ok now this is an algebra 2 problem... I have goten far down to hear
e^(ix) = (4i +/- 3)/5
now how do I solve for x I feel really stupid now as this is algebra 2...
can't use natural log right or the common log as
ln( e^(ix) )
does not equal ix
so what do I do?
figuered it out
To solve the equation e^(ix) = (4i +/- 3)/5 for x, you can use Euler's formula, which states that e^(ix) = cos(x) + i*sin(x). By equating the real and imaginary parts of the equation, you can find the value of x.
Let's consider the case where e^(ix) = (4i + 3)/5:
1. Equate the real parts:
Re(e^(ix)) = Re((4i + 3)/5)
cos(x) = 3/5
2. Equate the imaginary parts:
Im(e^(ix)) = Im((4i + 3)/5)
sin(x) = 4/5
Now, you have a system of equations with cosine and sine. To find x, you can use inverse trigonometric functions. Specifically, you can use the arccosine function (cos^(-1)) and arcsine function (sin^(-1)).
For the equation cos(x) = 3/5, you can take the arccosine of both sides to solve for x:
x = cos^(-1)(3/5)
Similarly, for the equation sin(x) = 4/5, you can take the arcsine of both sides:
x = sin^(-1)(4/5)
By solving these two equations, you can find the values of x.
Remember, also consider the case where e^(ix) = (4i - 3)/5 by changing the sign, and repeat the same steps to find another possible value of x.