Page 164 #2
The function fails to be analytic only at z=0, which is a distance 2 away from z_0. Therefore, the Taylor series will converge on the region 0 < |z-2| < 2. On this region, |(z-2)/2| < 1, so we have
n
1 1 1 1 inf n inf n (z-2)
- = - -------------- = - SUM [-(z-2)/2] = SUM (-1) ------.
z 2 1 - [-(z-2)/2] 2 n=0 n=0 2^(n+1)
Differentiating term by term, which is valid everywhere inside the circle of convergence of a power series, gives
n-1
1 inf n (z-2)
- --- = SUM (-1) n --------
z^2 n=0 2^(n+1)
n-1
inf n (z-2)
= SUM (-1) n -------- (the n=0 term is zero so it
n=1 2^(n+1) can be discarded)
n
inf n+1 (z-2)
= SUM (-1) (n+1) -------- (first re-indexing the sum using
n=0 2^(n+2) k=n-1, then renaming k to n)
n
1 inf n (z-2)
= - - SUM (-1) (n+1) ------.
4 n=0 2^n
Page 164 #5
When z != 0 we have
cz inf n
f(z) = (1/z)(e - 1) = (1/z)( SUM (cz) - 1)
n=0
0 inf n
= (1/z)( -1 + (cz) + SUM (cz) )
n=1
inf n n
= (1/z) SUM c z
n=1
inf n n-1
= SUM c z
n=1
inf n+1 n
= SUM c z
n=0
This sum converges for all non-zero z (because the original sum for e^(cz)
does), and converges when z=0 also, so it converges for all z and
hence converges to an entire function g(z). To show that f(z) is
entire, all we have to do is show that f(z)=g(z) for all z.
We have just seen that
f(z) = g(z) when z != 0. All that remains is to prove that
f(z) = g(z) also when z=0. This follows because by definition
f(0) = c, and when you plug z=0 into the series for g(z) you get
g(0) = c.Page 175 #1
(a) The only singular point is when z=0. To determine the principal part we will find the Laurent series in powers of z. We have
n 1-n
inf (1/z) inf z z 1 1 1 1 1
z exp(1/z) = z SUM ----- = SUM ---- = -- + -- + -- - + -- --- + ...
n=0 n! n=0 n! 0! 1! 2! z 3! z^2
inf 1 1
= z + 1 + SUM ------ --- .
n=1 (n+1)! z^n
The principal part is the part of the sum consisting of negative powers of
z. In this case, the principal part is
inf 1 1
= SUM ------ --- .
n=1 (n+1)! z^n
The principal part contains infinitely many terms, so z=0 is an
essential singular point. (And this wasn't asked, but the residue at z=0
is the coefficient of 1/z in the above sum, which is 1/2!).(b) The only singular point is when z=-1. To determine the principal part we will find the Laurent series in powers of z-(-1) (which is z+1). We have
2 2 2
z [(z+1)-1] (z+1) - 2(z+1) + 1 1
--- = ---------- = ------------------- = (z+1) - 2 + --- .
1+z z+1 z+1 z+1
The principal part is the part of the sum consisting of negative powers of
z+1. In this case, the principal part is 1/(z+1).
The principal part contains only finitely many terms,
with (z+1)^(-1) being the most negative power of z+1 that occurs
(in fact, it happens to be the only power of z+1 that occurs!)
Therefore, z=-1 is a pole of order 1.(c) The only singular point is when z=0. Writing (sinz)/z as a series in powers of z (take the series for sinz and divide it by z) gives 1 - (z^2)/(3!) + (z^4)/(5!) - .... There are no negative powers of z in this series, so the principal part is zero. The point z=0 is a removable singularity. (The formula (sinz)/z does not make sense there. However, if you were to define f(z) to equal 1 when z=0 and to equal (sinz)/z when z is non-zero, f(z) would then be an entire function.)
(d) Proceeding as in part (c), you get (cosz)/z = 1/z - z/(2!) + (z^3)/(4!) - .... The principal part is 1/z. The singularity at z=0 is a pole of order 1.
(e) The only singular point is at z=2. The function is already written as a power series in z-2 (after a change in sign), so the Laurent series is just (-1)/((z-2)^3). This also equals the principal part. The most negative exponent to which z-2 appears is -3, so the singularity at z=0 is a pole of order 3.
Page 175 #4
(a) There is only one singular point (z=0) inside this circle. The circle is a simple closed contour oriented positively. Therefore, the value of the integral is 2 pi i Res_0 exp(-z)/z^2.
