Taylor Series of Log(z) around z=-1+i

[HACR]

Homework Statement

Find the taylor series of Log(z) around z=-1+i.

Homework Equations

The Attempt at a Solution

So I have for the first few terms as
$$\frac{1}{2}*log(2)+\frac{3\pi i}{4}+\frac{z+1-i}{-1+i}-\frac{2(z+1-i)^{2}}{(-1+i)^{2}}+\frac{3(z+1-i)^{3}}{(-1+i)^{3}}-$$

But the correct series seems to be
$$\frac{1}{2}log(2)+\frac{3\pi i}{4}-\Sigma (\frac {-1+i}{2})^{n}*({z+1-i})^{n}$$

 

[Dickfore]

Hint:
$$\frac{d}{d z} \mathrm{Log}(z) = \frac{1}{z}$$

 

[SammyS]

Homework Statement

Find the taylor series of Log(z) around z=-1+i.

Homework Equations

The Attempt at a Solution

So I have for the first few terms as
$$\frac{1}{2}*log(2)+\frac{3\pi i}{4}+\frac{z+1-i}{-1+i}-\frac{2(z+1-i)^{2}}{(-1+i)^{2}}+\frac{3(z+1-i)^{3}}{(-1+i)^{3}}-$$

But the correct series seems to be
$$\frac{1}{2}log(2)+\frac{3\pi i}{4}-\Sigma (\frac {-1+i}{2})^{n}*({z+1-i})^{n}$$

Rationalize the denominator of $\displaystyle \frac{1}{-1+i}\,.$

 

[jackmell]

\frac{1}{2}log(2)+\frac{3\pi i}{4}-\Sigma (\frac {-1+i}{2})^{n}*({z+1-i})^{n}
[/tex]

Had some trouble with this. Sorry if you saw my earlier post but anyway, try and learn how to check these things directly. That's easy: plot Log(z) and your series, say 20 terms or so, the imaginary parts for example. Superimpose them. If they agree, then your series is likely correct. I don't think yours is although you've not stated what the index starts at but you run it if you can to check it like the plot below. Here's the Mathtematica code. See that lil' red thing in there? Yeah, that's what the series should do if it's correct.

pp1 = ParametricPlot3D[
     {Re[z], Im[z], Im[myseries[z]]} /. 
       z -> z0 + r*Exp[I*t], {r, 0, 0.9}, 
     {t, 0, 2*Pi}, PlotStyle -> Red]

pp2 = ParametricPlot3D[
     {Re[z], Im[z], Im[Log[z]]} /. z -> r*Exp[I*t], 
     {r, 0.01, 2}, {t, -Pi, Pi}]

Show[{pp2, pp1}, PlotRange -> All]

 

[vela]

Homework Statement

Find the taylor series of Log(z) around z=-1+i.

Homework Equations

The Attempt at a Solution

So I have for the first few terms as
$$\frac{1}{2}*log(2)+\frac{3\pi i}{4}+\frac{z+1-i}{-1+i}-\frac{2(z+1-i)^{2}}{(-1+i)^{2}}+\frac{3(z+1-i)^{3}}{(-1+i)^{3}}-$$

If you can't figure out your error from the hints above, show us how you got this result.

 

[HACR]

Thanks yes the index 2 and 3s should've been downstairs in fact. and rationalizing the -1+i terms do converge to the right series.