Two complex analysis questions

[strangequark]

Homework Statement

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

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

Homework Equations

none really...

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... I am not sure why?



any help would really be appreciated...

 

[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.

 

[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?

 

[Dick]

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

 

[strangequark]

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