# Homework Help: Complex analysis - Cauchy Theorem

1. Apr 13, 2006

### ElDavidas

Hi again. Can somebody help me out with this question?

"$$\int_{C_1(0)} \frac {e^{z^n + z^{n-1}+.....+ z + 1}} {e^{z^2}} \,dz$$

Where $C_r(p)$ is a circle with centre p and radius r, traced anticlockwise."

I'd be guessing that you have to compare this integral with the Cauchy integral formula. I have no idea how to go about doing this though!

Last edited: Apr 13, 2006
2. Apr 13, 2006

### HallsofIvy

I'm afraid you are going to have to explain what those "zn", "zn-1", etc. mean. Are those powers of z: "zn", "zn-1", etc.?
Oh, and is that "z" supposed to be ez?
That is, do you mean:
$$\int_{C_1(0)} \frac {e^{z^n} + e^{z^{n-1}}+.....+ e^z + 1} {e^{z^2}} \,dz$$
(Click on that to see the code and note, especially my use of "{ }".)

Since, looking at your LaTex code, it is clear that is what you mean, here is how I would do it: Go ahead and divide each term by $e^{z^2}$:
$$\int_{C_1(0)} \left({e^{z^{n-1}} + e^z^{n-3}+.....+ e^{z^{-1}} + e^{-2}}\right)dz$$

Now, which of those, if any, are singular inside C1(0)?

3. Apr 16, 2006

### ElDavidas

I assure you, this is what it says in the question:

$$\int_{C_1(0)} \frac {e^{z^n + z^{n-1}+.....+ z + 1}} {e^{z^2}} \,dz$$

Unless my lecturer made a mistake in printing the question. I doubt it though.

I think you're supposed to look at it as $e^w$ where $w = z^n + z^{n-1}+.....+ z + 1$

I suppose you could use the rule $e^{a+b} = e^a * e^b$ to try and start sorting this out.

Last edited: Apr 16, 2006
4. Apr 16, 2006

### HallsofIvy

Well, okay, then, do what I suggested with that integrand and you get
$$\int_{C_1(0)}\left(e^{z^{n-2}}+ e^{z^{n-3}}+ ...+ ze^{-2z}+ e^{-2z}\right)dz$$
Once again, which of those, if any, are singular inside the unit circle? Those that are not singular (the exponent of e is greater not negative) will give 0 integral. For those that are, you might try writing $z= e^{i\theta}$ and integrating from $\theta= 0 to 2\pi$.

5. Apr 16, 2006

### Hurkyl

Staff Emeritus
No, he gets

$$e^{z^n} \cdot e^{z^{n-1}} \cdot \ldots \cdot e^{z^3} \cdot e^{z^1} \cdot e^{z^0}$$

Last edited: Apr 16, 2006
6. Apr 17, 2006

### HallsofIvy

Hey, just because I can't do arithmetic!

7. Apr 17, 2006

### ElDavidas

Would I be right in saying that the answer is just zero?

I'm guessing this is because all the exponential powers are positive and so Cauchy's Theorem applies?