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Pawel Hitczenko Department of Mathematics Drexel University Philadelphia, PA 19104, U.S.A. email: phitczenko@math.drexel.edu

∗I

would like to thank Frederick Akalin for pointing out a couple of typos.

1

1

Introduction

These notes supplement a freely downloadable book Complex Analysis by George Cain (henceforth referred to as Cain’s notes), that I served as a primary text for an undergraduate level course in complex analysis. Throughout these notes I will make occasional references to results stated in these notes. The aim of my notes is to provide a few examples of applications of the residue theorem. The main goal is to illustrate how this theorem can be used to evaluate various types of integrals of real valued functions of real variable. Following Sec. 10.1 of Cain’s notes, let us recall that if C is a simple, closed contour and f is analytic within the region bounded by C except for ﬁnitely many points z0 , z1 , . . . , zk then k f (z)dz = 2πi

C j=0

Resz=zj f (z),

where Resz=a f (z) is the residue of f at a.

2

2.1

Evaluation of Real-Valued Integrals.

Deﬁnite integrals involving trigonometric functions

2π

We begin by brieﬂy discussing integrals of the form F (sin at, cos bt)dt.

0

(1)

Our method is easily adaptable for integrals over a diﬀerent range, for example between 0 and π or between ±π. Given the form of an integrand in (1) one can reasonably hope that the integral results from the usual parameterization of the unit circle z = eit , 0 ≤ t ≤ 2π. So, let’s try z = eit . Then (see Sec. 3.3 of Cain’s notes), cos bt = z b + 1/z b eibt + e−ibt = , 2 2 sin at = eiat − e−iat z a − 1/z a = . 2i 2i

Moreover, dz = ieit dt, so that dt = Putting all of this into (1) yields

2π

dz . iz

F (sin at, cos bt)dt =

0 C

F

z a − 1/z a z b + 1/z b , 2i 2

dz , iz

where C is the unit circle. This integral is well within what contour integrals are about and we might be able to evaluate it with the aid of the residue theorem.

2

It is a good moment to look at an example. We will show that

2π 0

cos 3t π dt = . 5 − 4 cos t 12

(2)

Following our program, upon making all these substitutions, the integral in (1) becomes (z 3 + 1/z 3 )/2 dz 5 − 4(z + 1/z)/2 iz = 1 i z6 + 1 dz z 3 (10z − 4z 2 − 4) z6 + 1 dz 3 2 C z (2z − 5z + 2) z6 + 1 dz. 3 C z (2z − 1)(z − 2)

C

C

1 = − 2i 1 = − 2i

The integrand has singularities at z0 = 0, z1 = 1/2, and z2 = 2, but since the last one is outside the unit circle we only need to worry about the ﬁrst two. Furthermore, it is clear that z0 = 0 is a pole of order 3 and that z1 = 1/2 is a simple pole. One way of seeing it, is to notice that within a small circle around z0 = 0 (say with radius 1/4) the function z6 + 1 (2z − 1)(z − 2) is analytic and so its Laurent series will have all coeﬃcients corresponding to the negative powers of z zero. Moreover, since its value at z0 = 0 is 06 + 1 1 = , (2 · 0 − 1)(0 − 2) 2 the Laurent expansion of our integrand is z6 + 1 1 1 = 3 3 (2z − 1)(z − 2) z z 1 + a1 z + . . . 2 = 1 1 a1 + 2 + ..., 3 2z z

which implies that z0 = 0 is a pole of order 3. By a similar argument (using a small circle centered at z1 = 1/2) we see that z1 = 1/2 is a simple pole. Hence, the value of integral in (2) is 2πi Resz=0 z6 + 1 − 1)(z − 2) + Resz=1/2 z6 + 1 − 1)(z − 2) .

z 3 (2z

z 3 (2z

The residue at a simple pole z1 = 1/2 is easy to compute by following a discussion preceding the second example in Sec. 10.2 in Cain’s notes: z6 + 1 z6 + 1 z6 + 1 = 3 2 = 5 z 3 (2z − 1)(z − 2) z (2z − 5z + 2) 2z − 5z 4 + 2z 3