You can calculate this residue by finding the Laurent series, as in the previous question (that's the only method the book had covered before these questions were asked). Or, you can do it the easy way: remember the fundamental result that when f(z) = (phi(z))/((z-z_0)^m) and the function phi(z) is analytic and non-zero at z_0, then f(z) has a pole of order m at z_0, and the residue there is phi^(m-1)(z_0)/(m-1)!
In our case we can let phi(z)=exp(-z^2) which is analytic and non-zero at z=0. Then we see that m=2, and the residue is phi'(0)/1! = (-exp(-0))/1 = -1. The integral is therefore -2 pi i.
(b) There is only one singular point (z=0) inside this circle. The circle is a simple closed contour oriented positively. Therefore, the value of the integral is 2 pi i Res_0 z^2 exp(1/z).
This time we have no other method for calculating the residue besides calculating the Laurent series, since the singular point is an essential singular point not a pole. The Laurent series is just z times the Laurent series found in question 1(a), so it is (skipping a few steps, which are the same as in question 1(a))
2 1 inf 1 1
z + z + - + SUM ------ --- .
2 n=1 (n+2)! z^n
The residue at z=0 is the coefficient of 1/z in this series, namely
1/3! = 1/6. The integral is therefore 2pii/6 = pii/3.(c) The integrand is f(z) = (z+1)/(z(z-2)) which is singular at z=0 and at z=2. Both of these singularities are inside the circle, and the circle is a simple closed contour oriented positively. Therefore, the integral equals 2 pii (Res_0 f(z) + Res_2 f(z)).
To calculate Res_0 f(z) we write f(z) = (1/z) phi(z) where phi(z) = (z+1)/(z-2) which is analytic and non-zero at z=0. Thus we see that f(z) has a pole of order 1 at z=0, and the residue there is phi(0)/0! = phi(0) = -1/2.
To calculate Res_2 f(z) we write f(z) = (1/z-2) phi(z) where phi(z) = (z+1)/z which is analytic and non-zero at z=2. Thus we see that f(z) has a pole of order 1 at z=2, and the residue there is phi(2)/0! = phi(2) = 3/2.
The integral is therefore 2pii(-1/2 + 3/2) = 2 pii.
Page 180 #1
(a) The only singular point is z=1. The function can be written as phi(z)/(z-1) where phi(z) = z^2+2 which is analytic and non-zero at z=1. Therefore, z=1 is a pole of order 1. The residue there is phi(0) = 2.
(b) The only singular point is z=-1/2. The function can be written as phi(z)/(z-1/2)^3 where phi(z) = z^3/2^3 = z^3/8 which is analytic and non-zero at z=1/2. Therefore, z=1/2 is a pole of order 3. The residue there is phi"(1/2)/2! = [(6)(-1/2)/8]/2 = -3/16.
(c) Since the function equals (sinhz)/(coshz), the singular points are where coshz = 0. (You can find these points if you like by writing (e^z + e^(-z))/2 = 0, multiplying both sides by 2e^z to get e^(2z) + 1 = 0, so e^(2z) = -1, so z = (1/2) log 1 = (1/2)[pii + 2 n pi i] = (n + 1/2) pi i. However, the question didn't ask you to find the singular points, just to prove they are poles and find out what the residue is at each).
This question is best tackled by the statement at the bottom of page 179, which we didn't have time to explicitly get to in class. The quotient is of the form p(z)/q(z) where p(z)=sinh z and q(z)=coshz. At each singular point z = (n + 1/2) pi i, we have p(z) = i != 0, q(z) = 0, and q'(z) = i != 0. This proves z is a pole of order 1 and the residue there is i/i = 1 (by the argument at the bottom of page 179).
(d) The singular points are where z^2 + pi^2 = 0, i.e., at z = +/- i pi. At z = i pi we can write the function as phi(z)/(z - i pi) where phi(z) = exp(z)/(z + i pi) which is analytic and non-zero at z = i pi. Therefore, z = i pi is a pole of order 1, and the residue there is phi(i pi) = exp(i pi)/(2 i pi) = -1 /(2 i pi) = i/(2 pi).
At z = -i pi we can write the function as phi(z)/(z + i pi) where phi(z) = exp(z)/(z - i pi) which is analytic and non-zero at z = - i pi. Therefore, z = -i pi is a pole of order 1, and the residue there is phi(-i pi) = exp(-i pi)/(-2 i pi) = -1 /(-2 i pi) = -i/(2 pi).
