Cool channel, post. Im a bit confused. When you use square roots in \sqrt{\pi^2-1}, is this a Real square root, or a Complex square root, where you have to start bringjng up branches and other machinery?
You found that the product of the two solutions for w is 1, which means that, once you've got the first solution, the other is 1/w. This in turn means that the ln of the second solution is -ln w. Of course, you knew that would have to happen from the start, because you got the formula in the first place by assuming that cos is an odd function.
@@mikecaetano Sorry, I meant even, obviously. If z is a solution at the end, -z is also a solution, because we used this property when extending the function to the complex numbers at the beginning.
2:09 1/2 is equal to cos(pi/3) not cos(pi/6)
true dude
Right
You mistakenly wrote the same solution twice in the end
👍
Right
Cool!
z = -i*ln(π ± √(π² - 1))
Cool channel, post. Im a bit confused. When you use square roots in \sqrt{\pi^2-1}, is this a Real square root, or a Complex square root, where you have to start bringjng up branches and other machinery?
pi^2-1 > 0, hence real root.
@@mikecaetano I meant if we are using a branch of the square root or the standard used with Real numbers.
You found that the product of the two solutions for w is 1, which means that, once you've got the first solution, the other is 1/w. This in turn means that the ln of the second solution is -ln w. Of course, you knew that would have to happen from the start, because you got the formula in the first place by assuming that cos is an odd function.
Cosine is an even function, so why would he assume it was odd?
@@mikecaetano Sorry, I meant even, obviously. If z is a solution at the end, -z is also a solution, because we used this property when extending the function to the complex numbers at the beginning.
z=arccos(pi) 😂
Ahah! Obviously.
Now, you have an expression of acos(pi)