ (\alpha_r + \alpha_c i ) }
(\beta_r + \beta_c i) } \\
~~~~ = ~~~ \lt \mbox{ conjugate } \alpha \gt \\
\overline{ (\alpha_r - \alpha_c i )
(\beta_r + \beta_c i) } \\
~~~~ = ~~~ \lt \mbox{ multiply out } \gt \\
\overline{ (\alpha_r \beta_r - \alpha_c \beta_r i +
\alpha_r \beta_c i + \alpha_c \beta_c ) }
\\
~~~~ = ~~~ \lt \mbox{ conjugate } \gt \\
\alpha_r \beta_r + \alpha_c \beta_r i -
\alpha_r \beta_c i + \alpha_c \beta_c \\
~~~~ = ~~~ \lt \mbox{ rearrange } \gt \\
\beta_r \alpha_r + \beta_r \alpha_c i -
\beta_c \alpha_r i + \beta_c \alpha_c \\
~~~~ = ~~~ \lt \mbox{ factor } \gt \\
( \beta_r - \beta_c i ) ( \alpha_r + \alpha_c i ) \\
~~~~ = ~~~ \lt \mbox{ definition of conjugation } \gt \\
\overline{( \beta_r + \beta_c i) } ( \alpha_r +
\alpha_c i ) \\
~~~~ = ~~~ \lt \alpha = \alpha_r + \alpha_c i ;
\beta = \beta_r + \beta_c i
\gt \\
\overline{\beta} \alpha .
\end{array}
\end{equation*}
Solution 3
Yet another way of presenting the proof uses an "equivalence style proof." The idea is to start with the equivalence you wish to prove correct:
\begin{equation*}
\overline{\overline \alpha \beta} = \overline
\beta \alpha
\end{equation*}
and through a sequence of equivalent statements argue that this evaluates to TRUE:
\begin{equation*}
\begin{array}{l}
\overline{\overline \alpha \beta} = \overline
\beta \alpha \\
~~~ \Leftrightarrow ~~~~
\lt \alpha = \alpha_r + \alpha_c i ;
\beta = \beta_r + \beta_c i
\gt \\
\overline{\overline{ ( \alpha_r + \alpha_c i)} ( \beta_r
+ \beta_c i )} = \overline
{( \beta_r + \beta_c i )}( \alpha_r + \alpha_c i ) \\
~~~ \Leftrightarrow ~~~~ \lt \mbox{ conjugate } \times
2 \gt \\
\overline{{ ( \alpha_r - \alpha_c i)} ( \beta_r
+ \beta_c i )} =
{( \beta_r - \beta_c i )}( \alpha_r + \alpha_c i ) \\
~~~ \Leftrightarrow ~~~~
\lt \mbox{ multiply out } \times 2 \gt \\
\overline{\alpha_r \beta_r + \alpha_r \beta_c i - \alpha_c
\beta_r i + \alpha_c \beta_c }
=
\beta_r \alpha_r + \beta_r \alpha_c i - \beta_c
\alpha_r i + \beta_c \alpha_c \\
~~~ \Leftrightarrow ~~~~
\lt \mbox{ conjugate } \gt \\
{\alpha_r \beta_r - \alpha_r \beta_c i + \alpha_c
\beta_r i + \alpha_c \beta_c }
=
\beta_r \alpha_r + \beta_r \alpha_c i - \beta_c
\alpha_r i + \beta_c \alpha_c \\
~~~ \Leftrightarrow ~~~~
\lt
\mbox{ subtract equivalent terms from left-hand side and right-hand side }
\gt \\
0 = 0 \\
~~~ \Leftrightarrow ~~~~ \lt \mbox{ algebra } \gt \\
\mbox{TRUE}.
\end{array}
\end{equation*}
By transitivity of equivalence, we conclude that \(\overline{\overline \alpha \beta} = \overline
\beta \alpha \) is TRUE.
Homework 1.2.1.4 .
