(FCA NN.MM-EE) is exercise EE in section NN.MM of the coursebook ``Fundamentals of Complex Analysis'' by E.B. Saff and A.D. Snider.

Marking scheme: every simple exercise yields 1 point. The total number of points will be rescaled to give at most 10% contribution to the total mark.

1. (FCA 4.5-4) Compute [Using Cauchy integral formula]

ó õ C   z+i z3+2z2 dz,

where C is

(a) the circle |z|=1 traversed once counterclockwise.

(b) the circle |z+2-i|=2 traversed once counterclockwise.

(c) the circle |z-2i|=1 traversed once counterclockwise.

Solution. Notice that we integrate function

f(z)=  z+i z3+2z2 =  z+i z2(z-(-2)) .

Singularities are: pole of order 2 at z=0, simple pole at z=-2. Let C₀ be a positively oriented circle of radius 1/2 around 0, C_-2 be a positively oriented circle of radius 1/2 around -2. Then, by Cauchy formula

1 2pi ó õ C-2   (z+i)/z2 z-(-2) dz = (z+i)/z2|z=-2=  -2+i 4 =-  1 2 +  i 4 ,

1 2pi ó õ C0   (z+i)/(z-(-2)) z2 dz = æ è  z+i z+2 ö ø ¢z=0 = æ è  z+2-(z+i) (z+2)2 ö ø z=0  =  2-i 4 =  1 2 -  i 4 .

Now notice that the domain D of definition of f(z) is the complex plain except points 0 and 2.

(In what follows it is reccomended to draw the contours discussed and singularities of function in complex plain).

(a) This circle C can be continuously deformed in D into C₀. Hence

ó õ C  f(z)dz= ó õ C0  f(z)dz=2pi(  1 2 -  i 4 ).

(b) This circle C can be continuously deformed in D into C_-2. Hence,

ó õ C  f(z)dz= ó õ C-2  f(z)dz=2pi(  1 2 -  i 4 ).

(c) This circle C containes no singularities of f inside and hence can be continuously deformed to a point in D. Hence

ó õ C  f(z)dz=0.

q.e.d.

2. (FCA 4.6-10) Find all functions f analytic in D:|z| < R that satisfy f(0)=i and |f(z)| £ 1 for all z in D [Hint: where does the maximum modulus occur?]

Solution. Notice that 0 Î D and |f(0)|=|i|=i. Hence the maximum of the modulus of f(z) (which by condition of the problem is not greater than 1) is achieved inside domain D. Therefore, by maximum modulus principle analytic f(z) can only be a constant, i.e., f(z₁)=f(z₂) foa all z₁, z₂ Î D. But f(0)=i, hence f(z)=f(0)=i for all z Î i, i.e., f(z)=i is the only analytic function which satisfy conditions of the question. q.e.d..

3. (FCA 4.6-18) Show that max_{|z| £ 1}|azⁿ+b|=|a|+|b|.

Solution 1. Since azⁿ+b is analytic for all z, it is analytic in the unit disk |z| < 1 and continuous on the closed unit disk |z| £ 1. Hence, maximum modulus principle implies that

max |z| £ 1  |azn+b|= max |z|=1  |azn+b|.

Applying parametrization z(t)=e^it, 0 £ t £ 2p, we get

max |z| £ 1  |azn+b| = max 0 £ t £ 2p  |aeint+b|.

What is ae^int+b? This vector can be thought of as a sum of vector a, rotated by angle nt and vector b. By triangle inequality the modulus of this sum is maximal when both vectors point in the same direction, i.e., arg(ae^int)=arg(b). Since the range of nt is [0,2pn], one can obviously find t^* such that arg(ae^nt*)=arg(b). But for t^* we have

|aent*+b|=|a|+|b|.

Hence max_{|z| £ 1}|azⁿ+b|=|a|+|b|.q.e.d.

Solution 2. Using triangle inequality we obtain that for |z| £ 1

|azn+b| £ |azn|+|b|=|a||z|n+|b| £ |a|+|b|.

(1)

Now one can guess that for z^*=e^int* as in previous solution we have

|a (z*)n+b|=|a|+|b|.

(2)

By estimate (1)

max |z| £ 1  |azn+b| £ |a|+|b|.

