Can you explain me how these equations were obtained?

[whuzzwhuzz]

ln(-2) = ln(2) + (1+2k)*pi*i
and
ln(16) = ln (16) + 2m*pi*i

I picked this equation from here: https://www.physicsforums.com/showthread.php?t=184129
Post #5 (D H's post.

Thanks a lot!

 

[jjmontero9]

Those equations are obtained by applying the "[URL identity[/URL] of the complex numbers:

$$\ln{(-2)}$$

$$= \ln{(2\cdot{}e^{(2k+1)\pi{}i})}$$

$$= \ln{(2)} + \ln{(e^{(2k+1)\pi{}i})}$$

$$= \ln{(2)} + (2k+1)\pi{}i$$

Where k = 0, 1, 2, ...


The second one is:

$$\ln{(16)}$$

$$= \ln{(16\cdot{}e^{2\pi{}mi})}$$

$$= \ln{(16)} + \ln{(e^{2\pi{}mi})}$$

$$= \ln{(16)} + 2\pi{}mi$$

Where m = 0, 1, 2, ...

 

[Mute]

Those equations are obtained by applying the "[URL identity[/URL] of the complex numbers:

$$\ln{(-2)}$$

$$= \ln{(2\cdot{}e^{(2k+1)\pi{}i})}$$

$$= \ln{(2)} + \ln{(e^{(2k+1)\pi{}i})}$$

$$= \ln{(2)} + (2k+1)\pi{}i$$

Where k = 0, 1, 2, ...The second one is:

$$\ln{(16)}$$

$$= \ln{(16\cdot{}e^{2\pi{}mi})}$$

$$= \ln{(16)} + \ln{(e^{2\pi{}mi})}$$

$$= \ln{(16)} + 2\pi{}mi$$

Where m = 0, 1, 2, ...

To clarify this a bit, when dealing with complex numbers the logarithm is a multivalued function, so you must choose a "branch" of the function that is single valued. However, all branches are related to each other by multiples of $2\pi$. If $\log(z)$, for complex numbers z, is the complex logarithm function (which is always base e), you can relate it to the real valued logarithm, ln(x), and the phase of the number z. In polar coordinates, $z = r\exp(i\theta)$, where r is the modulus of the complex number (and is purely real) and $\theta$ is the phase. Then, we may write

$$\log(z) = \log(re^{i\theta}) = \ln r + \log(e^{i\theta}) = \ln r + i(\theta + 2n\pi)$$
where n is an integer.

The $2\pi n$ term is there for the same reasons as when solving an equation like $y = \sin(\theta)$. You get $\theta = \arcsin y + 2n\pi$, because sin is $2\pi$ periodic. Similarly, the complex exponential function is $2i\pi$ periodic, so when taking the inverse you get this $2\pi i n$ term added on:

$$e^{i\theta} = e^{i\theta + 2n\pi i}.$$

In the example cited, $\theta$ was zero, but no branch of the complex logarithm was chosen, so there are infinitely many solutions, one for each integer n.

 

[whuzzwhuzz]

Thank you both both of you guys. Both helped me understand the question and the topic. Thank you again. :)

 

[CellCoree]

Sure, I'd be happy to explain how these equations were obtained. The equations you mentioned are the result of applying the properties of logarithms to complex numbers. Let's break down each equation separately to understand how they were obtained.

For the first equation, ln(-2) = ln(2) + (1+2k)*pi*i, we can start by writing -2 as a complex number in polar form: -2 = 2*e^(pi*i). Then, using the property of logarithms that states ln(a*b) = ln(a) + ln(b), we can rewrite ln(-2) as ln(2*e^(pi*i)). Next, we can use another property of logarithms, ln(e^x) = x, to simplify further and get ln(2) + ln(e^(pi*i)). Finally, we can use the fact that e^(pi*i) = -1 to get ln(2) + ln(-1), which can be rewritten as ln(2) + (1+2k)*pi*i.

For the second equation, ln(16) = ln (16) + 2m*pi*i, we can use similar steps. We can write 16 as a complex number in polar form: 16 = 16*e^(0*i). Then, using the property ln(a^b) = b*ln(a), we can rewrite ln(16) as 4*ln(2*e^(0*i)). Again, using the property ln(e^x) = x, we can simplify further to get 4*ln(2) + 4*ln(e^(0*i)). Since e^(0*i) = 1, we can rewrite this as 4*ln(2) + 4*0*i, which simplifies to 4*ln(2) + 0*i. Therefore, ln(16) = ln(16) + 2m*pi*i.

These equations were obtained by applying the properties of logarithms to complex numbers, which can help us simplify and manipulate expressions involving complex numbers. I hope this explanation helps! Let me know if you have any further questions.