Complex variable solvedproblems - Univerzita Karlova

1 Complex variablesolved problemsPavel Pyrih11:03 May 29, 2012( public domain )Contents1 Residue theorem problems22 Zero Sum theorem for residues problems763 Power series problems157 Acknowledgement. The following problems were solved using my own procedurein a program Maple V, release 5. All possible errors are my Residue theorem problemsWe will solve several problems using the following theorem:Theorem.(Residue theorem) SupposeUis a simply connected open subsetof the Complex plane, andw1, .. ,wnare finitely many points ofUand f is afunction which is defined and holomorphic onU\{w1,..,wn}. If is a simplyclosed curve inUcontaning the pointswkin the interior, then f(z)dz= 2 ik k=1res (f,wk).The following rules can be used for residue counting:Theorem.

2 (Rule 1) Iffhas a pole of orderkat the pointwthenres (f,w) =1(k 1)!limz w((z w)kf(z))(k 1).Theorem.(Rule 2) Iff,gare holomorphic at the pointwandf(w)6= 0. Ifg(w) = 0,g (w)6= 0, thenres(fg,w)=f(w)g (w).Theorem.(Rule 3) Ifhis holomorphic atwandghas a simple pole atw, thenres (gh,w) =h(w) res (g,w).2We will use special formulas for special types of problems:Theorem.( TYPE I. Integral from a rational function in sin and cos.) IfQ(a,b) is a rational function of two Complex variables such that for reala,b,a2+b2= 1 isQ(a,b) finite, then the functionT(z) :=Q(z+ 1/z2,z 1/z2i)/(iz)is rational, has no poles on the real line and 2 0Q(cost,sint)dt= 2 i |a|<1,T(a)= res (T,a).Example 2 012 + cos(x)dx3 Theorem.

3 ( TYPE II. Integral from a rational function. ) SupposeP,Qarepolynomial of orderm,nrespectively, andn m >1 andQhas no real for the rational functionf=PQholds + f(x)dx= 2 i kres (f,wk),where all singularities offwith a positive imaginary part are considered in theabove + 11 +x2dx4 Theorem.( TYPE III. Integral from a rational function multiplied by cos orsin ) IfQis a rational function such that has no pole at the real line and forz isQ(z) =O(z 1). Forb >0 denotef(z) =Q(z)eibz. Then + Q(x) cos(bx)dx= Re(2 i wres (f,w)) + Q(x) sin(bx)dx= Im(2 i wres (f,w))where onlywwith a positive imaginary part are considered in the above + cos(x)1 + the Residue theorem evaluate 2 0cos(x)213 + 12 cos(x) denotef(z) = I(12z+121z)2(13 + 6z+ 61z)zWe find singularities[{z= 0},{z= 32},{z= 23}]The singularityz= 0is in our region and we will add the following residueres(0,f(z)) =13144 IThe singularityz= 32will be skipped because the singularity is not in our singularityz= 23is in our region and we will add the following residueres( 23,f(z)) = 169720 IOur sum is2I ( res(z,f(z))) =1345 The solution is 2 0cos(x)213 + 12 cos(x)

4 Dx=1345 6We can try to solve it using real calculus and obtain the result 2 0cos(x)213 + 12 cos(x)dx=1345 the Residue theorem evaluate 2 0cos(x)41 + sin(x) denotef(z) = I(12z+121z)4(1 14(z 1z)2)zWe find singularities[{z= 0},{z= 2+1},{z= 1 2},{z= 2 1},{z= 1 2},{z= },{z= }]The singularityz= 0is in our region and we will add the following residueres(0,f(z)) =52 IThe singularityz= 2 + 1will be skipped because the singularity is not in our singularityz= 1 2is in our region and we will add the following residueres(1 2,f(z)) = 768I 2 + 1088I768 544 2 The singularityz= 2 1is in our region and we will add the following residueres( 2 1,f(z)) = 768I 2 + 1088I768 544 2 The singularityz= 1 28will be skipped because the singularity is not in our singularityz= will be skipped because only residues at finite singularities are singularityz= will be skipped because only residues at finite singularities are sum is2I ( res(z,f(z))) = 2I (52I+ 2 768I 2 + 1088I768 544 2)The solution is 2 0cos(x)41 + sin(x)2dx= 2I (52I+ 2 768I 2 + 1088I768 544 2)We can try to solve it using real calculus and obtain the result 2 0cos(x)41 + sin(x)2dx= (4 2 5)( 2 + 1) ( 2 1) the Residue theorem evaluate 2 0sin(x)254 cos(x) denotef(z) =14I(z 1z)2(54 12z 121z)

