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My first question is:

What is the proper definition of logarithmic function $f(z)=\ln{z}$. where $z\in \mathbb{C}$.

quoting Wikipedia.

a complex logarithm function is an "inverse" of the complex exponential function, just as the natural logarithm $\ln{x}$ is the inverse of the real exponential function $e^x$.

In a book Calculus Vol 1.By Tom M. Apostol, he tells the function

$\ln{x}$ is defined as $\ln{x}=\int_{1}^{x}{\frac1t\;dt}$ $\color{blue}{\star}$

and the function

$e^x$ is defined to be it's inverse

(rather than the opposite).

Some reasons why it is so as per the book and what I have understood is .

We can define what is $e^2$ $=$ $e\times e$ . But how can we give such a defintion to $e^{\sqrt{2}}$ or$\large e^{\sqrt{2+\sqrt[3]{3+\sqrt[5]{5}}}}$ or more generally the function $a^x$ when the domain is $\mathbb{R}$. Hence as the funciton $a^x$ is not properly defined for Real domain, how can we think about it's definition of it's inverse(The way Wikipedia and some other books define natural logarithm)

So if we are to define $\ln{x} $ as in $\color{blue}{\star}$, it solves all the problem( a proper definition of $\ln{x}$ in real domain , a definition for exponential function in real domain, getting rid of otherwise-circular proofs of some basic theorem in limits involving logarithm and exponential function)

Thinking in the same way just like $e^z$ have problems with definition.How can we define it's inverse?.

Doing a bit of research through internet I found some bits of information.

Wikipedia entry: Natural Logarithm says:

The first mention of the natural logarithm was by Nicholas Mercator in his work Logarithmotechnia published in 1668,2 although the mathematics teacher John Speidell had already in $\color{red}{1619}$ compiled a table on the natural logarithm.[3] It was formerly also called hyperbolic logarithm,[4] as it corresponds to the area under a hyperbola. It is also sometimes referred to as the Napierian logarithm, although the original meaning of this term is slightly different.

Wikipedia entry:Exponential Function

The exponential function arises whenever a quantity grows or decays at a rate proportional to its current value. One such situation is continuously compounded interest, and in fact it was this that led Jacob Bernoulli in $\color{red}{1683}$[4] to the number

now known as e. Later, in 1697, Johann Bernoulli studied the calculus of the exponential function

The dates(as per source) of discoveries suggests such($\color{blue}{\star}$) a definition.

So summing up I have two questions.

1.A proper definition of $\ln{z}$ when $z\in \mathbb{C}$

2.Is't the definition for logarithm in real domain, the one I have mentioned ($\color{blue}{\star}$) is the best/correct (Just because I have only seen a few places it is defined so).?

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Using the integral definition isn't a problem; just take any path from $1$ to $z$ that does not go through the branch cut, and you're golden. –  J. M. Apr 22 '13 at 14:06
@J.M. Nice!. I had thought about some definition like that. But if we follow the definition $\int_{1}^{z}{\frac{1}{t}\;dt}=\ln{z}$ How should I understand the fact that $\ln{z}$ is multivalued function?.(I don't have any idea what is $\int{f}$ where $f$ is a complex function and may be I am speaking foolish, $1/t$ is single valued and taking an integral as a sum how can sum of single valued function be a multivalued function?(I guess it may be because of we can choose any path, Am I right?.)).Thanking you in anticipation. –  boywholived Apr 22 '13 at 14:25
If you remember the treatment of complex numbers as points in a plane, consider for instance the line segment joining $1$ and $z$. That can be taken as an integration path. –  J. M. Apr 22 '13 at 14:29
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2 Answers

You can define the complex logarithm as $$ Lg(z)=lg(|z|)+iArg(z) $$ where $lg$ is the real logarithm and $Arg(z)$ is the principal branch of the argument (that is, $Arg(z)\in(-\pi ,\pi )$ ). The logarithm is defined for all $z$ in $U=\mathbb{C}-\{x\in\mathbb{R} | x\leq 0\}$.

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You can accept the answer if you enjoyed it, sir ;) –  Marra Apr 22 '13 at 11:47
Ok nice answer!. (Remember there is a time delay for accepting answers). And don't call me 'sir' I am a only 17 and is at school. –  boywholived Apr 22 '13 at 11:53
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The exponential function is usually defined as

$$\exp:= z\mapsto \sum\limits_{k=0}^{+\infty}\frac{z^k}{k!}$$

Then you find out it's a bijection from $\Bbb R$ to $\Bbb R _+^*$ and defined $\ln$ as its inverse function.

Then you define for $a>0$, $z\in \Bbb C$, $a^z=e^{z\ln a } = \exp ( z\ln a)$

So $a^x$ is properly defined.

Then to defined the logarithm on the complex numbers, it's a bit more complicated. You don't need $\cos$ and $\sin$ to define it but since those functions are introduced earlier, it might help you see what's happening so I'll use them.


$$\cos:=z \mapsto \Re \exp ( i z)$$

$$\sin := z \mapsto \Im \exp(iz)$$

You can then prove that

$$\exp(z) = \exp(\Re z)(\cos \Im z + i \sin \Im z)$$

And as you can see (because you know properties of $\cos$ and $\sin$ that can be proved from the definition above), $\exp(z+2\pi i)=\exp ( z)$.

And this is where the trouble starts to define the complex logarithm: You do not have an injection and therefore can't have a bijection either. So you have to restrict the domain to have a bijection which leads to different branches of the logarithm, much like when you search for an angle, you want it modulo $2\pi$.

For all $z\in \Bbb C^*$, you know you can find a unique $r\in \Bbb R_+^*$ and a unique $\theta \in [-\pi,\pi)$ so that $z = r e^{i\theta}$ (again, you "feel" it because of what you were taught but it can be proven). So we define

$$Arg:= z = re^{i\theta} \mapsto \theta$$

And then we can define

$$Log(z)=\ln |z|+i Arg(z)$$

The problem with doing that is that when we cross $\Bbb R_-^*$, $Arg(z)$ "jumps" from $-\pi$ to $\pi$, ie it is discontinuous. So we often define $Arg$ in $\Bbb C^* \setminus \Bbb R_-^* = \Bbb C \setminus \Bbb R_-$ to have a continuous function instead.

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