Sum of Logarithms/Complex Logarithm
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Theorem
Let $x, y \in \C$ where $x = r_1 e^{i \theta_1}$ and $y = r_2 e^{i \theta_2}$
Where:
- $r_1$ and $r_2$ are both (strictly) positive real numbers.
Then:
- $\ln x + \ln y = \map \ln {x y}$
where $\ln$ denotes the complex natural logarithm.
Proof
We have:
| \(\ds x\) | \(=\) | \(\ds r_1 e^{i \theta_1}\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \ln x\) | \(=\) | \(\ds \map \ln {r_1} + i \paren {\theta_1 + 2 k \pi}\) | Definition of Complex Natural Logarithm |
and:
| \(\ds y\) | \(=\) | \(\ds r_2 e^{i \theta_2}\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \ln y\) | \(=\) | \(\ds \map \ln {r_2} + i \paren {\theta_2 + 2 k \pi}\) | Definition of Complex Natural Logarithm |
Finally, we have:
| \(\ds x y\) | \(=\) | \(\ds \paren {r_1 e^{i \theta_1} } \paren {r_2 e^{i \theta_2} }\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x y\) | \(=\) | \(\ds \paren {r_1 r_2 e^{i \paren {\theta_1 + \theta_2} } }\) | Product of Powers | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map \ln {x y}\) | \(=\) | \(\ds \map \ln {r_1 r_2} + i \paren {\theta_1 + \theta_2 + 2 k \pi}\) | Definition of Complex Natural Logarithm |
Therefore:
| \(\ds \ln x + \ln y\) | \(=\) | \(\ds \map \ln {r_1} + i \paren {\theta_1 + 2 k \pi} + \map \ln {r_2} + i \paren {\theta_2 + 2 k \pi}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map \ln {r_1 r_2} + i \paren {\theta_1 + \theta_2 + 2 k \pi}\) | Sum of Logarithms/Natural Logarithm | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map \ln {x y}\) |
$\blacksquare$