5 ZWe find singularities[{z= 0},{z=12},{z= 2}]The singularityz= 0is in our region and we will add the following residueres(0,f(z)) = 54 IThe singularityz=12is in our region and we will add the following residueres(12,f(z)) =34 IThe singularityz= 2will be skipped because the singularity is not in our sum is2I ( res(z,f(z))) = 10 The solution is 2 0sin(x)254 cos(x)dx= We can try to solve it using real calculus and obtain the result 2 0sin(x)254 cos(x)dx= the Residue theorem evaluate 2 01sin(x)2+ 4 cos(x) denotef(z) = I( 14(z 1z)2+ 4 (12z+121z)2)zWe find singularities[{z= 13I 3},{z=13I 3},{z=I 3},{z= I 3}]The singularityz= 13I 3is in our region and we will add the following residueres( 13I 3,f(z)) = 14 IThe singularityz=13I 3is in our region and we will add the following residueres(13I 3,f(z)) = 14 IThe singularityz=I 3will be skipped because the singularity is not in our singularityz= I 3will be skipped because the singularity is not in our sum is2I ( res(z,f(z))) = 12 The solution is 2 01sin(x)2+ 4 cos(x)2dx= We can try to solve it using real calculus and obtain the result 2 01sin(x)2+ 4 cos(x)2dx= the Residue theorem evaluate 2 0113 + 12 sin(x) denotef(z) = I(13 6I(z 1z))zWe find singularities[{z= 23I},{z= 32I}]The singularityz= 23 Iis in our region and we will add the following residueres( 23I,f(z))

6 = 15 IThe singularityz= 32 Iwill be skipped because the singularity is not in our sum is2I ( res(z,f(z))) =25 The solution is 2 0113 + 12 sin(x)dx=25 We can try to solve it using real calculus and obtain the result 2 0113 + 12 sin(x)dx=25 the Residue theorem evaluate 2 012 + cos(x) denotef(z) = I(2 +12z+121z)zWe find singularities[{z= 2 + 3},{z= 2 3}]The singularityz= 2 + 3is in our region and we will add the following residueres( 2 + 3,f(z)) = 13I 3 The singularityz= 2 3will be skipped because the singularity is not in our sum is2I ( res(z,f(z))) =23 3 The solution is 2 012 + cos(x)dx=23 3We can try to solve it using real calculus and obtain the result 2 012 + cos(x)dx=23 the Residue theorem evaluate 2 01(2 + cos(x)) denotef(z) = I(2 +12z+121z)2zWe find singularities[{z= 2 + 3},{z= 2 3}]The singularityz= 2 + 3is in our region and we will add the following residueres( 2 + 3,f(z)) = 13I+13( 23I+13I 3) 3 The singularityz= 2 3will be skipped because the singularity is not in our sum is2I ( res(z,f(z))) = 2I ( 13I+13( 23I+13I 3) 3)The solution is 2 01(2 + cos(x))2dx= 2I ( 13I+13( 23I+13I 3) 3)We can try to solve it using real calculus and obtain the result 2 01(2 + cos(x))2dx=49 the Residue theorem evaluate 2 012 + sin(x) denotef(z) = I(2 12I(z 1z))