3

is of the form p(z)/q(z) with p(1/2) = 2−6 + 1 = 0 and q(1/2) = 0. Now, q (z) = 10z 4 − 20z 3 + 6z 2 , so that q (1/2) = 10/24 − 20/23 + 6/22 = −3/23 . Hence, the residue at z1 = 1/2 is p(1/2) (26 + 1) · 23 65 =− =− . q (1/2) 26 · 3 24 The residue at a pole of degree 3, z0 = 0, can be obtained in various ways. First, we can take a one step further a method we used to determine the degree of that pole: since on a small circle around 0, z6 1 z6 + 1 = + . (2z − 1)(z − 2) (2z − 1)(z − 2) (2z − 1)(z − 2) (3)

is analytic, the residue of our function will be the coeﬃcient corresponding to z 2 of Taylor expansion of the function given in (3). The ﬁrst term will not contribute as its smallest non-zero coeﬃcient is in front of z 6 so we need to worry about the second term only. Expand each of the terms 1/(2z − 1) and 1/(z − 2) into its Taylor series, and multiply out. As long as |2z| < 1 we get 1 (2z − 1)(z − 2) = = = = = 1 2 1 2 1 2 1 2 1 2 · 1 1 · 1 − 2z 1 − z/2 1 + 2z + 22 z 2 + . . . · 1 + z z2 + 2 + ... 2 2

1 1 1 + A1 z + 2 · z 2 + z 2 + 4z 2 + . . . 2 4 1 1 + A1 z + 5 + z2 + . . . 4 a1 21 + z + z2 + . . . 2 8

so that the residue at z0 = 0 is 21/8 (from the calculations we see that A1 = 2 + 1/2, but since our interest is the coeﬃcient in front of z 2 we chose to leave A1 unspeciﬁed). The same result can be obtained by computing the second derivative (see Sec. 10.2 of Cain’s notes) of z6 + 1 1 3 z 3 , 2! z (2z − 1)(z − 2) and evaluating at z = 0. Alternatively, one can open Maple session and type: residue((z^6+1)/z^3/(2*z-1)/(z-2),z=0); to get the same answer again. Combining all of this we get that the integral in (2) is − 1 2i z6 + 1 1 dz = − (2πi) z 3 (2z − 1)(z − 2) 2i 21 65 − 8 24 =π 65 − 63 π = , 24 12

C

as required. 4

2.2

Evaluation of improper integrals involving rational functions

∞

Recall that improper integral f (x)dx

0

is deﬁned as a limit

R→∞

R

lim

f (x)dx,

0

provided that this limit exists. When the function f (x) is even (i.e. f (x) = f (−x), for x ∈ R) one has

R

f (x)dx =

0

1 2

R

f (x)dx,

−R

and the above integral can be thought of as an integral over a part of a contour CR consisting of a line segment along the real axis between −R and R. The general idea is to “close”the contour (often by using one of the semi-circles with radius R centered at the origin), evaluate the resulting integral by means of residue theorem, and show that the integral over the “added”part of CR asymptotically vanishes as R → 0. As an example we will show that

∞ 0

dx π = . (x2 + 1)2 4

(4)

Consider a function f (z) = 1/(z 2 + 1)2 . This function is not analytic at z0 = i (and that is the only singularity of f (z)), so its integral over any contour encircling i can be evaluated by residue theorem. Consider CR consisting of the line segment along the real axis between −R ≤ x ≤ R and the upper semi-circle AR := {z = Reit , 0 ≤ t ≤ π}. By the residue theorem dz = 2πiResz=i (z 2 + 1)2 1 (z 2 + 1)2 .

CR

The integral on the left can be written as

R −R

dz + (z 2 + 1)2

AR

dz . (z 2 + 1)2

Parameterization of the line segment is γ(t) = t + i · 0, so that the ﬁrst integral is just R R dx dx =2 . 2 + 1)2 2 + 1)2 −R (x 0 (x Hence,

R 0

dx = πiResz=i (x2 + 1)2

1 (z 2 + 1)2 5

−

1 2

AR

dz . (z 2 + 1)2

(5)

Since

1 1 = , (z 2 + 1)2 (z − i)2 (z + i)2

and 1/(z + i)2 is analytic on the upper half-plane, z = i is a pole of order 2. Thus (see Sec. 10.2 of Cain’s notes), the residue is d dz (z − i)2 1 (z − i)2 (z + i)2 z=i =

−2 (z + i)3

=− z=i 2 1 1 =− 3 = 3 (2i) 4i 4i

which implies that the ﬁrst term on the right-hand side of (5) is πi π = . 4i 4 Thus the evaluation of (4) will be complete once we show that lim dz = 0. (z 2 + 1)2 (6)

R→∞

AR

But this is straightforward; for z ∈ AR we have |z 2 + 1| ≥ |z|2 − 1 = R2 − 1, so that for R > 2 1 1 ≤ . (z 2 + 1)2 (R2 − 1)2 Using our favorite inequality g(z)dz ≤ M · length(C),

C

(7)

where |g(z)| ≤ M for z ∈ C, and observing that length(AR ) = πR we obtain dz πR ≤ −→ 0, (z 2 + 1)2 (R2 − 1)2

AR

as R → ∞. This proves (6) and thus also (4).

