Question

e^y=x+c

Answer 1

question?

Answer 2

I need to solve for y. I know you take the natural log of both sides which gives you y = ln(x+c) or so I thought. The answer says ln(x+c) + 2(i)Pin. I'm not sure where the last part came from

Answer 3

For some reason you are talking about lines in the plane, meaning if you add 2pi to it, you'll continue to get the same line (radian coordinates)

Answer 4

I suppose by \(\ln\) you mean what most people refer to as \(\operatorname{Log}\) i.e. the principal branch of the complex logarithm. As it turns out, the complex exponential is periodic with a period if \(2\pi\), which follows readily from Euler's formula:$$\exp{ix}=\cos x+i\sin x\\\exp z=\exp(x+iy)=\exp x\exp iy=(\exp x)(\cos y+i\sin y)$$

Answer 5

As the complex logarithm is its inverse, it follows that it cannot be a function itself since there are multiply such complex \(z\) that yield the same \(\exp z\); this can be visualized with a Riemann surface as follows:

http://upload.wikimedia.org/wikipedia/commons/4/41/Riemann_surface_log.jpg

Just as for \(\arcsin\cdot\) we limit ourselves to the small domain \(\left(-\dfrac\pi2,\dfrac\pi2\right]\) of \(\sin\cdot\), called the principal branch, for the principal branch of \(\operatorname{Log}\cdot\) we pick that purple section corresponding to complex numbers with arguments in \([0,2\pi)\). This is no different than defining the principal square root \(\sqrt\cdot\) to yield only *non-negative* values, corresponding to only the section \([0,\infty)\) of the domain of \(\cdot^2\).

Just as \(\sin x=1\) yields the infinite real solutions \(x=\dfrac\pi2+2\pi n\) for integral \(n\), \(\exp y=x+c\) yields the infinite complex solutions \(y=\operatorname{Log}(x+c)+2\pi n\) for integral \(n\).

Answer 6

@FutureMathProfessor close but this has to do with functions of complex variables, and often covered in the beginning of a complex analysis class.

Answer 7

:/