By equality (2)

max |z| £ 1  |azn+b| ³ |a (z*)n+b|=|a|+|b|.

Therefore max_{|z| £ 1}|azⁿ+b|=|a|+|b|.q.e.d.

4. (FCA 2.5-1) Verify that the real and imaginary parts of the following analytic functions satisfy Laplace's equation [are harmonic functions]

(a) f(z)=z²+2z+1,        (b) g(z)=1/z,        (c) h(z)=e^z.

Solution. (a)

f(x+iy) =(x+iy)2+2(x+iy)+1=x2+2ixy-y2+2x+2iy+1= =(x2-y2+2x+1)+i(2xy+2y)=u(x,y)+iv(x,y).

Notice that

¶2 ¶x2 u(x,y)=2,    ¶2 ¶y2 u(x,y)=-2,

and therefore

¶2 ¶x2 u(x,y)+  ¶2 ¶y2 u(x,y)=2-2=0.

Similiar computations should be done to verify that v(x,y) is harmonic.

(b) Similiar to (a).

(c) Similiar to (a). q.e.d.

5. (FCA 5.1-2) Using the ratio test, show that the following series converge.

(a) å_j=1^¥[ 1/j!],        (b) å_k=1^¥[((3+i)^k)/k!],        (c) å_j=0^¥[(j²)/(4^j)],        (d) å_k=1^¥[ k!/(k^k)].

Solution. (a). In this case a_j=1/j!, hence

aj+1 aj =  1/(j+1)! 1/j! =  j! (j+1)! =  1 j+1 .

Hence

|  aj+1 aj | £  1 j £  1 2 < 1 for all j ³ 1.

Hence the sum converges by ratio test.

(b) In this case a_k=(3+i)^k/k!.

ak+1 ak =  (3+i)k+1/(k+1)! (3+i)k/k! =  k! (k+1)!  (3+i)k+1 (3+i)k =  3+i k+1 .

Therefore

|  ak+1 ak | £  |3+i| k+1 £  |3|+|i| k+1 £  4 k+1 £  4 5 for all k ³ k0=4.

(c) a_j=j²/4^j.

aj+1 aj =  (j+1)2/4j+1 j2/4j =  (j+1)2 j2  4j 4j+1 = æ è 1+  1 j ö ø 2    1 4 .

For j ³ 2 we have

1+  1 j £ 1+  1 2 =  3 2 .

æ è 1+  1 j ö ø 2   £ æ è  3 2 ö ø 2   =  9 4 .

1 4 æ è 1+  1 j ö ø 2   £  9 16 < 1.

Hence to j ³ j₀=2

|  aj+1 aj | £  9 16 < 1.

(d) a_k=k!/k^k.

ak+1 ak =  (k+1)!/(k+1)k+1 k!/kk =  (k+1)! k!  kk (k+1)k+1 =k  kk (k+1)k+1 = =  kk+1 (k+1)k+1 = æ è  k k+1 ö ø k+1   = æ è 1-  1 k+1 ö ø k+1

In order to verify that the last expression is less than some constant, which is less than one, we need to use the following well-known inequality:

Log(1+x) £ x for all x > -1.

Using this inequality, we obtain

ln æ è 1-  1 k+1 ö ø k+1   =(k+1)ln æ è 1-  1 k+1 ö ø £ (k+1)(-  1 k+1 )=-1.

Therefore

æ è 1-  1 k+1 ö ø k+1   £ e-1=  1 e < 1.

for all k ³ 1.q.e.d.

6. (FCA 5.6-2) What is the order of the pole of

f(z)=  1 (2cosz-2+z2)2

at z=0? [Hint: work with 1/f(z)].

Solution. The order of the pole of f(z) equals the doubled order of the zero for g(z)=2cosz-2+z², since f(z)=1/g(z)² and the order of zero for g(z)² is twice the order of zero for g(z). There are two approaches to find the order of zero.

Approach 1. The order of 0 os the order of the first non-zero derivative of g at z=0.

g(0) =2-2+0=0, g¢(0) =(-2sinz+2z)z=0=0, g¢¢(0) =(-2cosz+2)z=0=-2+2=0, g¢¢¢(0) =(2sinz)z=0=0, g¢¢¢¢(0) =(2cosz)z=0=2 ¹ 0.