(e) The singular points are where cos z = 0, i.e., where z = (n + 1/2)pi. To show these are poles and to find the residues there, we again use the method of page 179. The function can be written p(z)/q(z) where p(z)=z and q(z)=cos z. At z = (n + 1/2)pi we have p(z) = (n + 1/2)pi which is non-zero, q(z) = 0, and q'(z) = - sin(n + 1/2)pi= (-1)^(n+1) (to see this, note that when n=2k is even we have - sin(n + 1/2)pi= - sin( pi/2 + 2kpi) = - sin(pi/2) = -1, and when n=2k+1 is odd we have - sin(n + 1/2)pi= - sin( pi/2 + pi+ 2kpi) = - sin(3pi/2) = 1.)
Therefore, for each integer n, we see that (n + 1/2)pi is a pole of order 1, and the residue there is (n + 1/2)pi/ (-1)^(n+1).
(f) The singular points are at z=-1 (where the denominator is zero) and at all non-negative real numbers z (where the numerator fails to be analytic; for the chosen branch of z^(1/4) is analytic except at the origin and the positive real axis).
The singular point at z=-1 is isolated. The other singular points are not (since there is a whole ray of them), and hence they cannot be classified as poles or not; that is a notion that applies only to isolated singular points, which are the only kind of singular points for which Laurent series exist. Therefore, the question is incorrect; it should not have asked you to "show that the singular points are poles", but to "show that the isolated singular points are poles".
In this question, the only isolated singular point is z=-1. To show it is a pole, we note that we can write the function as phi(z)/(z-(-1)) where phi(z) is the given branch of z^(1/4), which is analytic and non-zero at z=-1. Thefore, z=-1 is a pole of order 1, and the residue there is phi(-1) = (-1)^(1/4) = exp(1/4 log_(0,2pi) (-1)) = exp(1/4 i pi) = cos (pi)/4 + i sin (pi)/4 = (1 + i)/sqrt(2).
Page 181 #3 (6th edition: page 197 #4)
(a) The integrand f(z) is singular at z=1, z=3i, and z=-3i. Only the first of these (z=1) is inside the circle |z-2|=2; the others are more than a distance 2 away from 2 (| +/- 3i - 2| = sqrt(13) > 2).
Therefore, the integral in question is 2pi i Res_1 f(z).
To calculate the residue we write f(z) = phi(z)/(z-1) where phi(z) = (3z^3 + 2)/(z^2 + 9) which is analytic and non-zero at z=1. Then we see that we have a pole of order 1 and the residue is phi(1) = 1/2, so the integral is (2 pi i)(1/2) = pi i.
(b) This time, all three singular points are inside the contour, so the integral equals 2pi i (Res_1 f(z) + Res_(3i) f(z) + Res_(-3i) f(z)).
To calculate the residue at 3i we write f(z) = phi(z)/(z-3i) where phi(z) = (3z^3 + 2)/((z-1)(z+3i)) which is analytic and non-zero at z=3i. Therefore, we have a pole of order 1 and the residue is
phi(3i) = (-81 i + 2)/(-18 - 6i) = ((-81i + 2)(-18+6i))/(18^2+6^2) = (15 + 49i)/(12).To calculate the residue at -3i we write f(z) = phi(z)/(z+3i) where phi(z) = (3z^3 + 2)/((z-1)(z-3i)) which is analytic and non-zero at z=-3i. Therefore, we have a pole of order 1 and the residue is phi(-3i) = (81 i + 2)/(-18 + 6i) = (15 - 49i)/(12).
The integral is therefore
2 pi i ( 1/2 + (15 + 49i)/(12) + (15 - 49i)/(12) ) = (2 pi i) (3) = 6 pi i.Page 190 #3(a) (6th edition: page 208 #3)
Using the fact that the integrand is even we have
/inf dx /R dx 1 /R dx | ----- = lim | ----- = - lim | ------ . /0 x^4+1 R->inf /0 x^4+1 2 R->inf /-R x^4+1Letting B_R be the contour consisting of the straight line segment from -R to R and C_R the upper semicircle from R to -R, we have
/inf dx 1 / dz
| ----- = - lim | -----
/0 x^4+1 2 R->inf /BR z^4+1
1 / dz / dz
= - lim | ----- - | -----
2 R->inf /BR+CR z^4+1 /CR z^4+1
On C_R we have |z^4+1| >= |z|^4 -1 = R^4 - 1, so |1/(z^4+1)| <= 1/(R^4 - 1), so
| / dz | 1 2 pi R | | ----- | <= (-----) (length of CR) = ------ | /CR z^4+1 | R^4-1 R^4-1which goes to zero as R -> infinity . Therefore,
/inf dx 1 / dz | ----- = - lim | ----- . /0 x^4+1 2 R->inf /BR+CR z^4+1This contour integral is over a closed contour oriented counterclockwise, and so can be evaluated using residues. The integrand f(z) is singular at the fourth roots of -1, namely ( +/- 1 +/- i)/sqrt(2). Only two of these, A = (1+i)/sqrt(2) and B = (-1+i)/sqrt(2), are inside the contour. The integral is therefore 2pii (Res_A f(z) + Res_B f(z)).