2.3

Improper integrals involving trigonometric and rational functions.

Integrals like one we just considered may be “spiced up”to allow us to handle an apparently more complicated integrals with very little extra eﬀort. We will illustrate it by showing that

∞ −∞

cos 3x 2π dx = 3 . 2 + 1)2 (x e

6

We keep the same function 1/(x2 + 1)2 , just to illustrate the main diﬀerence. This time we consider the function e3iz f (z), where f (z) is, as before 1/(z 2 +1)2 . By following the same route we are led to

R −R

e3xi dx = 2πiResz=i (f (z)e3zi − (x2 + 1)2

f (z)e3zi dz =

AR

2π − e3

f (z)e3zi dz.

AR

At this point it only remains to use the fact that the real parts of both sides must the same, write e3xi = cos 3x + i sin 3x, and observe that the real part of the left-hand side is exactly the integral we are seeking. All we need to do now is to show that the real part of the integral over AR vanishes as R → ∞. But, since for a complex number w, |Re(w)| ≤ |w| we have Re

AR

f (z)e3iz dz

≤

AR

f (z)e3iz dz ,

and the same bound as in (7) can be established since on AR we have |e3iz | = |e3i(x+iy) | = |e−3y e3xi | = e−3y ≤ 1, since AR is in the upper half-plane so that y ≥ 0. Remark. (i) This approach generally works for many integrals of the form

∞ ∞

f (z) cos azdz

−∞

and

−∞

f (z) sin azdz,

where f (z) is a rational function (ratio of two polynomials), where degree of a polynomial in the denominator is larger than that of a polynomial in the numerator (although in some cases working out the bound on the integral over AR may be more tricky than in the above example). The following inequality, known as Jordan’s inequality, is often helpful (see Sec. 2.4 for an illustration as well as a proof) π e−R sin t dt <

0

π . R

(8)

(ii) The integrals involving sine rather than cosine are generally handled by comparing the imaginary parts. The example we considered would give

∞ −∞

sin 3x dx = 0, (x2 + 1)2

but that is trivially true since the integrand is an odd function, and, clearly, the improper integral

∞ 0

sin 3x dx (x2 + 1)2

converges. For a more meaningful examples see Exercises 4-6 at the end. 7

2.4

One more example of the same type.

∞

sin x π dx = . (9) x 2 0 This integral is quite useful (e.g. in Fourier analysis and probability) and has an interesting twist, namely a choice of a contour (that aspect is, by the way, one more thing to keep in mind; clever choice of a contour may make wonders). Based on our knowledge form the previous section we should consider f (z) = eiz /z and try to integrate along our usual contour CR considered in Sections 2.2 and 2.3. The only problem is that our function f (z) has a singularity on the contour, namely at z = 0. To avoid that problem we will make a small detour around z = 0. Speciﬁcally, consider pick a (small) ρ > 0 and consider a contour consisting of the arc AR that we considered in the last two sections followed by a line segment along a real axis between −R and −ρ, followed by an upper semi-circle centered at 0 with radius ρ and, ﬁnally, a line segment along the positive part of the real exit from ρ to R (draw a picture to see what happens). We will call this contour BR,ρ and we denote the line segments by L− and L+ , respectively and the small semi-circle by Aρ . Since our function is analytic with BR,ρ its integral along this contour is 0. That is eiz dz = 0 + z

−ρ −R −ρ −R

Here we will show that

AR

eix dx + x

Aρ R

eiz dz + z

R ρ

eix dx, x

Since

eix dx = − x

ρ

e−ix dx, x eiz dz = 0. z

by combining the second and the fourth integral we obtain

R ρ

eix − e−ix dx + x

AR

eiz dz + z

Aρ

Since the integrand in the leftmost integral is 2i we obtain 2i ρ R

sin x eix − e−ix = 2i , 2ix x eiz dz − z eiz dz = 0, z

sin x dx = − x

AR

Aρ

or upon dividing by 2i

R ρ

i sin x dx = x 2

AR

eiz dz + z

Aρ

eiz dz z

.

