Where is Log(z-2i+1) analytic and differentiable?

[strangequark]

Homework Statement

Find where the function: $$f(z)=Log(z-2i+1)$$ is analytic and where it is differentiable.

Homework Equations

Cauchy-Riemann equations?

The Attempt at a Solution

Here's where I am so far:
$$Log(z-2i+1)=Log((x+1)+i(y-2))=ln(\sqrt{(x+1)^{2}+(y-2)^{2}})+iArg(z)$$

since I'm only looking at the principal value of the logarithm, $$0<\theta\leq2\pi$$ (this is the textbook's choice of principal argument), then
$$ln(\sqrt{(x+1)^{2}+(y-2)^{2}})$$ will be discontinuous at $$x=-1$$ and $$y=2$$... and the function is undefined everywhere on the positive real axis (because of the choice of argument). So, $$f(z)$$ is non-differentiable at $$z=-1+2i$$ and everywhere in the positive direction on the real axis extending from the point $$z=-1$$, because that is the "center" of my mapping.

Now, I'm not sure how to show that it is differentiable everywhere else... Am I supposed to apply the cauchy-riemann equations to $$u=ln(\sqrt{(x+1)^{2}+(y-2)^{2}})$$ and $$v=Arg(z)$$? If so, how do I take a partial derivative of $$Arg(z)$$?

 

[Dick]

Do you know where Log(z) is analytic? I think it's everywhere except along the branch cut you use to make it well defined, right? Then just translate by 2i-1. I don't think you need to get carried away with details.

 

[strangequark]

i was thinking something along the lines of using $$Arg(z)=arctan(\frac{y-2}{x+1})$$ but I'm unsure how to define the inverse tangent... but if I let $$Arg(z)=2arctan(\frac{y-2}{x+1})+C$$, where C is an arbitrary constant, then the cauchy-riemann equations will be satisfied... not sure why yet... my trig is not good...

but anyway, yeah, i like your way better... you're right in that it's probably sufficient for this exercise...

thanks!