Log of a negative number

1. Nov 12, 2004

Claire84

Hi there, I was hoping someone could check my solutions here to make sure I'm on the right track with these. We've been asked to find ALL values of z satisfying the following eqts -
z= ln(-e^2)

which I did by letting ln (-e^2)= ln|-e^2| + iarg(-e^2)
= e^2 + i(pi + 2m(pi)) where m = 0, +1, -1, +2, -2......

are these all the solutions or am I supposed to go further with this to show ALL the solutions?

I've another problem with ln(z)=1 because I have this = ln|z) +iarg(z) so 1=ln|z) when you compare real and imaginary parts. So then we get e^1=|z| so is z just 1/e? It just doesn't seem right to me, since we only get one solution and the question kinda hints at more than one (or maybe it' just the drink talking here!).

Any help would be much appreciated. Thanks!

2. Nov 12, 2004

NateTG

I don't think you have the second part quite right. I get something different for $$z$$

3. Nov 12, 2004

Gokul43201

Staff Emeritus
$$z = ln(-e^2) = ln(-1) + ln(e^2) = 2 + i(2m+1)\pi$$

That's what you wanted to write, I think ?

And I'm not clear where the 1/e came from, in the second part...

Last edited: Nov 12, 2004
4. Nov 13, 2004

Claire84

Yup, that's what I got for the first one. Was a bit tiddly poo when writing it up tho- sorry! Is that all I have to state for it when it just asks for all the solutions?

In the second one, it wa meant to be ln(z)=-1 (sorry, I'm not drinking again!) so I let -1=ln|z| so then I took exponentials of both sides so I had e^(-1)=|z| so 1/e would equal z. I wasn't sure about having z also equal to -1/e, because since ln(z) = ln|z| + iarg(z) then for z=-1/e then the arg would have to be pi or something.....

5. Nov 13, 2004

Claire84

Oh, and thanks for helping me before.

6. Nov 14, 2004

Claire84

Could someone maybe clarify if what I've done with the second part is okay now? Thanks, I'm just not sure if that's the only solution I can get. I think it is but I'm not 100% sure.