The plan now is to show that the integral over AR vanishes as R → ∞ and that ρ→0 lim

Aρ

eiz dz = −πi. z 8

(10)

To justify this last statement use the usual parameterization of Aρ : z = ρeit , 0 ≤ t ≤ π. Then dz = iρeit dt and notice (look at your picture) that the integral over Aρ is supposed to be clockwise (i.e. in the negative direction. Hence, −

Aρ

eiz dz = z

π 0

eiρe iρeit dt = i ρeit

it

π 0

eiρe dt.

it

Thus to show (10) it suﬃces to show that π ρ→0

lim

eiρe dt = π.

0

it

To this end let us look at π 0

eiρe dt − π =

0

it

π

eiρe dt −

0

it

π

π

dt =

0

eiρe − 1 dt .

it

We will want to use once again inequality (7). Since the length of the curve is π we only need to show that |eiρe − 1| −→ 0, But we have |eiρe − 1| = = ≤ ≤ it it

as

ρ → 0.

(11)

|eiρ(cos t+i sin t) − 1| = |e−ρ sin t eiρ cos t − 1| |e−ρ sin t eiρ cos t − e−ρ sin t + e−ρ sin t − 1| e−ρ sin t |eiρ cos t − 1| + |e−ρ sin t − 1| |eiρ cos t − 1| + |e−ρ sin t − 1|.

For 0 ≤ t ≤ π, sin t ≥ 0 so that e−ρ sin t ≤ 1 and thus the second absolute value is actually equal to 1 − e−ρ sin t which is less than ρ sin t, since for any real u, 1 − u ≤ e−u (just draw the graphs of these two functions). We will now bound |eiρ cos t − 1|. For any real number v we have |eiv − 1|2 = (cos v − 1)2 + sin2 v = cos2 v + sin2 v − 2 cos v + 1 = 2(1 − cos v). v2 2 If we knew that, then (11) would follow since we would have 1 − cos v ≤ it We will show that

(12)

|eiρe − 1| ≤ |eiρ cos t − 1| + |e−ρ sin t − 1| ≤ ρ(| cos t| + | sin t|) ≤ 2ρ → 0. But the proof of (12) is an easy exercise in calculus: let h(v) := v2 + cos v − 1. 2 9

Than (12) is equivalent to h(v) ≥ 0 for v ∈ R.

Since h(0) = 0, it suﬃces to show that h(v) has a global minimum at v = 0. But h (v) = v − sin v so that h (0) = 0, and h (v) = 1 − cos v ≥ 0.

That means that v = 0 is a critical point and h(v) is convex. Thus, it has to be the mininimum of h(v). All that remains to show is that

R→∞

lim

AR

eiz dz = 0. z

(13)

This is the place where Jordan’s inequality (8) comes into picture. Going once again through the routine of changing variables to z = Reit , 0 ≤ t ≤ π, we obtain eiz dz z π = =

i

0 π 0

eiRe dt ≤

0

it

π

eiR(cos t+i sin t) dt π AR

eiR cos t · e−R sin t dt ≤

0

e−R sin t dt ≤

π , R

where the last step follows from Jordan’s inequality (8). It is now clear that (13) is true. It remains to prove (8). To this end, ﬁrst note that π π/2

e−R sin t dt = 2

0 0

e−R sin t dt.

That’s because the graph of sin t and (thus also of e−R sin t ) is symmetric about the vertical line at π/2. Now, since the function sin t is concave between 0 ≤ t ≤ π, its graph is above the graph of a line segment joining (0, 0) and (π/2, 1); in other words 2t sin t ≥ . π Hence π/2 π/2 2R π π 1 − e−R ≤ e−R sin t dt ≤ e− π t dt = 2R 2R 0 0 which proves Jordan’s inequality.

