1.2 The Triangle Inequality
For any complex numbers \(z_1\) and \(z_2\), we have \(\quad \left |z_1 + z_2\right | \leq \left |z_1\right | + \left |z_2\right |\)
Proof \begin {align*} \left |z_1 + z_2\right |^2 & = \big (z_1 + z_2\big )\cdot \overline {\big (z_1 + z_2\big )}= \big (z_1 + z_2\big )\cdot \big (\overline {z_1} + \overline {z_2}\big )\\ & = z_1\overline {z_1} + z_2\overline {z_2} + \overline {z_1}z_2 + z_1\overline {z_2}\\ & = \left |z_1\right |^2 + \left |z_2\right |^2 + z_1\overline {z_2} + \overline {z_1\overline {z_2}}\\ & = \left |z_1\right |^2 + \left |z_2\right |^2 + 2 Re \big (z_1\overline {z_2}\big )\\ & \leq \left |z_1\right |^2 + \left |z_2\right |^2 + 2\left |z_1 \overline {z_2}\right |\\ & = \left |z_1\right |^2 + \left |z_2\right |^2 + 2\left |z_1\right |\left |\overline {z_2}\right |\\ & = \left |z_1\right |^2 + \left |z_2\right |^2 + 2\left |z_1\right |\left |z_2\right |\\ & = \Big (\left |z_1\right | + \left |z_2\right |\Big )^2 \end {align*}
\[\therefore \qquad \left |z_1 + z_2\right |\leq \left |z_1\right | + \left |z_2\right |\]
The triangle inequality can be extended to include any number of sumands that is we can write \begin {align*} \left |z_1 + z_2 + z_3\right | & \leq \left |z_1 + z_2\right | + \left |z_3\right |\\ & \leq \left |z_1\right | + \left |z_2\right | + \left |z_3\right |\\ \end {align*}
Which we can generalise \(\quad \left |\displaystyle {\sum ^n_{k =1} z_k}\right | \leq \displaystyle {\sum ^n_{k = 1} \left |z_k\right |}\)
\[\left |z_1 + z_2 + \cdots + z_n\right | \leq \left |z_1\right | + \left |z_2\right | + \cdots \left |z_n\right |\]
Prove using mathematical induction.
\(\implies \quad \left |z_1\right | = \left |z_1 - z_2 + z_2\right |\leq \left |z_1 - z_2\right | + \left |z_2\right |\)
\(\implies \quad \left |z_1\right | - \left |z_2\right | \leq \left |z_1 - z_2\right |\)
Similarly, \(\quad \left |z_1 - z_2\right | \geq \left |z_2\right | - \left |z_1\right |\). Thus \(\, \left |z_1 - z_2\right | \geq \left |\left |z_1\right | - \left |z_2\right |\right |\)
Lemma
For any complex numbers \(z_1\) and \(z_2\) the following holds
- i).
- \(\left |\left |z_1\right | - \left |z_2\right |\right | \leq \left |z_1 + z_2\right |\)
- ii).
- \(\left |\left |z_1\right | - \left |z_2\right |\right | \leq \left |z_1 - z_2\right |\)
Example 1.8. Find an upper bound for \(\,\displaystyle {\left |\frac {-1}{z^4 - 5z + 1}\right |}\,\) if \(z = 2\)
Solution
We need to find a number \(M\) such that
\[\left |\frac {-1}{z^4 - 5z + 1}\right |\leq M\]
Note that \begin {align*} \left |z^4 - 5z + 1\right | & = \left |z^4 - (5z -1)\right |\\ & \geq \left |\left |z\right |^4 - \left |5z - 1\right |\right | \end {align*}
We make the last expression as small as possible. This is only possible when \(\left |5z - 1\right |\) is large \[\left |5z - 1\right | \leq 5\left |z\right | + \left |-1\right | = 5\left |z\right | + 1\]
So we get \begin {align*} \left |z^4 - 5z + 1\right |& \geq \left |\left |z\right |^4 - \left |5z - 1\right |\right |\\ & \geq \left |\left |z\right |^4 - \big (5\left |z\right | + 1\big )\right |\\ & = \left |16 - 10 -1\right |\\ & = 5 \end {align*}
Hence for \(z = 2\), we get \(\,\displaystyle {\left |\frac {-1}{z^4 - 5z + 1}\right | \leq \frac {1}{5}}\)
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