So, another thought popped into my head today. This thought also involving complex numbers.

So we know what putting a complex number to another complex number is, but what about the opposite? What about complex logarithms?

\[\large\log_{z_1}{z_2}\]

So how do we go about solving this one?

Well first let's change the base to \(e\), we'll keep the complex numbers as they are for now.

\[\large \log_{z_1}{z_2} = \frac{\ln{z_2}}{\ln{z_1}}\]

Now we'll convert the complex numbers into a different form.

\[\large \frac{\ln{r_2e^{\theta_2 i}}}{\ln{r_1e^{\theta_1 i}}}\]

Then using the rules of logarithms we'll simplify.

\[\large \frac{\ln{r_2} + \ln{e^{\theta_2 i}}}{\ln{r_1} + \ln{e^{\theta_1 i}}}\]

\[\large \frac{\ln{r_2} + \theta_2 i}{\ln{r_1} + \theta_1 i}\]

Now we need to multiply the top and bottom halves by \(\boxed{\ln{r_1} - \theta_1 i}\) in order to make the bottom of the fraction a real number rather than a complex one.

\[\large \frac{(\ln{r_2} + \theta_2 i)(\ln{r_1} - \theta_1 i)}{(\ln{r_1} + \theta_1 i)(\ln{r_1} - \theta_1 i)}\]

\[\large \frac{\ln{(r_2)}\ln{(r_1)} - \ln{(r_2)}\theta_1 i + \ln{(r_1) \theta_2 i + \theta_2\theta_1}}{(\ln{(r_1)})^2 + \theta_1^2}\]

\[\large \frac{(\ln{(r_2)\ln{(r_1)} + \theta_2\theta_1) + i(\ln{(r_1)}\theta_2 - \ln{(r_2)}\theta_1)}}{(\ln{(r_1)})^2 + \theta_1^2}\]

And now the messy bit, substitution.

\[\large \log_{z_1}{z_2} = \frac{\left(\ln{\sqrt{a_2^2 + b_2^2}}\ln{\sqrt{a_1^2 + b_1^2}} + \arctan{\frac{b_2}{a_2}}\arctan{\frac{b_1}{a_1}}\right) + i\left(\ln{\sqrt{a_1^2 + b_1^2}}\arctan{\frac{b_2}{a_2}} - \ln{\sqrt{a_2^2 + b_2^2}}\arctan{\frac{b_1}{a_1}}\right)}{\left(\ln{\sqrt{a_1^2 + b_1^2}}\right)^2 + \arctan{\left(\frac{b_1}{a_1}\right)}^2}\]

So that's that, not as difficult as calculating this but still pretty tedious.

Hope you enjoyed the note.

## Comments

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TopNewestWow! That's a gigantic formula. I'm not sure if it's applicable or not because I seldom (never) see such forms of \( \log_{z_1} z_2 \) before.

Why don't you put your working into one of these wikis here? – Pi Han Goh · 1 year ago

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