Hence the order of 0 is 4 and the order of the pole for f(z) at z=0 is 8. q.e.d.

Approach 2. Since we know power series expansion for cosz around z=0, we can simply substitute series into g(z) and the order of 0 would be the order of first term with nonzero coefficient:

g(z)=2(1-  z2 2! +  z4 4! -  z6 6! +¼)-2+z2 =  2z4 4! -  2z6 6! +¼

Hence the order of 0 is 4 and the order of the pole for f(z) at z=0 is 8. q.e.d.

7. (Similar to 6.1-2) Explain why Cauchy's integral formula can be regarded as a special case of the residue theorem, compute residues for the function f(z)=(z+i)/(z³+2z²), and use them to verify results of Problems 1(a), 1(b) and 1(c).

Solution. If f(z) is analytic in simply connected domain D, z₀ lies inside D, then, by Taylor series expansion

f(z)=f(z0)+  f¢(z0) 1! (z-z0)+  f¢¢(z0) 2! (z-z0)2+¼

in some neighbourhood of z₀ and

f(z) z-z0 =  f(z0) z-z0 +  f¢(z0) 1! +  f¢¢(z0) 2! (z-z0)+¼

(3)

in some punctured neighbourhood of z₀. If z₀ lies inside simple closed contour G, belonging to D, then residues theorem states that

ó õ G   f(z) z-z0 dz=2pi a-1,

(4)

where a_-1=f(z₀) is a coefficient of (z-z₀)^-1 in Laurent expansion (3) of f(z)/(z-z₀). To get Cauchy's theorem from (4), one only needs to divide both sides of identity by 2pi. Similarly,

ó õ G   f(z) (z-z0)2 dz=2pi f¢(z0),

and, for general n ³ 0,

ó õ G   f(z) (z-z0)n+1 dz=2pi  f(n)(z0) n! ,

Now, if g(z)=(z+i)/(z²(z+2)) as in problem 2, then

Res(g;0)= lim z®0  æ è z2  z+i z2(z+2) ö ø ¢=  1 2 -  i 4 .

Res(g;-2)= lim z®-2  æ è (z-(-2))  z+i z2(z+2) ö ø =-  1 2 +  i 4 .

Now we can directly apply residues theorem to contours in problems 1(a), 1(b), 1(c)

1(a) ó õ C  g(z)dz =2piRes(g;0) = 2pi æ è  1 2 -  i 4 ö ø =pi +p/2. 1(b) ó õ C  g(z)dz =2piRes(g;-2) = 2pi æ è -  1 2 +  i 4 ö ø =-pi -p/2. 1(b) ó õ C  g(z)dz =0, since no singularities of g lie inside Cq.e.d..

Let us introduce notations which would be common to the following 7 question. We shall need 3 simple contours in complex plain, which depend on positive r > 0:

• g_r - a line segment, connecting -r to r with parametrization z(t)=(1-t)(-r)+tr, t Î [0,1].
• C⁺_r - a semicircle of radius r around 0, which lie in the upper half plain with parametrization z(t)=re^it, t Î [0,p], i.e., it goes from r to -r.
• C^-_r - a semicircle of radius r around 0, which lie in the lower half plain with parametrization z(t)=re^-it, t Î [0,p], i.e., it goes from r to -r.

We also will need combined contours:

Gr=gr+C+r and G-r=gr+C-r.

Notice that G_r is positively oriented, while G^-_r is negatively oriented.

Verify the integral formulas in Problems 8-10 with the aid of residues.

8. (FCA 6.3-2) p.v.ò_-¥^¥ [(x²)/((x²+9)²)]dx=[(p)/6].

Solution. Denote f(z)=[(z²)/((z²+9)²)].

p.v. ó õ ¥ -¥  f(x) dx = lim r®¥  ó õ r -r  f(x) dx = lim r®¥  ó õ gr  f(z)dz= = lim r®¥  ó õ gr  f(z)dz+ ó õ C+r  f(z)dz- ó õ C+r  f(z)dz= = lim r®¥  ó õ Gr  f(z)dz- ó õ C+r  f(z)dz= = ó õ G100  f(z)dz- lim r®¥  ó õ C+r  f(z)dz.

