Taylor Series Expansion for z^i at z=1+i: First Three Terms

[xorbie]

I need to find the first three terms of this series.

Am I correct in saying z^i = exp(i*Log(z)), then using the taylor series for e^z, giving me:

(i*Log(z)) - 1/2*(Log(z))^2 - i/6*(Log(z))^3 + 1/24*(Log(z))^4 + ...

I haven't worked it out, but this seems to mean that the coefficients for every term in the Taylor series for z^i is actually an infinite series, found by combining terms in the Log(z) Taylor series.

This seems like I could be way off though, so I don't want to do the calculations if I don't have to.

Thanks in advance.

 

[HallsofIvy]

I do not understand this at all! A "Taylor's series" of a function f(z) is, by definition,
a power series in z. You said in your title that this was to be "at z= i" so your series should have (z- i), (z- i)², etc. The coefficients would be, of course,
f(i), f '(i)/2, etc. which are f(i)= iⁱ, f '(i)= i(i^i-1)= iⁱ,
f"(i)= i(i-1)i^i-2= iⁱ-2i^i-1 etc so that the Taylor's series is iⁱ+ iⁱ(z-i)+ ((iⁱ- 2i^i-1)/2)(z-i)²+ . . .

 

[mathman]

zⁱ=(i+w)ⁱ, where w=z-i
Use the binomial expansion to get
iⁱ+i*i^(i-1)(z-i)+0.5*i(i-1)i^(i-2)(z-i)²

 

[Hurkyl]

Why not just apply the definition of Taylor series?

f(z) = z^i
f'(z) = i z^{-1} z^i
f''(z) = i (i-1) z^{-2} z^i
...

<br /> f(z) = f(i) + f&#039;(i) (z-i) + f&#039;&#039;(i) (z-i)^2 + \cdots<br />
<br /> f(z) = i^i + i i^{-1} i^i (z-i) + i (i-1) i^{-2} i^i (z - i)^2 + \cdots<br />

...

 

[xorbie]

Why not just apply the definition of Taylor series?

f(z) = z^i
f&#039;(z) = i z^{-1} z^i
f&#039;&#039;(z) = i (i-1) z^{-2} z^i
...

<br /> f(z) = f(i) + f&#039;(i) (z-i) + f&#039;&#039;(i) (z-i)^2 + \cdots<br />
<br /> f(z) = i^i + i i^{-1} i^i (z-i) + i (i-1) i^{-2} i^i (z - i)^2 + \cdots<br />

...

Firstly, I made a mistake in my OP. It should be at z = 1 + i, but that shouldn't make a huge difference. The problem is that terms such as (z)^i aren't allowed. We need to restate them as e^i Log(z). So in your last line, it would have to be:

f(z) = e^{i Log(i))} + i e^{i Log(i)} (z- i- 1) +\cdots

Does that sound right? When my professor asks for the "first three terms" is it just the first three of these?

 

[Hurkyl]

i^i isn't a term of the form z^i. And, as you point out, one usually defines i^i = \exp(i \mathop{Log} i), so one doesn't really need to convert. By the way, you can simplify that expression quite a bit!

Anyways, you can't replace (z-i) with (z-i-1) in this Taylor series any more than you can replace z with (z-1) to get the (wrong) Taylor series: e^z = 1 + (z-1) + (z-1)^2/2! + (z-1)^3/3! + \cdots.

 

[xorbie]

Ok, basically I'm clearly being an idiot today. I've just not been thinking this through. The first three terms of the correct Tayler series expansion for z^i about z = 1+i should be (I hope) something along the lines of:

f(z) = e^{i Log(i+1)} + i (1+i)^{-1} e^{i Log(i+1)} (z-1-i) + i (i-1) (1+i)^{-2} e^{i Log(i+1)} (z-1-i)^2

That look ok? I realize it can be simplified, I can do that on my own (hopefully).