3

Summation of series. π2 1 = , n2 6 n=1 10

∞

As an example we will show that (14)

but as we will see the approach is fairly universal. Let for a natural number N CN be a positively oriented square with vertices at (±1 ± i)(N + 1/2) and consider the integral cos πz dz. z 2 sin πz (15)

CN

Since sin πz = 0 for z = 0, ±1, ±2, . . . the integrand has simple poles at ±1, ±2, . . . and a pole of degree three at 0. The residues at the simple poles are lim (z − k) cos πz π(z − k) cos πk 1 lim . = = z 2 sin πz k 2 z→k π sin(π(z − k)) πk 2

z→k

The residue at the triple pole z = 0 is −π/3. To see that you can either ask Maple to ﬁnd it for you by typing: residue(cos(Pi*z)/sin(Pi*z)/z^2,z=0); in your Maple session or compute the second derivative (see Sec. 10.2 of Cain’s notes again) of 1 z cos πz 1 3 cos πz z = , 2! z 2 sin πz 2 sin πz and evaluate at z = 0, or else use the series expansions for the sine and cosine, and ﬁgure out from there, what’ s the coeﬃcient in front of 1/z in the Laurent series for cos πz z 2 sin πz around z0 = 0. Applying residue theorem, and collecting the poles within the contour CN we get π cos πz dz = 2πi − + 2 sin πz z 3

N

CN

k=1

1 + πk 2

−N

k=−1

1 πk 2

= 2πi −

π +2 3

N

k=1

1 πk 2

.

The next step is to show that as N → ∞ the integral on the left converges to 0. Once this is accomplished, we could pass to the limit on the right hand side as well, obtaining N 2 1 π lim = , 2 N →∞ π k 3 k=1 which is equivalent to (14). In order to show that the integral in (15) converges to 0 as N → ∞, we will bound cos πz dz z 2 sin πz CN using our indispensable inequality (7).

11

First, observe that each side of CN has length 2N + 1 so that the length of the contour CN is bounded by 8N + 4 = O(N ). On the other hand, for z ∈ CN , |z| ≥ N + so that 1 ≥ N, 2

1 1 ≤ 2, 2 z N

so that it would be (more than) enough to show that cos πz/ sin πz is bounded on CN by a constant independent of N . By Exercise 9 from Cain’s notes | sin z|2 = sin2 x + sinh2 y, where z = x + iy. Hence cos πz sin πz

2

| cos z|2 = cos2 x + sinh2 y,

= ≤

cos2 πx + sinh2 πy cos2 πx sinh2 πy = + sin2 πx + sinh2 πy sin2 πx + sinh2 πy sin2 πx + sinh2 πy cos2 πx + 1. 2 sin πx + sinh2 πy

On the vertical lines of the contour CN x = ±(N + 1/2) so that cos(πx) = 0 and sin(πx) = ±1. Hence the ﬁrst term above is 0. We will show that on the horizontal lines the absolute value of sinh πy is exponentially large, thus making the ﬁrst term exponentially small since sine and cosine are bounded functions. For t > 0 we have et − e−t et − 1 sinh t = ≥ . 2 2 Similarly, for t < 0 sinh t = so that, in either case e|t| − 1 . 2 Since on the upper horizontal line |y| = N + 1/2 we obtain | sinh t| ≥ | sinh πy| ≥ eπ(N +1/2) − 1 → ∞, 2 et − e−t 1 − e−t e−t − 1 ≤ =− , 2 2 2

as N → ∞. All of this implies that there exists a positive constant K such that for all N ≥ 1 and all for z ∈ CN cos πz | | ≤ K. sin πz Hence, we conclude that cos πz 8N + 4 dz ≤ K , z 2 sin πz N2 12

CN

which converges to 0 as N → ∞. This establishes (14). The above argument is fairly universal and applies generally to the sums of the form

∞

f (n). n=−∞ Sums from 0 (or 1) to inﬁnity are then often handled by using a symmetry of f (n) or similarly simple observations. The crux of the argument is the following observation: Proposition 3.1 Under mild assumptions on f (z),

∞

f (n) = −π n=−∞ Res f (z)

cos πz , sin πz

where the sum extends over the poles of f (z). In our example we just took f (z) = 1/z 2 . Variations of the above argument allow us to handle other sums. For example, we have Proposition 3.2 Under mild assumptions on f (z),

∞

(−1)n f (n) = −π n=−∞ Res

f (z) sin πz

,

where the sum extends over the poles of f (z).

13

4

Exercises.