Since f(z)=P(z)/Q(z), where P(z)=z², Q(z)=(z²+9)² and degP+2 £ degQ,

lim r®¥  ó õ C+r  f(z)dz=0.

Notice that the only singularity of f(z) inside G₁₀₀ is at z=3i (in fact, we chosen G₁₀₀ so that all singularities of f in upper half plane were included; deformation theorem ensures us then that the integrals along G_r is the same as integrals along G₁₀₀ for all r > 100).

By residue theorem

ó õ G100  f(z)dz=2pi Res(f(z);3i).

Notice that f(z) has the pole of second order at 3i (since 1/f(z) has the zero of 2nd order at 3i). Hence

Res(f;3i) = lim z® 3i  ((z-3i)2f(z))¢ = lim z®3i  æ è (z-3i)2  z2 (z2+9)2 ö ø ¢ = lim z®3i  æ è  z2 (z+3i)2 ö ø ¢= = lim z®3i   2z(z+3i)-2z2(z+3i) (z+3i)3 = = lim z®3i   2z(z+3i-z) (z+3i)3 = = lim z®3i   6iz (z+3i)3 =  6i×3i (6i)3 =  1 2×6i .

Hence

ó õ G100  f(z)dz=2pi×  1 2×6i =  p 6 .

q.e.d.

9. (FCA 6.3-4) p.v.ò_-¥^¥ [ dx/((x²+1)(x²+4))]=[(p)/6].

Solution. Notice that f(z)=1/((z²+1)(z²+4)) has simple poles at z=i and 2i in the upper halfplain. Using the same argument as we did in the previous Problem 8, we obtain

p.v. ó õ ¥ -¥   dx (x2+1)(x2+4) = ó õ G100   dz (z2+1)(z2+4)        = 2pi(Res(  1 (z2+1)(z2+4) ;i) +Res(  1 (z2+1)(z2+4) ;2i)).

Now

Res(  1 (z2+1)(z2+4) ;i) = lim z® i  (z-i)  1 (z2+1)(z2+4) = lim z® i   1 (z+i)(z2+4 =  1 6i . Res(  1 (z2+1)(z2+4) ;2i) = lim z® i  (z-2i)  1 (z2+1)(z2+4) = lim z® i   1 (z2+1)(z+2i =  1 (-4+1)4i =-  1 12i .

Hence

p.v. ó õ ¥ -¥   dx (x2+1)(x2+4) =2pi (  1 6i -  1 12i )=  p 6 .

q.e.d.

10. (FCA 6.3-6) ò₀^¥[(x²)/((x²+1)(x²+4))]dx=[(p)/6].

Solution. Let f(z)=x²/((x²+1)(x²+4)). Since f(-x)=f(x),

ó õ ¥ 0   x2 (x2+1)(x2+4) dx = ó õ 0 -¥   x2 (x2+1)(x2+4) dx =  1 2 ó õ ¥ -¥   x2 (x2+1)(x2+4) dx

We shall compute the value of the integral in r.h.s. Let f(z)=z²/((z²+1)(z²+4)). Then, using method similiar to the one used in Problem 8, we get

1 2 ó õ ¥ -¥   x2 (x2+1)(x2+4) dx =  1 2 2pi(Res(f;i)+Res(f;2i)).

Res(f;i) = lim z® i  ((z-i)  z2 (z2+1)(z2+4) )= lim z®i   z2 (z+i)(z2+4) =  i2 (2i)(-1+4) =-  1 6i . Res(f;2i) = lim z® 2i  ((z-2i)  z2 (z2+1)(z2+4) )= lim z®2i   z2 (z2+1)(z+2i) =  -4 (-4+1)(4i) =  1 3i .

Hence

ó õ ¥ 0  f(x) dx=pi(-  1 6i +  1 3i =  p 6

q.e.d.

11. (FCA 6.4-2) Using the method of residues, verify the integral formula

ó õ ¥ -¥   xsinx x2-2x +10 dx =  p 3e3 (3cos1+sin1).

Solution. Notice that

ó õ ¥ -¥   xsinx x2-2x +10 dx = Á ó õ ¥ -¥   xeix x2-2x +10 dx.

