# Branch points

1. Aug 25, 2011

### Gh0stZA

Hi everyone,

2. Aug 25, 2011

### HallsofIvy

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

3. Aug 29, 2011

### Gh0stZA

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

4. Aug 29, 2011

### HallsofIvy

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

5. Aug 29, 2011

### 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 $$\log(2-\sqrt{2})$$ and $$\log(2+\sqrt{2})$$

6. Aug 29, 2011

### jackmell

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.