What are the branch points for \log (z^2 + 2z + 3)?

[Gh0stZA]

Hi everyone,

Could someone please help me calculate the branch points?

Find a branch of [tex]\log (z^2 + 2z + 3)[/tex] that is analytic at -1, and compute the derivative at -1.

 

[HallsofIvy]

The "branch point" of ln(z) itself is z= 0. So you need to solve [itex]z^2+ 2z+ 3= 0[/itex].

 

[Gh0stZA]

The "branch point" of ln(z) itself is z= 0. So you need to solve [itex]z^2+ 2z+ 3= 0[/itex].

I did. My answer is [tex]-1 \pm i[/tex] but Wolfram Alpha gives it as [tex]-1 \pm \sqrt{2} i[/tex]

 

[HallsofIvy]

Well, since you don't say how you got [itex]-1\pm i[/itex], I can only say that Wolfram Alpha is correct.

 

[Gh0stZA]

I'm sorry, I made a stupid mistake with my quadratic formula. I now have the same answer as Wolfram.

So do I basically substitute the values back into the expression within the logarithm? In that case, I get [tex]\log(2-\sqrt{2})[/tex] and [tex]\log(2+\sqrt{2})[/tex]

 

[jackmell]

I'm sorry, I made a stupid mistake with my quadratic formula. I now have the same answer as Wolfram.

So do I basically substitute the values back into the expression within the logarithm? In that case, I get [tex]\log(2-\sqrt{2})[/tex] and [tex]\log(2+\sqrt{2})[/tex]

No. You need do nothing more to identify the branch points and sides, if you back-substituted the zeros of that quadratic back into the quad, you should get zero.