7 ZWe find singularities[{z= 2I+I 3},{z= 2I I 3}]The singularityz= 2I+I 3is in our region and we will add the following residueres( 2I+I 3,f(z)) = 13I 3 The singularityz= 2I I 3will be skipped because the singularity is not in our sum is2I ( res(z,f(z))) =23 3 The solution is 2 012 + sin(x)dx=23 3We can try to solve it using real calculus and obtain the result 2 012 + sin(x)dx=23 the Residue theorem evaluate 2 015 + 4 cos(x) denotef(z) = I(5 + 2z+ 21z)zWe find singularities[{z= 2},{z= 12}]The singularityz= 2will be skipped because the singularity is not in our singularityz= 12is in our region and we will add the following residueres( 12,f(z)) = 13 IOur sum is2I ( res(z,f(z))) =23 The solution is 2 015 + 4 cos(x)dx=23 We can try to solve it using real calculus and obtain the result 2 015 + 4 cos(x)dx=23 the Residue theorem evaluate 2 015 + 4 sin(x) denotef(z) = I(5 2I(z 1z))zWe find singularities[{z= 2I},{z= 12I}]The singularityz= 2 Iwill be skipped because the singularity is not in our singularityz= 12 Iis in our region and we will add the following residueres( 12I,f(z)) = 13 IOur sum is2I ( res(z,f(z))) =23 The solution is 2 015 + 4 sin(x)dx=23 We can try to solve it using real calculus and obtain the result 2 015 + 4 sin(x)dx=23 the Residue theorem evaluate 2 0cos(x)54 cos(x) denotef(z) = I(12z+121z)(54 12z 121z)

8 ZWe find singularities[{z= 0},{z=12},{z= 2}]The singularityz= 0is in our region and we will add the following residueres(0,f(z)) =IThe singularityz=12is in our region and we will add the following residueres(12,f(z)) = 53 IThe singularityz= 2will be skipped because the singularity is not in our sum is2I ( res(z,f(z))) =43 The solution is 2 0cos(x)54 cos(x)dx=43 21We can try to solve it using real calculus and obtain the result 2 0cos(x)54 cos(x)dx=43 the Residue theorem evaluate x2+ 1x4+ denotef(z) =z2+ 1z4+ 1We find singularities[{z= 12 2 12I 2},{z=12 2+12I 2},{z= 12 2+12I 2},{z=12 2 12I 2}]The singularityz= 12 2 12I 2will be skipped because the singularity is not in our singularityz=12 2 +12I 2is in our region and we will add the following residueres(12 2 +12I 2,f(z)) =1 +I2I 2 2 2 The singularityz= 12 2 +12I 2is in our region and we will add the following residueres( 12 2 +12I 2,f(z)) =1 I2I 2 + 2 2 The singularityz=12 2 12I 2will be skipped because the singularity is not in our sum is2I ( res(z,f(z))) = 2I (1 +I2I 2 2 2+1 I2I 2 + 2 2)23 The solution is x2+ 1x4+ 1dx= 2I (1 +I2I 2 2 2+1 I2I 2 + 2 2)

9 We can try to solve it using real calculus and obtain the result x2+ 1x4+ 1dx= 2 the Residue theorem evaluate x2 1(x2+ 1) denotef(z) =z2 1(z2+ 1)2We find singularities[{z= I},{z=I}]The singularityz= Iwill be skipped because the singularity is not in our singularityz=Iis in our region and we will add the following residueres(I,f(z)) = 0 Our sum is2I ( res(z,f(z))) = 0 The solution is x2 1(x2+ 1)2dx= 0We can try to solve it using real calculus and obtain the result x2 1(x2+ 1)2dx= the Residue theorem evaluate x2 x+ 2x4+ 10x2+ denotef(z) =z2 z+ 2z4+ 10z2+ 9We find singularities[{z= 3I},{z= 3I},{z= I},{z=I}]The singularityz= 3 Iis in our region and we will add the following residueres(3I,f(z)) =116 748 IThe singularityz= 3 Iwill be skipped because the singularity is not in our singularityz= Iwill be skipped because the singularity is not in our singularityz=Iis in our region and we will add the following residueres(I,f(z)) = 116 116 IOur sum is2I ( res(z,f(z))) =512 26 The solution is x2 x+ 2x4+ 10x2+ 9dx=512 We can try to solve it using real calculus and obtain the result x2 x+ 2x4+ 10x2+ 9dx=512 the Residue theorem evaluate 1(x2+ 1) (x2+ 4) denotef(z) =1(z2+ 1) (z2+ 4)

10 2We find singularities[{z= 2I},{z= I},{z=I},{z= 2I}]The singularityz= 2 Iis in our region and we will add the following residueres(2I,f(z)) =11288 IThe singularityz= Iwill be skipped because the singularity is not in our singularityz=Iis in our region and we will add the following residueres(I,f(z)) = 118 IThe singularityz= 2 Iwill be skipped because the singularity is not in ou