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Understanding logarithmic functions for complex numbers

Imaginary numbers are not 'imaginary'.
Real numbers are not 'real'.
In fact, they're both in our heads...

We are quite familiar with the logarithmic function. We define it as the inverse of the exponential function.

That is, we say:
$$\log{e^x} = e^{\log{x}} = x$$
(In fact, this is how the inverse of a function is even defined.)

Now, we're all comfortable with the natural log. We use it for various results, but we always keep in mind that the number we're evaluating is POSITIVE.

Why though?

Let's consider some negative number $a$. What does it mean to take it's natural logarithm?

It means to find a number such that $e$ raised to that number gives $a$. And surely since $e$ raised to any power always gives a positive value, it makes no sense to take the log of a negative number, right?

Right?

Complex Numbers: Function Domains

Remember how we extended the existence of real numbers to the complex number system? What did we do?

The square root of number of a negative number does not exist, obviously, because any number multiplied with itself gives a positive number. But then, we extended the domain of this square root function to include negative numbers too.

And for this, we defined:
$$i = \sqrt{-1}$$
And now we know that it makes sense to take the square root of a negative number.

Question: Can we do such extension of domain for other functions too, with the help of complex numbers? (Spoiler: Yes)

Euler's Identity

I hope we're all familiar with the famed Euler's Identity, which goes like this:
$$e^{ix} = \cos{x} + i\sin{x}$$
where $x$ is any real number you like.

(If you have no idea why this true, or what this even means, I'll write a post on this some other time.)

Now, if we substitute for $x = \pi$, we get:
$$e^{i\pi} = -1$$
This, according to many, is the most beautiful equation in the world, because they think it matters that the '5 fundamental constants' ended up in a single equation, when they write it as:
$$e^{i\pi} + 1 = 0$$
Sure, it might be aesthetically pleasing, but what do we get from it, in our previous efforts to understand the log of negative numbers? Consider:
$$e^{i\pi} = -1$$
Taking log on both sides:
$$i\pi = log(-1)$$
And almost magically, we've ended up with not only somewhat defining negative logs, but also finding it's 'value'.

Next, note that this can be extended to any real negative number.
$\log(-69)$, say, is equal to, $\log(69) +  \log(-1) = \log(69) + i\pi$

Onward:

Any complex number can be represented by both it's rectangular and polar forms, or symbolically:
$$z = a + ib = r[\cos\theta + i\sin\theta]$$
Note that in the above expression, $r = \sqrt{a^2+b^2}$ is the modulus, and $\theta = \tan^{-1}{\frac{b}{a}}$, is the argument.

From Euler's Identity, we have:
$$z = r(e^{i\theta})$$
Taking log on both sides, we get:
$$\log{z} = \log{r} + i\theta$$
And that's it. We're done.

Let's try an example to get ourselves familiar with this.

Q. Evaluate $\log{(1+\sqrt{3}i)}$

You should immediately be able to see that this complex number makes an angle of $\frac{\pi}{6}$ with the Real Axis.
Also, note that the modulus of this number is $\sqrt{1^2+\sqrt{3}^2} = 2$

Thus, $1+\sqrt{3}i = 2(e^{i\large\frac{\pi}{6}})$

And so: $\log(1+\sqrt{3}i) = \log{2} + i\large\frac{\pi}{6} = 0.6931 + 0.5235i$, approximately.

As simple as that.

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