has anyone noticed that the magnitude of iz + sqrt(1-z^2) = 1^2 + (1-z^2) is 1? And so is mag of e^(iPI/3). Therefore their a's and b's are seperately equal. In particular, their b's: sqrt(...) + iz = cos(PI/3) + isin(PI/3), so z = sin(PI/3) = sqrt(3)/2.
![The most beautiful equation in math, explained visually [Euler’s Formula]](http://i.ytimg.com/vi/f8CXG7dS-D0/mqdefault.jpg)
Alternatively you could do:
ln(iz+sqrt(1-z^2)) = iPI/3
iz+sqrt(1-z^2) = e^(iPI/3)
sqrt(1-z^2) = e^(iPI/3) - iz
1-z^2 = (e^(iPI/3))^2 - 2 e^(iPI/3) iz - z^2
2 e^(iPI/3) iz = (e^(iPI/3))^2 - 1
z = (e^(iPI/3) - e^(-iPI/3))/(2i)
z = sin(PI/3)
z = sqrt(3)/2
My first move - convert ln to e
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Nice!
Thanks!
has anyone noticed that the magnitude of iz + sqrt(1-z^2) = 1^2 + (1-z^2) is 1?
And so is mag of e^(iPI/3). Therefore their a's and b's are seperately equal.
In particular, their b's:
sqrt(...) + iz = cos(PI/3) + isin(PI/3), so z = sin(PI/3) = sqrt(3)/2.
You can set z = sin y and end up getting z = sin pi/3
Ur right
Yup, that's what I immediately thought after seeing the √(1-z²) (cuz it's related to circles)
De moivre and simplify.
I think, there are also two complex solutions.
WA only gives z = ½√3. No complex solutions.