Complex logarithms

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?


So how do we go about solving this one?

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

logz1z2=lnz2lnz1\large \log_{z_1}{z_2} = \frac{\ln{z_2}}{\ln{z_1}}

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

lnr2eθ2ilnr1eθ1i\large \frac{\ln{r_2e^{\theta_2 i}}}{\ln{r_1e^{\theta_1 i}}}

Then using the rules of logarithms we'll simplify.

lnr2+lneθ2ilnr1+lneθ1i\large \frac{\ln{r_2} + \ln{e^{\theta_2 i}}}{\ln{r_1} + \ln{e^{\theta_1 i}}}

lnr2+θ2ilnr1+θ1i\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 lnr1θ1i\boxed{\ln{r_1} - \theta_1 i} in order to make the bottom of the fraction a real number rather than a complex one.

(lnr2+θ2i)(lnr1θ1i)(lnr1+θ1i)(lnr1θ1i)\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)}

ln(r2)ln(r1)ln(r2)θ1i+ln(r1)θ2i+θ2θ1(ln(r1))2+θ12\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}

(ln(r2)ln(r1)+θ2θ1)+i(ln(r1)θ2ln(r2)θ1)(ln(r1))2+θ12\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.

logz1z2=(lna22+b22lna12+b12+arctanb2a2arctanb1a1)+i(lna12+b12arctanb2a2lna22+b22arctanb1a1)(lna12+b12)2+arctan(b1a1)2\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.

Note by Jack Rawlin
5 years, 4 months ago

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Wow! That's a gigantic formula. I'm not sure if it's applicable or not because I seldom (never) see such forms of logz1z2 \log_{z_1} z_2 before.

Why don't you put your working into one of these wikis here?

Pi Han Goh - 5 years, 4 months ago

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