There is a mistake in 6:40 you write cos(z)= (e^iz -e^-iz)/2. But before you write cos (a)= (e^ia+e^-ia)/2 , {+, not minus }, which is the right expression.
@Don-Ensley But symmetry of function cos is for real number, I am not sure that this symmetry holds in imaginary field. I am not a mathematician, only a physicist.
You are correct. He made a mistake and wrote a minus sign in front of the e^(-iz) term. As a consequence, under the square root it says pi^2-1, which avoids the problem of having a negative log, arising from the plus sign he got.
![The most beautiful equation in math, explained visually [Euler’s Formula]](http://i.ytimg.com/vi/f8CXG7dS-D0/mqdefault.jpg)
I love how euler's formula can be derived using Taylor series of each function(e^x, cos(x), and sin(x)).
that's how it was derived in the first place. Infinite series provide the foundation to complex analysis.
Very cool!
There is a mistake in 6:40 you write cos(z)= (e^iz -e^-iz)/2. But before you write cos (a)= (e^ia+e^-ia)/2 , {+, not minus }, which is the right expression.
Not the first there is a mistake in these videos. It has to be fixed.
The correct solution, for n=o, according to my calculations are: z1=-i ln(pi +((pi^2-1)^.5) ; z2= -i ln(pi -((pi^2-1)^.5). Of course I cold be wrong.
@Don-Ensley But symmetry of function cos is for real number, I am not sure that this symmetry holds in imaginary field. I am not a mathematician, only a physicist.
You are correct. He made a mistake and wrote a minus sign in front of the e^(-iz) term. As a consequence, under the square root it says pi^2-1, which avoids the problem of having a negative log, arising from the plus sign he got.
z = -i*ln[π ± √(π²-1)]
I'd like to see a general proof that cos(cos(z)) = cos(w).