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Logarithm of 4i, ln(4i)
Logarithm of 4i, ln(4i)Now, let's calculate a very simple example \(ln(4i)\i. In the previous lecture we already derive the equation of the logarithm of a complex value, and now we will use it. Just to remind you: \[ln(a + ib) = ln(\sqrt{a^2 + b^2}) + i(arctan(\frac{b}{a}) + 2\pi n)\] from previous lectures you can easily see, that \(a = 0; b = 4\) and therefore: \(\sqrt{a^2 + b^2} = 4; \theta = \frac{\pi}{2}\). Then we can use these values: \[ln(4i) = ln(4) + i(\frac{\pi}{2} + 2\pi n)\] and the principal value is (when \(n = 0\)): \[ln(4i) = ln(4) + i(\frac{\pi}{2})\] Check our further explanations in the video:
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