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Contour integral, residue theorem?

I'm going over a past paper that has this question:

Evaluate

∫C1∣z∣dz\int_C\frac{1}{|z|} dz

Where C is a circle in the complex plane with centre origin and radius 1.

So the only pole of 1∣z∣\dfrac{1}{|z|} is just z=0.

I'm guessing I should use the Cauchy residue theorem but I can't find the residue of 1∣z∣\dfrac{1}{|z|} at z=0.

Or is there another way of evaluating it?
HINT 1: C=eit C = e^{it}

HINT 2: ∫C(t)f(z)dz=∫abf(C(t))⋅C′(t)dt \displaystyle \int_{C(t)} f(z) dz = \int_a^b f(C(t)) \cdot C'(t) dt
(edited 11 years ago)
Reply 2
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jamie092
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Original post by claret_n_blue
HINT 1: C=eit C = e^{it}

HINT 2: ∫C(t)f(z)dz=∫abf(C(t))⋅C′(t)dt \displaystyle \int_{C(t)} f(z) dz = \int_a^b f(C(t)) \cdot C'(t) dt


doh! ok thanks a lot
Reply 3
Avatar for jamie092
jamie092
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Original post by claret_n_blue
HINT 1: C=eit C = e^{it}

HINT 2: ∫C(t)f(z)dz=∫abf(C(t))⋅C′(t)dt \displaystyle \int_{C(t)} f(z) dz = \int_a^b f(C(t)) \cdot C'(t) dt


Although this gives me 0 => residue of that pole =0 which I can't find by CRT =(
Reply 4
Avatar for Slowbro93
Slowbro93
20
Original post by jamie092
Although this gives me 0 => residue of that pole =0 which I can't find by CRT =(


Show all of our working out that you've done so far :smile:
Original post by jamie092
Although this gives me 0 => residue of that pole =0 which I can't find by CRT =(


f(z) = 1/|z| isn't holomorphic (anywhere in C\mathbb{C}, let alone around 0) so you cannot apply Cauchy's residue theorem.
Reply 6
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DFranklin
18
Bigger hint:

Spoiler

Reply 7
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jamie092
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I substituted e^it for z and ended up with e^i2pi - e^0 between 2pi and 0 which is 0.

Wait is that wrong? ;O
Reply 8
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jamie092
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Original post by Glutamic Acid
f(z) = 1/|z| isn't holomorphic (anywhere in C\mathbb{C}, let alone around 0) so you cannot apply Cauchy's residue theorem.


Doh!

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