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

[opticaltempest]

Homework Statement

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)}

Homework Equations

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

 

[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|.

 

[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} ?

 

[Dick]

The first choice looks just fine.

 

[opticaltempest]

Ok, I see why. Thank you very much!