# Order of Zero of Complex Logarithm

1. Jul 19, 2013

### Tsunoyukami

I'm not entirely sure if my reasoning is correct here...

"[G]ive the order of each of the zeroes of the given function.

6. $Log(1 - z)$, $|z| < 1$" (Complex Variables, 2nd edition by Stephen D. Fisher; pg. 133)

The complex logarithm function is defined in the following manner:

$log(z) = ln|z| + iarg(z)$
$Log(z) = ln|z| + iArg(z)$, $Arg(z) \in [0, 2\pi)$ (this is known as the principal branch)

As a result of this definition, the principal logarithm can only have a zero when both $ln|z| = 0$ and $iArg(z) = 0$. This is true if and only if $|z| = 1$ and $Arg(z) = 0$; that is to say, for the complex number $z = x + iy = 1 + i \cdot 0 = 1$. However, there is an issue with this as the problem explicitly states that $|z| < 1$ and so the function given in the question is never zero - therefore this function has no zeroes. Is this correct?

My next question is suppose I had a similar function..say $g(z) = log(z-1)$, $|z| < 2$ - this function would have an infinite number of zeroes (because of the way "little log" is defined) of order one corresponding to each $z = 1 + 2 \pi in$ where n is an integer. Is that correct?

2. Jul 19, 2013

### jackmell

Your reasoning is not correct. First, always start with the multi-valued $\log(u)$ function. Now, learn what this function looks like (draw it): the real part looks like a funnel, and the imaginary part, a twisted helix sheet. If we now want to study a single-valued part of it, we cut out a single-valued "determination" such as the principal-valued logarithm which we sometimes designate with capital letters:

$$Log(u)=\ln|u|+i\theta,\quad -\pi<\theta\leq \pi$$

or excise any other single-valued component of $\log$ such as your particular determination:

$$Log(u)=\ln|u|+i\theta,\quad 0<\theta\leq 2\pi$$

Now, no matter which determination or part of $\log(u)$ we set up, there is still only one value when it's zero:

$$\log(1)=\ln|1|+0\theta=0$$

Now take your $g(z)=\log(z-1)$. Where is the only place that function is zero no matter what the radius of definition is?

3. Jul 20, 2013

### Tsunoyukami

Okay, let me try this again. I must not have been thinking clearly earlier.

My first question is the textbook problem assigned in the original post. For the given function, $f(z) = Log(1-z)$ there is a zero of order one when $z = 0$ because $g(u) = Log(u) = 0$ iff $u = 1$ (by the definition of the complex logarithm. We can then write this as $f(z) = Log(1-z) = 0$ iff $1 - z = 1$ iff $z = 0$. Is this correct now?

As for my second question...I'm not really sure why I was thinking what I was thinking when I asked this bit. There can still only be a single zero because $h(z) = log(z)$ is only zero when both its real and imaginary parts are zero - this means the argument must be zero for $log(z)$ to be zero. Is this better? For the function $j(z) = log(z -1)$ to be zero, $z - 1 = 1$ iff $z = 2$.

Thanks for your response - hopefully you'll be able to look at this and let me know if this is better. Thanks again!

4. Jul 20, 2013

### jackmell

Ok, that's good. You don't mention why the zero is order one and not say three for example. Also, good idea to keep up with precisely which part of the multivalued $\log(u)$ function you're talking about. Usually we do that by using $\log$ for the whole thing and then make a specific branch like $\log(z)=\ln(r)+i\theta,\quad 0<\theta\leq 2\pi$ when we're referring to a part of it.

5. Jul 20, 2013

### Tsunoyukami

Ah, in my class we have used the convention that the principal branch of logarithm is for $\theta \in [0, 2\pi)$ and is denoted $Log(z)$.

The reason why the zero is order one as opposed to any other number is that it "occurs" only once. A function $f(z) = [Log(1-z)]^{n}$ would have a zero of order n at $z=0$.

6. Jul 20, 2013

### jackmell

That's not a good reason. The order represents how fast it's going to zero when it's near zero. What's holding it up the most from going to zero when it's close? Well, the lowest power in z is since the other ones with higher power go to zero really fast. So for the function:

$$Log(1-z)=\sum_{n=0}^{\infty}a_n z^n$$

the order is the lowest power of z and:

$$Log(1-z)=-\frac{z}{1}-\frac{z^2}{2}-\frac{z^3}{3}+\cdots$$

so the order is one, contrasted with $Log(1-z^2)$ which is of order two and so forth.