Question

Complex logarithms: Give examples for when the logarithm Log(z1/z2) \neq Log(z1)-Log(z2) if Log(z1), Log(z2) and Log(z1/z2) are all defined. Log assumes the principal branch.

Answer 1

@pitamar

Answer 2

\(z_1=z_2= -3i\)
\(log(\dfrac{z_1}{z_2})= log 1=0\) while \(logz_1-logz_2= (ln 3-\dfrac{\pi i}{2})-(ln 3-\dfrac{\pi i}{2})=\pi i\)

Answer 3

But not sure. Honestly, I don't know what is complex logarithms, just make a short research about it and give you what you need: a counter example of log (z1/z2) is not = log z1-log z2.

Answer 4

Although it seems that your answer is incorrect;
\[log(z_1)-log(z_2) = (ln(3)-\pi i) - (ln(3)-\pi i ) = ln(3)-\pi i - ln(3) + \pi i = 0\]
Effectively making this answer equal to the first.

Answer 5

Hi. sorry I didn't answer yesterday, I was very tired.
However, I did get to do some thinking and here's how I see it:

Well first, as basic identity in complex numbers there is Euler's formula:
$$
e^{i \varphi} = cos(\varphi) + i \cdot sin(\varphi)
$$Though tempting, I'll not go through the proof to keep it shorter, but you can find proofs here:
http://en.wikipedia.org/wiki/Euler's_formula#Using_power_series

As you can see, Euler's formula does not get a complex number, it gets only the imaginary part of it.
Notice that this 'imaginary part' is used as an angle for the trig functions, so you can add or subtract \(2 \pi\) from it and get the same result. I'll talk about that shortly.

Now, here is a picture from Wikiedia's article "Complex plane":
http://en.wikipedia.org/wiki/Complex_plane#mediaviewer/File:Complex_conjugate_picture.svg
If we look on \(z\) then it has a real part \(x\) and imaginary part \(iy\) and we can say:
$$
|z| = r\\
\arg(z) = \varphi \\
x = r \cdot cos(\varphi) \\
y = r \cdot sin(\varphi) \implies iy = ir \cdot sin(\varphi)
$$And therefore we can write z as:
$$
z = x + iy \\
z = r\cdot cos(\varphi) + ir \cdot sin(\varphi) = r \cdot (cos(\varphi) + i \cdot sin(\varphi)) = r \cdot e^{i \varphi}
$$
That means, that every complex number can be represented by the distance from the origin (the absolute value) and the angle with the \(x\) axis (the argument).
This form of representation is called 'polar form'. you can find more information here:
http://en.wikipedia.org/wiki/Complex_number#Polar_form

Now, as you can see, the representation contains a real part \(r\) and an imaginary part \(i \varphi\) which implies we can combine the parts and transform it to work with a complex number:
$$
z = r \cdot e^{i \varphi} = e^{ln(r)} \cdot e^{i \varphi} = e^{ln(r) + i \varphi}
$$ Let \(ln(r)\) be a real part of some complex number \(c\) and \(i \varphi\) the imaginary and we have:
$$
c = ln(r) + i \varphi \\
z = e^c
$$That means that for every complex number \(z\) we can get another complex number \(c\) which represents it in polar form!
But, as we've seen above, adding or subtracting \(2 \pi\) of the imaginary part (the angle) wouldn't change the result, which means we can get infinite number of complex numbers that represent the same complex number in polar form.
Ah, we're getting somewhere =)

We can also get the sense now of complex numbers multiplication and division, because for 2 complex numbers \(z_1\) and \(z_2\) we can say:
$$
c_1 = ln(|z_1|) + i \cdot \arg(z_1) \\
c_2 = ln(|z_2|) + i \cdot \arg(z_2) \\
\large z_1 \cdot z_2 = e^{c_1} \cdot e^{c_2} = e^{c_1 + c_2} = e^{ln(|z_1|) + ln(|z_2|) + i(\arg(z_1) + \arg(z_2))}
$$
We can simplify it even more using log rules and such, but I just wanted to show the idea.
Likewise, division would be:
$$
\large \frac{z_1}{ z_2} = z_1 \cdot z_2^{-1}= e^{c_1} \cdot e^{-c_2} = e^{c_1 - c_2} = \\
\large = e^{ln(|z_1|) - ln(|z_2|) + i(\arg(z_1) - \arg(z_2))}

$$

Let's now try and define \(ln(z)\) for complex numbers. Basically \(\ln(x)\) is the inverse of \(\exp(x)\). So we could say:
$$
e^{ln(z)} = z
$$So \(ln(z)\) will be defined to return the value which makes this work.
The only problem is that as we already know, this value has to be a polar representation of \(z\).
But there are infinite possible values for that...
That's where the `principal branch` comes in. By limiting the imaginary part of what \(ln(z)\) returns to, say \((-\pi, \pi]\) we assure a single defined value that \(ln(z)\) could return to represent \(z\) in polar form.

But now we can see the 'problem'.
Say we have 2 complex numbers \(z_1\) and \(z_2\). and say
$$
\arg(z_1) = -\frac{2\pi}{3} \\
\arg(z_2) = \frac{2\pi}{3}
$$
So we can say that:
$$
c_1 = ln(z_1) = ln(|z_1|) + i \cdot \arg(z_1) = ln(|z_1|) - \frac{2 \pi i}{3} \\
c_2 = ln(z_2) = ln(|z_2|) + i \cdot \arg(z_2) = ln(|z_2|) + \frac{2 \pi i}{3} \\
c1 - c2 = ln(z_1) - ln(z_2)= ln(|z_1|) - ln(|z_2|) + i(- \frac{2 \pi i}{3} - \frac{2 \pi i}{3}) =\\
= ln(|z_1|) - ln(|z_2|) - \frac{4\pi i}{3}\\
$$
Now, we're tempted to say, hey! I can say
$$
\frac{z_1}{z_2} = e^{c_1 - c_2} \implies ln\bigg( \frac{z_1}{z_2} \bigg) = c_1 - c_2
$$It is true that \((c_1 - c_2)\) represents \(\frac{z_1}{z_2}\) in a polar form, but as we already know, so do many other complex numbers.
If we examine \((c_1 - c_2)\) we can see that its imaginary part exceeds the limits of the principal branch of \(ln(x)\) so it cannot be the number returned by it.
We can see however that we can add \(2 \pi\) to its imaginary part and get a new number that still represents \(\frac{z_1}{z_2}\) in a polar form, but lays within the boundaries of the principal branch of \(ln(x)\):
$$
\large e^{c_1 - c_2} = e^{ ln(|z_1|) - ln(|z_2|) - \frac{4\pi i}{3}} = \\
\large = e^{ln(|z_1|) - ln(|z_2|) + i\bigg(-\frac{4\pi}{3} + 2\pi\bigg)} = \\
\large = e^{ln(|z_1|) - ln(|z_2|) + i\bigg(-\frac{4\pi}{3} + \frac{6\pi}{3}\bigg)} = \\
\large = e^{ln(|z_1|) - ln(|z_2|) + \frac{2\pi i}{3}} \\

ln\bigg( \frac{z_1}{z_2} \bigg) = ln(|z_1|) - ln(|z_2|) + \frac{2\pi i}{3} = c_1 - c_2 + 2\pi i
$$
And hence, in our example:
$$
ln\bigg( \frac{z_1}{z_2} \bigg) \neq ln(z_1) - ln(z_2)
$$