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:

Published: 2023-05-09 02:05:48
Updated: 2023-05-09 22:09:39