Let f(z)=ze^iz/(z²-2z+10). Notice that

p.v. ó õ ¥ -¥  f(z) dz = lim r®¥  ó õ Gr  f(z)dz - ó õ Cr+  f(z)dz = ó õ G100  f(z)dz- lim r®¥  ó õ Cr+  f(z)dz.

By Jordan's Lemma,

ó õ Cr+  f(z)dz®0 as r®¥.

Notice that f(z) has singularities where z²-2z+10=0. Using formulas for quadratice equation we find solution z_±=1±3i. Hence z²-2z+10=(z-(1+3i))(z-(1-3i)). The only singularity of f(z) inside G₁₀₀ is 1+3i. Hence

ó õ G100  f(z)dz =2piRes(f;1+3i)= =2pi lim z® 1+3i  (z-(1+3i))  z eiz (z-(1+3i))(z-(1-3i)) = =2pi lim z® 1+3i   z eiz z-(1-3i) = 2pi  (1+3i)ei(1+3i 1+3i-(1-3i) = =2pi  (1+3i)ei-3 6i =  p 3e3 (1+3i)(cos1+isin1) =  p e3 (cos1+3isin1+isin1-3sin1)=I.

It stays to recall that

ó õ ¥ -¥   xsinx x2-2x +10 dx = ÂII =  p 3e3 (3cos1+sin1).

q.e.d.

Compute each of the integrals in Problems 12-14.

12. (FCA 6.4-4) p.v. ò_-¥^¥[(e^3ix)/(x-2i)]dx.

Solution. Denote f(z)=e^3iz/(z-2i). Then

ó õ ¥ -¥  f(x)dx= ó õ G100  f(z)dz - lim r®¥  ó õ C+r  f(z)dz.

By Jordan's Lemma,

lim r®¥  ó õ C+r  f(z)dz=0.

So

ó õ G100  f(z)dz =2piRes(f;2i)= =2pi lim z®2i  ((z-2i)  e3ix x-2i )=2pi e-6 =  2p e6 i,

and the answer is 2pi/e⁶. q.e.d.

13. (FCA 6.4-6) p.v. ò_-¥^¥[(e^-2ix)/(x²+4)]dx.

Solution. Let f(z)=e^-2iz/(z²+4). Then

p.v. ó õ ¥ -¥  f(x)dx = lim r®¥  ó õ G-r  f(z) dz - ó õ C-r  f(z)dz= ó õ G-100  f(z) dz- lim r®¥  ó õ C-r  f(z)dz,

where G^-_r and C^-_r were defined on page pageref. Using Jordan's Lemma for negative exponent we get

lim r®¥  ó õ C-r  e-2iz/(z2+4) dz = 0.

Since G^-₁₀₀ is negatively oriented,

ó õ G-100  f(z) dz =- ó õ -G-100  f(z)dz = -2pi Res(f;-2i) =-2pi lim z®-2i  (z-(-2i))  e-2i (z-2i)(z+2i) = =-2pi lim z®-2i   e-2i z-2i = -2pi  e-2i(-2i) -4i =  p 2e4 .

Hence,

p.v. ó õ ¥ -¥   e-2ix x2+4 dx =  p 2e4 .

q.e.d.

14. (FCA 6.4-8) p.v. ò_-¥^¥[ cosx/((x²+1)(x²+4))]dx.

Solution. Solution is similiar to problems 11 and 12. We have

p.v. ó õ ¥ -¥   cosx (x2+1)(x2+4) dx = Âp.v. ó õ ¥ -¥   eiz (z2+1)(z2+4) dx

Using Jordan's Lemma (or estimate |e^iz| £ 1 for Áz ³ 0), one can reduce the problem to computing residues of f(z)=e^iz/((z²+1)(z²+4)) in the upper halfplane, which are

Res(f;i)=  1 6ie ,

Res(f;2i)=-  1 12ie2 ,

Hence

ó õ ¥ -¥   eiz (z2+1)(z2+4) dx = 2pi(Res(i)+Res(2i))=  p(2e-1) 6e2 .

Hence we enjoy a real number in the r.h.s., the real part of which is itself, and therefore

ó õ ¥ -¥   cosx (x2+1)(x2+4) dx =  p(2e-1) 6e2 .

q.e.d.