The residue at A is found by factoring z^4+1 = (z-A)(z-B)(z-C)(z-D) (with C = (1-i)/sqrt(2) and D = (-1-i)/sqrt(2)), enabling us to write f(z) = phi(z)/(z-A) where phi(z) = 1/(z-B)(z-C)(z-D) which is analytic and non-zero at A. The residue is therefore phi(A) = (2 sqrt(2))/((2)(2i)(2+2i)) = (-1-i)sqrt(2)/8.
The residue at B is found by writing f(z) = phi(z)/(z-B) where phi(z) = 1/(z-A)(z-C)(Z-D), which is analytic and non-zero at B. The residue is therefore phi(B) = (2 sqrt(2))/((-2)(-2+2i)(2i)) = (1-i)sqrt(2)/8.
The contour integral over B_R+C_R is therefore (2 pi i)(((-1-i)sqrt(2))/8 + ((1-i)sqrt(2))/8) = (2 pi i)((-2 i sqrt(2))/8) = pisqrt(2)/2 = pi/sqrt(2).
Our final answer is therefore
/inf dx 1 / dz 1 pi pi | ----- = - lim | ----- = - ------- = ---------- /0 x^4+1 2 R->inf /BR+CR z^4+1 2 sqrt(2) 2 sqrt(2)
Page 190 #2(a) (6th edition: page 208 #4).
The solution begins exactly as in the previous problem. The bound for the integral over C_R is
| / z^2 dz | R^2 2 pi R^3 | | -------------- | <= (--------------) (2 pi R) = ------------ | /CR (z^2+1)(z^2+4) | (R^2-1)(R^2-4) (R^2-1)(R^2-4)which goes to zero as R -> infinity .
To evaluate the contour integral over B_R + C_R, observe that the singularities are +/- i and +/- 2i. Only two of these, i and 2i, are inside the contour. The contour integral therefore equals (2pi i)(Res_i f(z) + Res_(2i) f(z).
For the residue at i, write f(z) = phi(z)/(z-i) where phi(z) =(z^2)/((z+i)(z^2+4)) which is analytic and non-zero at z=i. The residue there is phi(i) = -1/(6i).
For the residue at 2i, write f(z) = phi(z)/(z-2i) where phi(z) =(z^2)/((z^2+1)(z+2i)) which is analytic and non-zero at z=2i. The residue there is phi(2i) = 1/(3i).
The contour integral over B_R+C_R is therefore (2pii)(1/(3i) - 1/(6i)) = pi/3. Our original integral is therefore
/inf x^2 dx 1 / z^2 dz | -------------- = - lim | -------------- = pi/6. /0 (x^2+1)(x^2+4) 2 R->inf /BR+CR (z^2+1)(z^2+4)
Page 190 #5 (6th edition: page 214 #2)
Because the integrand is even,
/inf cos ax /R cos ax 1 /R cos ax
| ------ dx = lim | ------ dx = - lim | ------ dx
/0 x^2+1 R->inf /0 x^2+1 2 R->inf /-R x^2+1
1 / exp(iaz)
= - lim Re [ | -------- dz ]
2 R->inf /BR z^2 + 1
1 / exp(iaz) / exp(iaz)
= - lim Re [ | -------- dz - | --------]
2 R->inf /BR+CR z^2 + 1 /CR z^2+1
On C_R we have |exp(iaz)| = |exp(iax-ay)| = e^(-ay) <= 1
since y >= 0 on C_R and we are assuming a >=0
(if we had a < 0, we
would have let
C_R be the lower semicircle so that y <= 0 on C_R).
Therefore,
| / exp(iaz) | 1 2 pi R | | -------- dz | <= (-----) (2 pi R) = ------ | /CR z^2 + 1 | R^2-1 R^2-1which goes to zero as R -> infinity .
We are left with just the integral over the closed contour B_R + C_R. The integrand is singular at z= +/- i; only the z=i singularity is inside the contour. The integral is therefore 2 pi i Res_i f(z). We find the residue by writing f(z) = phi(z)/(z-i) where phi(z) = exp(iaz)/(z+i) which is analytic and non-zero at i. Then we obtain that the residue at i is phi(i) = exp(-a)/2i, so the contour integral over B_R+C_R is (2 pi i)e^(-a)/(2i) = pie^(-a).
Our final answer for the original real-valued integral is therefore (pi)/2 e^(-a).