My Coding > Mathematics > Complex numbers > Logarithm of complex number

Logarithm of complex number

To calculate the logarithm of a complex number, we need to express the complex number in polar form. If we have a complex number in the form \(z = a + bi\), where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit, we can express \(z\) in polar form as \(z = r(cos \theta + i sin \theta)\), where \(r\) is the modulus of \(z\) and (\theta\) is the argument of \(z\).

The modulus of \(z\) is given by \(\left|z\right| = \sqrt{a^2 + b^2}\), and the argument of \(z\) is given by \(\theta = arg(z) = arctan(\frac{b}{a}) + 2\pi k\), where \(k\) is an integer such that \(\theta\) is in the desired range (usually taken to be \((-\pi, \pi]\)).

Once we have expressed \(z\) in polar form, we can apply the formula for the logarithm of a complex number. The logarithm of a complex number \(z\) in polar form is given by \(log(z) = ln(r) + i\theta\), where \(ln(r)\) is the natural logarithm of the modulus of \(z\), and \(i\theta\) is the argument of \(z\) expressed as a multiple of \(i\).

It is important to note that the logarithm of a complex number is not uniquely defined. In particular, there are infinitely many values of the logarithm of a complex number, differing by multiples of \(2\pi i\). This is because the complex exponential function is periodic with period \(2\pi i\), and so the logarithm of a complex number can be shifted by any multiple of this period and still be valid.

After all these introductory explanations we can write:

\[ln(z) = ln(re^{i(\theta + 2\pi n)}) = ln(r) + ln(e^{i(\theta + 2\pi n)})\]

Now we can combine everything into final equation:

\[ln(a + ib) = ln(\sqrt{a^2 + b^2}) + i(arctan(\frac{b}{a}) + 2\pi n)\]

and when \(\theta + 2\pi n \in (-\pi, \pi]\) this is a principal value

Check our further explanations in the video:


Published: 2023-05-09 01:52:33
Updated: 2023-05-09 01:54:46

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