∞

1. Evaluate

0 ∞

2x2 − 1 dx. x4 + 5x2 + 4 x dx. (x2 + 1)(x2 + 2x + 2) x4 dx . +1 sin x dx. + 4x + 5

Answer: π/4. Answer: −π/5. Answer: Answer: − Answer: π √ . 2 2

2. Evaluate

−∞ ∞

3. Evaluate

0 ∞

4. Evaluate

−∞ ∞

x2

π sin 2. e

5. Evaluate

−∞ ∞

x sin x dx. x4 + 4 x sin πx dx. x2 + 2x + 5 dt . 1 + sin2 t cos2 3x dx. 5 − 4 cos 2x

π sin 1. 2e

6. Evaluate

−∞ π

Answer: −πe−2π . Answer: √ 2π.

7. Evaluate

−π 2π

8. Evaluate

0 ∞

Answer:

3 π. 8

9. Show that

1 π4 = . n4 90 n=1 (−1)n+1 π2 = . 2 n 12 n=1

∞

10. Use a suggestion of Proposition 3.2 to show that

14…...

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...Holistic Account of Care for a Patient or Client with Complex Needs What are complex needs? Who has them? It could be argued that everyone's needs are complex and as such, complex needs have been difficult to define (Rosengard et al., 2007). This essay intends to use literature to outline a systematic and holistic approach to care of a patient's complex needs. The nursing process and its framework of assessment, planning, implementation and evaluation will be utilised to provide a patient centred approach (Castledine, 2011). Rankin and Regan (2004) described complex needs as being a framework rather than a label that determines service eligibility. The needs of people include breadth (range of need) and depth (severity); these factors have to be taken into account when providing holistic care. Holistic care is defined as an approach which takes into consideration the physical, mental and social factors in an illness, rather than just the disease itself (Martin, 2010). The person this essay will centre around gave informed consent to have their notes consulted with the intention to write this essay, in line with the Nursing and Midwifery Council's (NMC) The Code (2008). There are seven elements to informed consent; competence to decide, voluntariness to decide, disclosure of information, recommendation of a plan, understanding of the disclosure and recommendation, decision of the plan and authorising of the plan (Beauchamp & Childress, 1994 pp. 145-146). In line with......

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...Complex Interdependence In 1970s, a new perspective known as transnational relations arose to question realism key assumptions of state as main and the only actor of world politics. This theory gave a new analytical framework to understand the international relations which was contrary to the state centric paradigm of realism. With complex interdependence as its central concept, this theoretical analysis advanced “synthesis of liberal + realist perspectives by constructing a way of looking at world politics and also the idea of institutionalism to foster cooperation”. It is asked how under conditions of complex interdependence world politics would be different than under realist conditions means this analysis ask to see “ what realism overlooked”. The phenomenon of Interdependence is not new; states have always been interdependent on each other for their economic functions and especially for security purposes. We have seen the complicated interdependence among states in the alliance system to achieve their political ends before First World War. The earliest example of interdependence comes from Norman Angell’s “The great illusion (1910) when he said the economic interdependence will minimize the occurrence of war on the basis of cost-benefit analysis.” Countries are dependent on each other for trade and transaction, they interact globally e.g. flow of money, goods, people and messages across international boundaries but all transactions among countries are not characterized......

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...A PASTORAL APPROACH TO THE PROBLEM OF AN INFERIORITY COMPLEX FOR WORKERS IN THE KOREAN CONTEXT BY EUN-MI, HAN THE DEGREE OF MASTER IN THEOLOGY (PRACTICAL THEOLOGY) STUDY-LEADER: PROF. JOHAN J VAN RENSBURG UNIVERSITY OF THE FREE STATE FACULTY OF THEOLOGY DECEMBER 2011 1 CONTENTS INTRODUCTION ....................................................................... 4 1. Introduction ................................................................................................................................. 4 2. Research Problem...................................................................................................................... 5 3. Research Hypothesis ................................................................................................................ 8 4. Research Objectives................................................................................................................11 5. Research Methodology .........................................................................................................13 5.1 The General Psychological Approach ...................................................................13 5.2 The Biblical Principle Approach...............................................................................14 5.3 The Systematic Theology Approach ......................................................................14 5.4 The Pastoral Approach ........................................................................

