How to Solve a Complex Integration Problem with a Unit Circle?

[malawi_glenn]

Homework Statement

evaluate

$$\int_{c} | z - 1 | |dz|$$

where c is the positive oriented unit circle.

Homework Equations

The Attempt at a Solution

$$| z - 1 | = \left[ ( z-1)( \overline{z} - 1 ) \right] ^{1/2} = \left[ |z|^{2} - z - \overline{z} +1 \right] ^{1/2}$$

$$c : z(t) = e^{it} ; 0 \leqslant t \leqslant 2\pi$$

$$|dz| = dt$$

$$|z| = 1$$

$$\int_{c} | z - 1 | |dz| = \int_{0} ^{2 \pi} (2 - e^{it} - e^{-it}) ^{1/2} dt$$

$$\int_{0} ^{2 \pi} \sqrt{2-2 \cos t } dt$$

Is this right so far? What to do next? :S

 

[D H]

Use some appropriate half-angle formula.

 

[malawi_glenn]

Use some appropriate half-angle formula.

you mean I should express 2cos(t) as something else?

 

[mjsd]

try $$2 \sin^2 (x) = 1 - \cos (2x)$$

 

[malawi_glenn]

try $$2 \sin^2 (x) = 1 - \cos (2x)$$

aha ok, thanks a lot! cheers

 

[D H]

try $$2 \sin^2 (x) = 1 - \cos (2x)$$

Yep. That's the half-angle formula I was talking about.

$$\sin\left(\frac x 2\right) = \sqrt{1-\cos 2x}$$

 

[malawi_glenn]

Yep. That's the half-angle formula I was talking about.

$$\sin\left(\frac x 2\right) = \sqrt{1-\cos 2x}$$

got the right answer now, thanks a lot !