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

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SUMMARY

The branch points for the function \log(z^2 + 2z + 3) are located at the roots of the quadratic equation z^2 + 2z + 3 = 0, which are -1 ± i. This was confirmed through the quadratic formula, with the correct roots being -1 ± \sqrt{2} i as verified by Wolfram Alpha. To find the branch points, one does not need to substitute these values back into the logarithmic expression, as the zeros of the quadratic already indicate the branch points directly.

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Gh0stZA
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Hi everyone,

Could someone please help me calculate the branch points?

Find a branch of \log (z^2 + 2z + 3) that is analytic at -1, and compute the derivative at -1.
 
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The "branch point" of ln(z) itself is z= 0. So you need to solve z^2+ 2z+ 3= 0.
 
HallsofIvy said:
The "branch point" of ln(z) itself is z= 0. So you need to solve z^2+ 2z+ 3= 0.

I did. My answer is -1 \pm i but Wolfram Alpha gives it as -1 \pm \sqrt{2} i
 
Well, since you don't say how you got -1\pm i, I can only say that Wolfram Alpha is correct.
 
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 \log(2-\sqrt{2}) and \log(2+\sqrt{2})
 
Gh0stZA said:
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 \log(2-\sqrt{2}) and \log(2+\sqrt{2})

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.
 

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