1 Errata group 1: Complex integration [correction] z = x + iy dz = - - PowerPoint PPT Presentation

Errata group 1: Complex integration

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Errata group 1: Complex integration [correction]

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Z

L

zdz = Z −1

3

(2 + iy)idy =

• 2iy − y2

2

• −1

3 = −8i + 4

(2, 3) (2, −1)

L

z = x + iy dz = idy

• Theorem. (cf. Stein, Theorem 3.2 ) If a continuous function f has an

antiderivative F and 𝛿 is a curve from w1 to w2 then

Much better approach. Complex integration works exactly as expected.

Z

γ

f(z)dz = F(w2) − F(w1)

= 1 2

• (2 − i)2 − (2 + 3i)2

Z

L

zdz = z2 2

• 2−i

2+3i

= −8i + 4

Errata group 1: Complex integration (continued)

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Errata group 1: Complex integration [improved]

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L1 L3 L4

(−4, 3) (−4, −1) (2, 3) (2, −1)

L2

Ex 1. Integrate f (z) = z on a rectangle

R

b a c d

Z

L1

zdz = z2 2

• b

a

= b2 2 − a2 2 Z

L2

zdz = z2 2

• c

b

= c2 2 − b2 2 Z

L3

zdz = z2 2

• d

c

= d2 2 − c2 2 Z

L4

zdz = z2 2

• a

d

= a2 2 − d2 2 Z

R

zdz = Z

L1+L2+L3+L4

zdz = 0

A new question was posted by Eric Neyman (April 2017). Residues Calculated Incorrectly I believe that every residue in the lectures is calculated incorrectly (off by a sign). The formula given for the residue is $$-\frac{f(\alpha)} {g'(\alpha)}$$, but this is incorrect. It should be $$\frac{f(\alpha)} {g'(\alpha)}$$. The error seems to be traceable back to Slide 59 of the Poles lecture slides, where instead of writing $$\frac{h_{-1}}{z - \alpha} $$, $$\frac{h_{-1}}{\alpha - z}$$ was written. But in fact the way residues are defined in Slide 51 (as in $$\frac{h_{-1}}{z - z_0}$$). I think that the reason that the asymptotics in the examples are right is because the constant $$c$$, which is claimed to be $$\frac{h_{-1}}{\alpha} $$, really should be $$(-1)^M \frac{h_{-1}}{\alpha}$$, and that these two mistakes cancel out.

Errata group 2: Residues

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similar error on p. 256 in book should be z-𝜷 everywhere

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leads to numerous sign errors in later slides

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f(z) rational with a single dominant pole α where β = 1/α and M is the multiplicity of α

Bottom line 1

[zN]f(z) = βNNM−1 (M − 1)!αM lim

z→α(α − z)M f(z)

should be -n-r

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should be 𝜷1-z

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where β = 1/α and M is the multiplicity of α

Bottom line 2

h(z) meromorphic with a single dominant pole α

[zN]h(z) = (−1)MMf(α) αMg(M)(α) βNN M−1