Problem with Logarithms solutions

  • Context: Undergrad 
  • Thread starter Thread starter Jacopo
  • Start date Start date
  • Tags Tags
    Logarithms
Click For Summary
SUMMARY

The discussion focuses on the properties of logarithms, specifically the relationship between natural logarithms and logarithms of different bases. It establishes that if \( \omega = z(\cos \theta + i \sin \theta) \), then \( \log_{e} \omega = \log_{e} z + i \theta \). Furthermore, it clarifies that \( \log_{b} \omega \) can be expressed as \( \frac{\log_{e} \omega}{\log_{e} b} \), allowing the addition of \( \frac{2 \pi i}{\log_{e} b} \) to \( \log_{b} \omega \) due to the properties of logarithmic functions.

PREREQUISITES
  • Understanding of complex numbers and Euler's formula
  • Familiarity with natural logarithms (logarithm base e)
  • Knowledge of change of base formula for logarithms
  • Basic trigonometry involving sine and cosine functions
NEXT STEPS
  • Study the properties of complex logarithms in depth
  • Learn about the change of base formula for logarithms
  • Explore Euler's formula and its applications in complex analysis
  • Investigate the implications of adding multiples of \( 2\pi i \) in logarithmic functions
USEFUL FOR

Students of mathematics, particularly those studying complex analysis, as well as educators and anyone seeking to deepen their understanding of logarithmic properties and their applications in complex numbers.

Jacopo
Messages
2
Reaction score
0
I know why I can add [tex]2 \pi i[/tex] to [tex]log_{e} \omega[/tex]
([tex]log_{e} \omega = log_{e}z + i \theta[/tex] if [tex]\omega = z(cos \theta +i sen \theta )[/tex]) but I can't understand why I can add [tex]\frac{2 \pi i}{log_{e}b}[/tex] to [tex]log_{b} \omega[/tex].
Does anyone have the answer for me?
Thanks!
 
Physics news on Phys.org
Jacopo said:
I know why I can add [tex]2 \pi i[/tex] to [tex]log_{e} \omega[/tex]
([tex]log_{e} \omega = log_{e}z + i \theta[/tex] if [tex]\omega = z(cos \theta +i sen \theta )[/tex]) but I can't understand why I can add [tex]\frac{2 \pi i}{log_{e}b}[/tex] to [tex]log_{b} \omega[/tex].
Does anyone have the answer for me?
Thanks!
If you know the first, the second is easy! Because
[tex]log_b(\omega)= \frac{log_e(\omega)}{log_e(b)}= \frac{log_e(\omega)+ i2\pi}{log_e(b)}= \frac{log_e(\omega)}{log_e(b)}+ \frac{2\pi i}{log_e(b)}[/tex]
 
Oh yes :D Thanks
 

Similar threads

Undergrad Question about branch of logarithms
  • · Replies 14 ·
Replies
14
Views
4K
Undergrad Why is this function not ##L^1(\mathbb{R} \times \mathbb{R})##?
  • · Replies 6 ·
Replies
6
Views
3K
Undergrad On deriving the (inverse) Fourier transform from Fourier series
  • · Replies 4 ·
Replies
4
Views
4K
Find the Value of z in z^{1+i}=4 using Logarithms
  • · Replies 1 ·
Replies
1
Views
2K
Undergrad Bounding modulus of complex logarithm times complex power function
  • · Replies 4 ·
Replies
4
Views
4K
Undergrad ##SU(2, \mathbb C)## parametrization using Euler angles
  • · Replies 15 ·
Replies
15
Views
4K
Graduate Question about covering map of punctured unit disk
  • · Replies 3 ·
Replies
3
Views
3K
Using Properties of Logarithms
  • · Replies 22 ·
Replies
22
Views
3K
Undergrad Running through a complex math derivation of plasma frequency
  • · Replies 1 ·
Replies
1
Views
1K
Block sliding on vertical track with friction (Solved problem)
  • · Replies 4 ·
Replies
4
Views
3K