Part A: The result is complex.
Problem: Let's find what log(i) is. After that, we'll try to find loga(b+ci).
If we want to do this, we first need to understand what a log means, and what i is.
log(100)=2 because without putting anything for a in the above statement, we assume the base is 10.
This basically means a to the power of what gives the inside of the parenthesis?
So let's also look at what i is. i is a complex number, which is the square root of -1. This means the log-1(i)=1/2. So how would we take the log(i)?
My first instinct is to rewrite i as 11/2 since that would mean I get to use a few logarithmic rules on it. Here's how it plays out.
However, because we can't raise something to a real power (and I will not be doing research here), we get stuck at log(-1). So let's take a look at this in another way.
All complex numbers can be written in Polar Form, as r(cosθ+isinθ). In the case of just i, we can rewrite this as the following:
So let's try and take the log of this. (Which I can already see how it plays out, but let's try it anyways.)
Because there is no "log of a sum," we had to simplify the inside. However, it loops. We'll just get 0+0+...+0+log(i).
There's one more form in which we can write a complex number. It's called Euler's Form, and it's in a similar form to Polar form.
This should actually work, considering we have i in the exponent immediately. In order to write a number in Euler's form, we need to find its argument (preferably its principle argument) and its modulus (absolute value).
The argument is the angle formed with the origin of that point in the Argand Diagram. This would be found by taking the arctan of the triangle formed. Because we are looking at a pure imaginary number, it is on the imaginary axis, giving us an Argument of π/2. Furthermore, the modulus is easy to obtain because we are looking at a pure imaginary number, and it would just be the value attached to i, or 1 in this case.
This should be much easier to take the log of.
And that is perfect. We have our answer. I could have left it as log(e), but I (with no actual reason) decided to change it to use ln(10) instead.
So how about all complex numbers in the form of a+bi? Well, let's apply the same logic.
First, we need the modulus.
Next, we ned the argument.
Therefore, we can write this in Euler's Form using the following (which might look a bit ugly):
And when we take the logc(...), we can write it in the following way.
Part B: The base is complex
Problem: Well, now I'm curious. We have logc(a+bi). But what if we were to switch that? Say, logi(x) or loga+bi(x).
ix ∈ {i, -1, -i, 1} for x ∈ {0, 1, 2, 3, ...}
So how would we get logi(10)? I'm thinking we could reverse this. Because we're writing one as an imaginary number and the other as a real, this is limiting us. How about we write them in Euler's Form, and in the form of ax=b just so we can see this a bit more visually?
Being completely honest here, this doesn't exactly help my case. So I'm going to rewrite this again by taking the ath root of 10 to equal i, still in the form above.
This might actually help because now we're back in the last section's exploration, and we can simplify the right side so we're back to a real number.
So how about the log of any complex number raised to any regular number? We can apply the same thinking.
Part C: Both are complex
Problem: We have a complex in both the base and the result. How do we solve this?
My mind has actually been thinking about this since I started the second part of the exploration, which makes this significantly easier. There's a logarithm rule we need to do this much easier.
We can use this when we have a complex number in each. Let's look at the following example.
Sorry, it starts to look very packed towards the end, that's the exponent in its exact form. I don't think I can simplify it further than that. But this looks a bit packed to make a formula out of it already. So I'll restart in order to find a final formula.
Because there are no actual values in this apart from the 10 and i, this is as far as we can get.
And that's all there is to having complex numbers within logarithms! It's not as complicated as I thought it would be.
Problem: Let's find what log(i) is. After that, we'll try to find loga(b+ci).
If we want to do this, we first need to understand what a log means, and what i is.
log(100)=2 because without putting anything for a in the above statement, we assume the base is 10.
This basically means a to the power of what gives the inside of the parenthesis?
So let's also look at what i is. i is a complex number, which is the square root of -1. This means the log-1(i)=1/2. So how would we take the log(i)?
My first instinct is to rewrite i as 11/2 since that would mean I get to use a few logarithmic rules on it. Here's how it plays out.
However, because we can't raise something to a real power (and I will not be doing research here), we get stuck at log(-1). So let's take a look at this in another way.
All complex numbers can be written in Polar Form, as r(cosθ+isinθ). In the case of just i, we can rewrite this as the following:
So let's try and take the log of this. (Which I can already see how it plays out, but let's try it anyways.)
Because there is no "log of a sum," we had to simplify the inside. However, it loops. We'll just get 0+0+...+0+log(i).
There's one more form in which we can write a complex number. It's called Euler's Form, and it's in a similar form to Polar form.
This should actually work, considering we have i in the exponent immediately. In order to write a number in Euler's form, we need to find its argument (preferably its principle argument) and its modulus (absolute value).
The argument is the angle formed with the origin of that point in the Argand Diagram. This would be found by taking the arctan of the triangle formed. Because we are looking at a pure imaginary number, it is on the imaginary axis, giving us an Argument of π/2. Furthermore, the modulus is easy to obtain because we are looking at a pure imaginary number, and it would just be the value attached to i, or 1 in this case.
This should be much easier to take the log of.
And that is perfect. We have our answer. I could have left it as log(e), but I (with no actual reason) decided to change it to use ln(10) instead.
So how about all complex numbers in the form of a+bi? Well, let's apply the same logic.
First, we need the modulus.
Next, we ned the argument.
Therefore, we can write this in Euler's Form using the following (which might look a bit ugly):
And when we take the logc(...), we can write it in the following way.
Part B: The base is complex
Problem: Well, now I'm curious. We have logc(a+bi). But what if we were to switch that? Say, logi(x) or loga+bi(x).
ix ∈ {i, -1, -i, 1} for x ∈ {0, 1, 2, 3, ...}
So how would we get logi(10)? I'm thinking we could reverse this. Because we're writing one as an imaginary number and the other as a real, this is limiting us. How about we write them in Euler's Form, and in the form of ax=b just so we can see this a bit more visually?
Being completely honest here, this doesn't exactly help my case. So I'm going to rewrite this again by taking the ath root of 10 to equal i, still in the form above.
This might actually help because now we're back in the last section's exploration, and we can simplify the right side so we're back to a real number.
So how about the log of any complex number raised to any regular number? We can apply the same thinking.
Part C: Both are complex
Problem: We have a complex in both the base and the result. How do we solve this?
My mind has actually been thinking about this since I started the second part of the exploration, which makes this significantly easier. There's a logarithm rule we need to do this much easier.
We can use this when we have a complex number in each. Let's look at the following example.
Sorry, it starts to look very packed towards the end, that's the exponent in its exact form. I don't think I can simplify it further than that. But this looks a bit packed to make a formula out of it already. So I'll restart in order to find a final formula.
Because there are no actual values in this apart from the 10 and i, this is as far as we can get.
And that's all there is to having complex numbers within logarithms! It's not as complicated as I thought it would be.
