742 CHAPTER 36. ISOLATED SINGULARITIES AND ANALYTIC FUNCTIONS

It is natural to try and use the contour in the following picture in which the small circlehas radius r and the large one has radius R.

x

y

However, this will create problems with the log since the usual version of the log is notdefined on the negative real axis. This difficulty may be eliminated by simply using anotherbranch of the logarithm as in Example 36.1.3. Leave out the ray from 0 along the negativey axis and use this example to define L(z) on this set. Thus L(z) = ln |z|+ iarg1 (z) wherearg1 (z) will be the angle θ , between −π

2 and 3π

2 such that z = |z|eiθ . Then the functionused is f (z)≡ L(z)

1+z4 . Now the only singularities contained in this contour are

12

√2+

12

i√

2,−12

√2+

12

i√

2

and the integrand f has simple poles at these points. Thus res(

f , 12

√2+ 1

2 i√

2)=

limz→ 1

2√

2+ 12 i√

2

(z−(

12

√2+ 1

2 i√

2))

(ln |z|+ iarg1 (z))

1+ z4

= limz→ 1

2√

2+ 12 i√

2

(ln |z|+ iarg1 (z))+(

z−(

12

√2+ 1

2 i√

2))

(1/z)

4z3

=

ln(√

12 +

12

)+ i π

4

4(

12

√2+ 1

2 i√

2)3 =

(1

32− 1

32i)√

Similarly

res(

f ,−12

√2+

12

i√

2)=

332

√2π +

332

i√

2π.

Of course it is necessary to consider the integral along the small semicircle of radius r. Thisreduces to ∫ 0

π

ln |r|+ it

1+(reit)4

(rieit)dt

which clearly converges to zero as r→ 0 because r lnr→ 0. Therefore, taking the limit asr→ 0, ∫

large semicircle

L(z)1+ z4 dz+ lim

r→0+

∫ −r

−R

ln(−t)+ iπ1+ t4 dt+

limr→0+

∫ R

r

ln t1+ t4 dt = 2πi

(332

√2π +

332

i√

2π +1

32

√2π− 1

32i√

).

742 CHAPTER 36. ISOLATED SINGULARITIES AND ANALYTIC FUNCTIONSIt is natural to try and use the contour in the following picture in which the small circlehas radius r and the large one has radius R.Xx7However, this will create problems with the log since the usual version of the log is notdefined on the negative real axis. This difficulty may be eliminated by simply using anotherbranch of the logarithm as in Example 36.1.3. Leave out the ray from 0 along the negativey axis and use this example to define L(z) on this set. Thus L(z) = In|z| + iarg, (z) wherearg, (z) will be the angle 6, between —4 and 3% such that z = |z|e’®. Then the functionused is f(z) = Tee Now the only singularities contained in this contour are124 tiv2,-1v24 tia2Ver give aver a!and the integrand f has simple poles at these points. Thus res ( Sf, 5v2 + siv2) =(2— ($v2+ giv2) ) (In|z| + éarg, (2))lim z249524 giVv2 1+z: 1 1:(In |c| +éarg, (2) + (2—($v2+ $iv2)) (1/2)= lim 5zo4hV24+hiv2 Azn(/S+h) 44 14= z= ( _ ') V2n4(jv2+$iv2) 97%?Similarlyres (1. v245iv2) =3 3,3g V 28+ 32! 20.Of course it is necessary to consider the integral along the small semicircle of radius 7. Thisreduces to 0In|jr|+it | .;| Jnlrlie — (rie") dtm 1+(re")which clearly converges to zero as r — 0 because rlnr — 0. Therefore, taking the limit asr—>0,L — In(—t) +i| act lim / DIElarge semicircle 1+24 r30+ J_R 1+tR int 3 3 1 1li —— dt = 2mi | —-V 204+ iV 2474+ —V 20 - =iv27 |}.sim | i+A i( gp V20+ gpiV2a + 55 V2 iV)