# Question integrating |f(z)||dz| over a contour C

1. Feb 28, 2009

### opticaltempest

1. The problem statement, all variables and given/known data

I want to compute $$\int_{C}^{}{|f(z)||dz|}$$ along the contour C given by the curve $$y=x^2$$ using endpoints (0,0) and (1,1). I am to use $$f(z)=e^{i\cdot \texrm{arg}(z)}$$

2. Relevant equations

3. The attempt at a solution
I know that for all complex numbers z, $$|e^{i\cdot \texrm{arg}(z)}}|=1$$. So now I am looking at the integral $$\int_{C}^{}{1|dz|}$$

Is the approach I take below correct?

A complex representation of $$C$$ can be given by $$\gamma (t)=t+it^2$$ for $$0 \leq t \leq 1$$. Then $$\gamma^{'}(t)=1+i2t$$. We have

$$\int_{C}^{}{|f(z)||dz|}$$

$$=\int_{0}^{1}{\gamma^{'}(t)\bigg|\frac{dz}{dt}\bigg||dt|}$$

$$=\int_{0}^{1}{(1+i2t)|dt|}$$

$$=[t+it^2]_{t=0}^{t=1}$$

$$=1+i$$

2. Feb 28, 2009

### Dick

It's a real integral. The result can't have an i in it. dz=d(t+it^2)=dt*(1+i2t). |dz|=dt*|1+i2t|.

3. Feb 28, 2009

### opticaltempest

OK, I should have had $$|\gamma^{'}(t)|$$, instead of $$\gamma^{'}(t)$$.

So the original integral becomes $$\int_{0}^{1}{\sqrt{1+4t^2}}dt$$ or would I have
$$\int_{0}^{1}{\sqrt{1+4t^2} \cdot \sqrt{1+4t^2}dt}$$ ?

Last edited: Feb 28, 2009
4. Feb 28, 2009

### Dick

The first choice looks just fine.

5. Feb 28, 2009

### opticaltempest

Ok, I see why. Thank you very much!