Definition 1.3.1.
The number \(x-iy\) is the (complex) conjugate of \(x+iy\text{.}\) We denote the conjugate by
\begin{equation*}
\overline{x+iy} \ := \ x-iy \,\text{.}
\end{equation*}
Section 1.3 Geometric Properties
From the chain of basic inequalities
\begin{equation*}
- \sqrt{ x^2 + y^2 } \le - \sqrt{ x^2 } \le x \le \sqrt{ x^2 }
\le \sqrt{ x^2 + y^2 }
\end{equation*}
(or, alternatively, by arguing with basic geometric properties of triangles), we obtain the inequalities
\begin{equation}
-|z| \ \leq \ \Re(z) \ \leq \ |z| \qquad \text{ and } \qquad
-|z| \ \leq \ \Im(z) \ \leq \ |z| \, \text{.}\tag{1.16}
\end{equation}
The square of the absolute value has the nice property
\begin{equation*}
\left| x+iy \right|^2 \ = \ x^2 + y^2 \ = \ (x+iy) (x-iy) \,\text{.}
\end{equation*}
This is one of many reasons to give the process of passing from \(x+iy\) to \(x-iy\) a special name.
Geometrically, conjugating \(z\) means reflecting the vector corresponding to \(z\) with respect to the real axis, as shown in Figure 1.3.2.
The following proposition collects some basic properties of the conjugate.
Proposition 1.3.3.
For any \(z, z_1, z_2 \in \C\text{,}\)
-
\(\displaystyle \conj{z_1 \pm z_2} = \conj{z_1}\pm\conj{z_2}\)
-
\(\displaystyle \conj{z_1 \cdot z_2} = \conj{z_1}\cdot\conj{z_2}\)
-
\(\displaystyle \conj{ \left( \frac{ z_1 }{ z_2 } \right) } = \frac{ \ \conj{z_1} \ }{ \conj{z_2} }\)
-
\(\displaystyle \conj{\, \conj{z} \, }=z\)
-
\(\displaystyle \abs{\conj{z}}=\abs z\)
-
\(\displaystyle \abs{z}^2=z\conj z\)
-
\(\displaystyle \Re(z) = \frac 1 2 \left( z+\conj z \right)\)
-
\(\displaystyle \Im(z) = \frac{ 1 }{ 2i } \left( z-\conj z \right)\)
-
\(\conj{e^{i\phi}}=e^{-i\phi}\text{.}\)
The proofs of these properties are straightforward (see Exercise 1.5.22); once more we give a sample (b).
Proof.
Let \(z_1 = x_1 + i y_1\) and \(z_2 = x_2 + i y_2\text{.}\) Then
\begin{align*}
\conj{z_1 \cdot z_2} \amp \ = \ \conj{ (x_1 x_2 - y_1 y_2)
+ i ( x_1 y_2 + x_2 y_1 ) }\\
\amp \ = \ (x_1 x_2 - y_1 y_2) - i ( x_1 y_2 + x_2 y_1 )\\
\amp \ = \ (x_1 - i y_1) (x_2 - i y_2)\\
\amp \ = \ \conj{z_1}\cdot\conj{z_2} \, .
\end{align*}
We note that the property \(\abs{z}^2=z\conj z\) yields a neat formula for the inverse of a nonzero complex number, which is implicit already in (1.14):
\begin{equation*}
z^{-1} \ = \ \frac1z \ = \ \frac{\conj z}{\abs{z}^2} \,\text{.}
\end{equation*}
A famous geometric inequality (which holds, more generally, for vectors in \(\R^n\)) goes as follows.
Proposition 1.3.4. Triangle inequality.
For any \(z_1, z_2 \in \C\) we have \(\ds \left| z_1 + z_2 \right| \leq \left| z_1 \right| + \left| z_2 \right| \text{.}\)
By drawing a picture in the complex plane, you should be able to come up with a geometric proof of the triangle inequality. Here we proceed algebraically:
Proof.
We make extensive use of Proposition 1.3.3:
\begin{align*}
\left| z_1 + z_2 \right|^2 \ \amp = \ \left( z_1 + z_2
\right) \o{ \left( z_1 + z_2 \right) }\\
\amp = \ \left( z_1 + z_2
\right) \left( \o{z_1} + \o{z_2} \right)\\
\amp = \ z_1 \o{z_1} + z_1
\o{z_2} + z_2 \o{z_1} + z_2 \o{z_2}\\
\amp = \ \left| z_1 \right|^2 + z_1 \o{z_2} + \o{ z_1 \o{z_2}
} + \left| z_2 \right|^2\\
\amp = \ \left| z_1 \right|^2 + 2 \Re
\left( z_1 \o{z_2} \right) + \left| z_2 \right|^2\\
\amp \leq \ \left| z_1 \right|^2 + 2 \left| z_1 \o{z_2}
\right| + \left| z_2 \right|^2\\
\amp = \ \left| z_1 \right|^2 + 2
\left| z_1 \right| \left| \o{z_2} \right| + \left| z_2
\right|^2\\
\amp = \ \left| z_1 \right|^2 + 2 \left| z_1 \right|
\left| z_2 \right| + \left| z_2 \right|^2\\
\amp = \ \left( \left| z_1 \right| + \left| z_2 \right|
\right)^2\text{,}
\end{align*}
where the inequality follows from (1.16). Taking square roots on the left- and right-hand sides proves our claim.
For future reference we list several useful variants of the triangle inequality:
Corollary 1.3.5.
For \(z_1, z_2, \dots, z_n \in \C\text{,}\) we have the following relations:
-
The triangle inequality:\begin{equation*} \abs{\pm z_1\pm z_2} \le \abs{z_1}+\abs{z_2} \,\text{.} \end{equation*}
-
The reverse triangle inequality:\begin{equation*} \abs{\pm z_1\pm z_2} \ge \bigl| \abs{z_1}-\abs{z_2} \bigr| \,\text{.} \end{equation*}
-
The triangle inequality for sums:\begin{equation*} \abs{\sum_{k=1}^n z_k} \ \le \ \sum_{k=1}^n \abs{z_k} \, \text{.} \end{equation*}
The first item (a) is just a rewrite of the original triangle inequality, using the fact that \(\abs{\pm z}=\abs{z}\text{,}\) and (c) follows by induction. The proof of the reverse triangle inequality is left as Exercise 1.5.25.
