Why is this function continuous on $ x \in \mathbb{R} $ where $ x < 1 $?

Asked May 15 '13 at 06:43

Active May 15 '13 at 07:37

Viewed 77 times

The function is

$$ \exp(\frac{1}{3}\text{log}[(z-1)(z-2)(z-3)]) $$

where

$$ \log[(z-1)(z-2)(z-3)] = \int_4^z{\frac{[(z-1)(z-2)(z-3)]'}{[(z-1)(z-2)(z-3)]}dz} + \log 6 $$

This function is clearly continuous on $ \mathbb{C} \backslash \{ x \in \mathbb{R} : x < 3 \} $ since this is a simply connected domain and $ [(z-1)(z-2)(z-3)]' $ and $ [(z-1)(z-2)(z-3)] $ don't equal zero anywhere in the domain. I don't see how to to show this is continuous on the real axis for $ x \lt 1 $ though. Any thoughts?

edited May 15 '13 at 07:37

Seirios

33,157

asked May 15 '13 at 06:43

Max

I don't understand your question. Those $x < 1$ are not even part of the domain of your function (the path integral does not give a unique solution)... – Lord_Farin May 15 '13 at 07:47
According to my textbook, this function is continuous on x < 1. I can't seem to find a way to prove or disprove this. – Max May 15 '13 at 07:49
A related problem. – Mhenni Benghorbal May 15 '13 at 08:01
For $Re(z)=x<1$ the function is single-valued and continuous. – Mhenni Benghorbal May 15 '13 at 08:03
I asked that related problem, and I still don't understand any of the answers on the continuity of the function. Perhaps you could elaborate your answer a bit more. – Max May 15 '13 at 08:26
You need to make a correction $ \mathbb{C} \backslash { x \in \mathbb{R} : x < 3 } $. It should be $x<1$. – Mhenni Benghorbal May 15 '13 at 08:31
Was also asked and answered here, with two different answers. – 75064 May 17 '13 at 02:28

Why is this function continuous on $ x \in \mathbb{R} $ where $ x < 1 $?

0 Answers0