When thinking of complex numbers, most people think about the complex solutions to quadratic equations. However, there is much more to the complex world than that. In quadratics, we get a complex solution when there is a negative number inside a square root. But what else can get us imaginary numbers. In other words, what else doesn't exist. Well, you can't have a number raised to any power and get a negative number. But what if you wanted to solve an exponential equation where this was the case. Euler's identity is the most famous example of this. We know that if we raise e to the pi*i power, we get -1. Therefore, if we take the natural log (ln) of both sides, we get than pi*i = ln(-1). This is a new type of complex numbers: the log of negative numbers. The log of -1 was easy. We just manipulated Euler's formula to get the answer of pi*i. But what about the log of i? Well, by manipulating Euler's formula in a more clever way, we get find ln(i) as well. Here's how its' done:
or, we could do this:
The Natural Log of Negative Numbers
Using Euler's identity, you can easily prove that the natural log of -1 is pi*i. But what about the natural log of -2, -3, etc? Well, using Euler's identity that ln(-1) = pi*i, we can do this:
Using Euler's identity, you can easily prove that the natural log of -1 is pi*i. But what about the natural log of -2, -3, etc? Well, using Euler's identity that ln(-1) = pi*i, we can do this:
What is arcsin(i) and arccos(i)?
The first thing we need to do to find arcsin(i) and arccos(i) is find a general formula for the arcsin and arcos functions that includes complex numbers. A good place to start would be to find a general formula for the sine and cosine function that includes imaginary numbers. Using Euler's identity, this can be done like this:
The first thing we need to do to find arcsin(i) and arccos(i) is find a general formula for the arcsin and arcos functions that includes complex numbers. A good place to start would be to find a general formula for the sine and cosine function that includes imaginary numbers. Using Euler's identity, this can be done like this:
Now, if we set sinx=a, then arcsin(a)=x. Likewise, if we set cosx=b, then arccos(b)=x. Then if we set a and b equal to the formulas for sinx and cosx, we will get the formula for arcsinx and arccosx.
Note that I did not include the negative square root of 1-x^2. This is because if you graph the arcsine function with the negative square root of 1-x^2, you get a very small domain and range. However, if you graph the arcsine function with the positive square root of 1-x^2, the domain is all real numbers.
Now that we have the formula for arcsinx, we can use the identity arcsinx+arccosx=pi/2 to find the formula for arccosx without doing tons of math.
Now, using those two formulas, we can evaluate arcsin(i) and arccos(i).
Here is a list of the formulas derived on this page:
Note: (I manipulated the arcsinx formula so that there wouldn't be a negative before the i the same way I did for arcsin(i).)
What are sin(i) and cos(i)? This problem is very easy. Using the sin(x) and cos(x) formulas from earlier, we can easily find sin(i) and cos(i): Very cool, right. The cosine of an imaginary number is a real number! Proof that arcsin(x)+arccos(x)=pi/2 What is ln(0)? If you look up what ln(0) is online, you will always get that it is either undefined or infinity. Even Wolfram Alpha identifies ln(0) as infinity. However, using the inverse tangent function, we can find a better answer. The first thing that we need to do is use the formulas for sin(x) and cos(x) to get the formula for tan(x). Next, we will use the tan(x) formula to find the formula for arctan(x). Now that we have the formula for arctan(x), we can attempt to find arctan(i) to find ln(0) instead. The above answer is true, but using the laws of logs, we can simplify this even more. Proof that (i+x)/(i-x)=(xi-1)/(xi+1) |
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Therefore:
Proof that this is true:
What is the integral of e^(2iarctanx)?
If you write a general function for ln(0) = 2iarctan(i), you get ln(y) = 2iarctanx, so y = e^(2iarctanx). Now, if you add 1 to this function, you get that f(x) = 1 + e^(2iarctanx). If you integrate this function from x = -1 to x = 1, you get a very strange, yet cool result.
We will start by moving the 1 out and evaluating it from -1 to 1:
Now, we will do a u-substitution. However, note from above that:
If we substitute e^(2iarctanx) for u and the result from above for dx, we can do partial fractions:
Finally, we will put in the limits of integration and add the 2 from earlier:
So overall:
Or, we could have the fact that e^2iarctanx = (i-x)/(i+x) and the integration becomes much easier.
Now, we will plug in the limits of integration to get:
The Sigma Function
In case your not familiar with it, the Sigma Function is a function that I recently (June 2018) made up. It is an extension of what we did previously on this page and can be written in several forms. Also, the function has a lot of cool properties. Here are the two integral definitions:
Now, we will evaluate the above integrals and do a bit of algebra to get the Sigma Function as a logarithm:
The fourth and final form of the Sigma Function is the coolest. To simplify the logarithmic expression into the final form, we will set the expressions inside the two integrals equal to each other and do a bit of algebra:
This is very cool: you can write this function as 4arctan(n). Now, when we found pi earlier by integrating the function, we could have just plugged in 1 to 4arctan(n) to get 4arctan(1) which equals pi. Therefore:
What is arctan(1)+arctan(2)+arctan(3)?
We know that arctan(1) is pi/4, but what about the arctan2+arctan3 part? How do we evaluate this? Algebraically, we will have to use the complex definition of arctanx from above. However, before plugging in any numbers, we should manipulate the expression a bit to make the algebra easier.
Now it is time to plug in the numbers and do a lot of algebra.
How cool is this? arctan1+arctan2+arctan3 = pi !!!