# Taylor series of z^i at z=i

1. Apr 26, 2005

### 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.

2. Apr 26, 2005

### HallsofIvy

Staff Emeritus
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(ii-1)= ii,
f"(i)= i(i-1)ii-2= ii-2ii-1 etc so that the Taylor's series is ii+ ii(z-i)+ ((ii- 2ii-1)/2)(z-i)2+ . . .

3. Apr 26, 2005

### 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

4. Apr 26, 2005

### Hurkyl

Staff Emeritus
Why not just apply the definition of Taylor series?

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

$$f(z) = f(i) + f'(i) (z-i) + f''(i) (z-i)^2 + \cdots$$
$$f(z) = i^i + i i^{-1} i^i (z-i) + i (i-1) i^{-2} i^i (z - i)^2 + \cdots$$

...

5. Apr 26, 2005

### xorbie

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?

6. Apr 26, 2005

### Hurkyl

Staff Emeritus
$i^i$ 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: $e^z = 1 + (z-1) + (z-1)^2/2! + (z-1)^3/3! + \cdots$.

7. Apr 26, 2005

### 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).