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...John J. Hawkins Rapp 630: N. Africa and the Middle East Dr. Johnathan Pidlunzy A complex man Ruhollah Mostafavi Moosavi or ayatollah Khomeini for short. His name brings to mind a different man in every mind. To those who are aware of who or what he was they just knew that he had control over Iran as its leader till his death. They didn’t know that his title wasn’t president but was supreme leader something that echoes of the Soviet Era although Iran was not part of the USSR. To some a feeling dread and fear come to life. After taking power in Iran Khomeini ordered the death of thoughts using mostly the Quran as his justification finding something about that them that didn’t support the revolution the he started. In 1979 Shah Reza Palavi fled Iran to America for that they may have better health care for the shah who was dying cancer. In 1979 protests and anger filled the streets I Iran. Students mostly took the streets in anger that the United States would give refuge in their nation. This was a perfect time for Rohollah Khomenini to return to Iran and that’s what he exactly did. The Ayatollah didn’t use the Quran as a source of peace or general calmness but instead that Ayatollah felt that the Quran could be used for something much greater, forming a pure Islamic political state. On the command of the ayatollah the United States embassy was stormed and prisoners taken for more than a year. Eventually those captives were taken out of Iran by a CIA......

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...Exercise 31.6: Identify each sentence simple, compound, complex, or compound-complex. 1. Rock and roll originated in the 1950’s - Simple 2. Chuck Berry, Jerry Lee Lewis, and Elvis Presley were early rock-and-roll greats. - Simple 3. Teenagers loved the new music, but it disturbed many parents. - Compound 4. As much as the music itself, it was the sexually suggestive body language of the performers that worried the older generation. – Complex 5. The social turmoil that marked the 1960’s influenced many performers, and some began to use their music as a vehicle for protest. - Compound Exercise 31.7: Circle the simple subjects and verbs in the following passage. Place one line under each independent clause and two lines under each dependent clause. (Recall that as independent clause can stand on its own as a complete sentence.) Many argue that the blues and jazz are the first truly American musical forms. With its origins in slave narratives, the blues took root during the 1920’s and 1930’s as African-America composers, musicians and singers performed in the cabarets and clubs of Harlem. Jazz, however, has is origins in New Orleans. Today, we can still appreciate the music of Bessie Smith, Duke Ellington, and B.B. King. Rock and roll is also a distinctively American form of music. Our country’s first “rock star” was without a doubt Elvis Presley, who emerged in the nation’s airwaves in the mid-1950’s with such hits as Heartbreak Hotel, Don’t Be Cruel, and All Shook...

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...Religion as a response to repressed guilt Freud came up with a physiological metaphor called the Oedipal complex. This complex is where Freud believes the feeling of guilt to stem from and also believes that we respond to this with religion. The oedipal complex is derived from a play called Oedipus Rex in which the son Laius kills his father, Oedipus , in order to have a relationship with his mother . The Oedipal complex basically refers to the unconscious desire of a son to have a relationship with his mother. Freud believed that this complex began at the phallic stage of a child’s physco –sexual development in which the source of libido pleasure was in a different erogenous zone of the infant’s body. At the phallic stage Freud believes that boys feel an unconscious desire to displace the father and therefore feel ambivalent towards him, as they fear genital mutilation due to several reasons such as the loss of the mothers breasts. At this stage boys have an unconscious love affair with their mother and it becomes the duty of the parents to act upon this incestuous urge. Parents must act on the incestuous urge as if they do not it could cause society to break down and threatens the family unit. A balance must be found and the urge must be controlled. The way in which parents do this is by introjection of their egos into the child; this means that the child absorbs the parents values, personality and characteristics and mold themselves to be similar. When a parent......

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...Astronomy 101 Midterm Professor Gonzalez 2/10/16 Complex Life What led to complex life as we know it? Scientists have always been baffled with how it happened, because there are so many things that contributed in both big and small ways. The earth had to let its Primordial soup boil for millions of years, just for the slight chance that life would be created. Many things contributed to the creation of complex life but there are a few ways that really shaped how we know life today. The most treasured source of life today is pretty obvious. We are all made of water and we need it to live our daily lives. Meanwhile while the Earth was cooling, asteroids and comets collided with earth, transferring materials that they were made of. Luckily enough one of those common materials happened to be Ice, which eventually thawed and became the oceans that we know today, which over time with a combination of sunlight and other organic materials that asteroids brought. This gave rise to probiotic cells only after about a billion years of simmering in the Primordial soup. Now after that, how did the cells become complex life? Well another abundant resource we treasure today is good old oxygen. Oxygen did not show on Earth until about 2.5 Ga. Cyanobacteria introduced oxygen to the ocean, which reacted with the iron and created iron oxide. Once the iron was used up and could no longer create iron oxide, oxygen was released, forming a huge portion of our atmosphere....

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...Date: 21 February 2012 Write an essay about the Oedipus complex. