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?

logz1z2\large\log_{z_1}{z_2}

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.

#Algebra

Note by Jack Rawlin
5 years, 3 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

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, 3 months ago
×

Problem Loading...

Note Loading...

Set Loading...