Expert: Sherry Wallin - 5/1/2010

Question
QUESTION: hi
so euler says e^(i*pi)+1=0
so ln (e^(i*pi)+1) = -infinity

so I don't understand
why ln (e^(i*pi)+1) = -infinity
can this be proven I mean I see no reason for it to be - infinity

ANSWER: Hi Hamad~
    

ln[e^(i*pi)+1] = ln 0 but since you can't take the ln of 0 you can examine what happens as you get closer and closer to 0 and the ln function of a very small number becomes very negative so you might say the limit is -00.

Math Prof



---------- FOLLOW-UP ----------

QUESTION: ok but I don't understand  e^(i*pi)+1 is a very small number that's close to 0 I mean I was taught that > and < are meaningless in complex numbers

Answer
Hamad~
   Really? You think that e^(ipi)+ 1 is a very small number? Well, it is certainly at least 1, right? And remember that the ln of a number is smaller than the number. When you take the ln of e^(ipi) you would just get ipi which is purely complex right? That +1 is what fouls the whole thing up. Since you have no rules for taking the ln of a sum, all you can do is try to bound it with inequalities. It is true in complex numbers the inequalities have little meaning and that is because you can't order complex numbers so there is no way to bound them. Keep in mind what Euler's really gives you: e^(ipi) = cos pi +isin pi = -1 + 0 = -1 -> e^(ipi) + 1 = 0, hence you are taking the ln of a REAL number, and not a complex number. Instead of looking at the
ln[e^(ipi) + 1] why not look at the ln(cos pi + isin pi + 1)= ln (-1 + 0 + 1) = ln 0 which of course you can't really do but you could examine what happens as you get closer and closer to 0. Does this make more sense?

Math Prof