Let \(\alpha \in \mathbb C \text{.}\)
ALWAYS/SOMETIMES/NEVER: \(\overline{\alpha}
\alpha \in \mathbb R \)
Answer Solution
Let \(\alpha = \alpha_r + \alpha_c i \text{.}\) Then
\begin{equation*}
\begin{array}{l}
\overline{\alpha} \alpha \\
~~~ = ~~~~ \lt \mbox{ instantiate } \gt \\
\overline{( \alpha_r + \alpha_c i )}(\alpha_r +
\alpha_c i ) \\
~~~ = ~~~~ \lt \mbox{ conjugate } \gt \\
{( \alpha_r - \alpha_c i )}(\alpha_r +
\alpha_c i ) \\
~~~ = ~~~~ \lt \mbox{ multiply out } \gt \\
\alpha_r^2 + \alpha_c^2,
\end{array}
\end{equation*}
which is a real number.
Homework 1.2.1.5 .
Prove that the absolute value function is homogeneous: \(\vert \alpha \beta \vert = \vert \alpha
\vert \vert \beta \vert \) for all \(\alpha, \beta
\in \mathbb C \text{.}\)
Solution
\begin{equation*}
\begin{array}{l}
\vert \alpha \beta \vert = \vert \alpha \vert \vert
\beta \vert \\
~~~ \Leftrightarrow ~~~~ \lt
\mbox{ squaring both sides simplifies }\gt \\
\vert \alpha \beta \vert^2 = \vert \alpha \vert^2 \vert
\beta \vert^2 \\
~~~ \Leftrightarrow ~~~~ \lt \mbox{ instantiate }\gt \\
\vert (\alpha_r + \alpha_c i ) (\beta_r + \beta_c i )
\vert^2 =
\vert \alpha_r + \alpha_c i \vert^2 \vert \beta_r +
\beta_c i
\vert^2 \\
~~~ \Leftrightarrow ~~~~ \lt \mbox{ algebra }\gt \\
\vert (\alpha_r \beta_r - \alpha_c \beta_c ) + (
\alpha_r \beta_c + \alpha_c \beta_r ) i
\vert^2 =
(\alpha_r^2 + \alpha_c^2)
(\beta_r^2 + \beta_c^2)
\\
~~~ \Leftrightarrow ~~~~ \lt \mbox{ algebra }\gt \\
(\alpha_r \beta_r - \alpha_c \beta_c )^2 + (
\alpha_r \beta_c + \alpha_c \beta_r )^2
=
(\alpha_r^2 + \alpha_c^2)
(\beta_r^2 + \beta_c^2)
\\
~~~ \Leftrightarrow ~~~~ \lt \mbox{ algebra }\gt \\
\alpha_r^2 \beta_r^2 - 2 \alpha_r \alpha_c \beta_r
\beta_c + \alpha_c^2 \beta_c^2 +
\alpha_r^2 \beta_c^2 + 2 \alpha_r \alpha_c \beta_r
\beta_c + \alpha_c^2 \beta_r^2 \\
~~~~~~~~~~~~~~~~
=
\alpha_r^2 \beta_r^2 + \alpha_r^2 \beta_c^ + \alpha_c^2 \beta_r^2 +
\alpha_c^2 \beta_c^2
\\
~~~ \Leftrightarrow ~~~~ \lt
\mbox{ subtract equivalent terms from both sides }\gt \\
0 = 0
\\
~~~ \Leftrightarrow ~~~~ \lt \mbox{ algebra }\gt \\
T
\end{array}
\end{equation*}
Homework 1.2.1.6 .
Let \(\alpha \in \mathbb C \text{.}\)
ALWAYS/SOMETIMES/NEVER: \(\vert \overline \alpha \vert
= \vert \alpha \vert \text{.}\)
Answer Solution
Let \(\alpha = \alpha_r + \alpha_c i \text{.}\)
\begin{equation*}
\begin{array}{l}
\vert \overline \alpha \vert \\
~~~~ = ~~~ \lt \mbox{ instantiate } \gt \\
\vert \overline { \alpha_r + \alpha_c i } \vert \\
~~~~ = ~~~ \lt \mbox{ conjugate } \gt \\
\vert { \alpha_r - \alpha_c i } \vert \\
~~~~ = ~~~ \lt \mbox{ definition of } \vert \cdot
\vert \gt \\
\sqrt { \alpha_r^2 + \alpha_c ^2 } \\
~~~~ = ~~~ \lt \mbox{ definition of } \vert \cdot
\vert \gt \\
\vert \alpha_r + \alpha_c i \vert \\
~~~~ = ~~~ \lt \mbox{ instantiate } \gt \\
\vert \alpha \vert \\
\end{array}
\end{equation*}
