# Two complex analysis questions

1. Sep 26, 2007

### strangequark

1. The problem statement, all variables and given/known data

1) Where is $$f(z)=\frac{sin(z)}{z^{3}+1}$$ differentiable? Analytic?

2) Solve the equation $$Log(z)=i\frac{3\pi}{2}$$

2. Relevant equations

none really...

3. The attempt at a solution

For #1 I started out trying to expand this with $$z=x+iy$$, but it got extremely messy... so, I simply said that because $$sin(z)$$ is everywhere analytic, then $$f(z)$$ will only be non-diff'able were $$f'(z)$$ (which I got by simply differentiating wrt z) has poles... ie, at $$z=-1$$, $$z=\frac{1}{2}+i\frac{\sqrt{3}}{2}$$, and $$z=\frac{1}{2}-i\frac{\sqrt{3}}{2}$$.

I find my reasoning a little flimsy, is there something i;m missing?

For #2... this looked easy, I did this:
$$exp(Log(z))=exp(i\frac{3\pi}{2})$$
so...
$$z=-i$$

but if i take $$Log(-i)$$ it's equal to $$-\frac{\pi}{2}$$...
now, this seems like the same thing to me... but my text says no solution... im not sure why?

any help would really be appreciated...

Last edited: Sep 26, 2007
2. Sep 27, 2007

### Dick

On the first one, it's not flimsy at all. sin(z) is entire. The only place things can go wrong is where the denominator vanishes. On the second one it depends entirely on where you put the branch cut for Log(z). Think about it.

3. Sep 27, 2007

### strangequark

ok, so say i define $$0<Arg(z)\leq2\pi$$, then I'm thinking that $$Log(-i)=i\frac{3\pi}{2}$$, as I'm not crossing any branch cuts.... and then on the same note if I define $$-\pi<Arg(z)\leq\pi$$, then $$Log(-i)=i\frac{\pi}{2}$$... is that the right idea?

4. Sep 27, 2007

### Dick

I think so, but you mean Log(-i)=-i*pi/2, right?

5. Sep 27, 2007

### strangequark

yeah, sorry, that's what i meant... ok, i think i'm on track with this one...thanks again!