Is There a Complex Number Equal to Its Own Exponential?

[SebastianG]

Hello, everybody!

I'm a Maths/Physics student at Ecuador. Sorry if my English sucks, i'll try to do my best... Some fellow Physics buddy asked me if there was a way to fin a complex number that would be equal to its exponential... it is a very simple question to understand, but not to easy to prove or disprove (at least not for me, and I've tried it for a day).

So, this is what I got:

z = Ln (z)

and we would have to solve these equations:

ln r = rcos O
O = rsen O

that would be hard to solve algebraically, I guess... I wouldn't want a numeric aproach, so maybe there is a more ellegant way to find an example of disprove the hypothesis. (I already tried Taylor expansions)

Well, please try to give some ideas on the subject.

Have a good one!

 

[benorin]

It looks harmless enough, but it's not.

It looks harmless enough, but it's not.

The equation $$exp(z)=z$$ has a solution given by $$z=-W(-1)$$ and, if you put $$z=x+iy$$, then the solution is $$\left\{\begin{array}{cc}x=-iy-W(-1)\\y=y\end{array}\right.$$.

Where $$W(\cdot)$$ is the Lambert W function; for an excellent reference, see

http://mathworld.wolfram.com/LambertW-Function.html

 

[SebastianG]

hmmm...

Hmmm ... Matlab gave me this answer: W(-1) = -0.3181... + i 1.3372...

Which doesn't seem to work. Maybe I made a calculation mistake (because I only typed it once on a computer at my school, and wrote the result), but it seems to make sense, since you can work z = exp (z) to

z = exp (z)
z*exp(-z) = 1
(-z)*exp(-z) = -1

Which indicates that W(-1) should be the value for Z (on the complex form of the Lambert-W function, which I assumed was W(z)*exp[W(z)] = F(z) )

Anyway, this function has been quite a discovery for me! I'll see if I can find an answer, you can still give me some more ideas.

 

[SebastianG]

oops

Nevermind... just a silly mistake...

Z = - W(-1) is the answer.