complex-analysislogarithmsnotation
A definition for complex logarithm that I am looking at in a book is as follows –
$\log z = \ln r + i(\theta + 2n\pi)$
Why is it $\log z = \ldots$ and not $\ln z = \ldots$? Surely the base of the log will make a difference to the answer.
It also says a few lines later $e^{\log z} = z$.
Yet again I don't see how this makes sense. Why isn't $\ln$ used instead of $\log$?
The way we define the Logarithm is the inverse of the exponential.
If $z= re^{i\theta}$,then $ln(z)=ln(r)+i\theta$.
Now , a problem that is there is, with this equation alone , logarithm is multi-valued i.e for every z there are infinitely many values for the logarithm . For $\theta$ differing by a value of $2\pi$ , one will get the same value. It isn't injective. This leads us restricting its domain so that each point(z) corresponds to only one value. This is referred to as a branch cut. Check the image from Wiki,
So, for log(z), simply remove any ray joining 0 and infinity.(Restricting complete rotation(0 to $2\pi$!). Also, see in the graph what happens. The principle log removes one particular ray, the negative real axis, I guess. This is only a convention, I guess.
Now, coming back to your question, you want log(iz)(PS:-The $iz$ only changes the labels on the axes) to be analytic in some given region. All you need to do is to remove a ray(not in the given region) to make it analytic in the given region.
Recall the definition of $f(n) \in O(g(n))$: there exists some $k$ and $C > 0$ such that
$$n \ge k \implies |f(n)| \le C|g(n)|.$$
We have $n < 2^n$, as in the question, for $n \ge 1$. Therefore, taking $k = 1$ and $C = 1$, we have $$n \ge 1 \implies |n| = n \le 2^n = |2^n|$$ so $n \in O(2^n)$.
But, you're also right that $n \in O(n)$ too. These are not mutually exclusive. We have $$n \ge 1 \implies |n| \le |n|,$$ which shows $n \in O(n)$. In fact, if $f(n) \in O(g(n))$ and $g(n) \in O(h(n))$, then it's a good exercise to show that $f(n) \in O(h(n))$.
You said, in the comments, that your understanding of $n < 2^n$ implies that at some point $k$, $2^n$ will grow faster than $n$. It's not entirely clear what you mean here, but I wouldn't sweat it too much. The point is, $2^n$ starts bigger than $n$, and stays that way forever afterwards, fulfilling the requirement for $n \in O(2^n)$.
Now, consider the case where $b \neq 2$. This is one situation where I think the book needs a little bit more detail. In particular, I think we should also assume $b > 1$. The case for $0 \le b \le 1$ should be handled separately (and the $b < 0$ case handled separately again). I will assume $b > 1$.
Since $\log_2 n < n$ for all $n \ge 1$, we can use the change of base formula. Recall that $$\log_b n = \frac{\log_2 n}{\log_2 b}.$$ (If you don't accept this formula, try looking it up online.) But, since $b > 1$, it follows that $\frac1{\log_2 b} > 0$, so $$\log_2 n < n \implies \log_b n = \frac{\log_2 n}{\log_2 b} < \frac{1}{\log_2 b} n.$$ This proves that $\log_b n \in O(n)$, using $C = \frac{1}{\log_2 b}$.
Best Answer
Often in math books the base of $\log$ is just assumed to be $e$.
In this case it looks like the reason they are using $\log z$ instead of $\ln$ is to differentiate between when it is a complex function versus when it is a real function.
http://en.wikipedia.org/wiki/Complex_logarithm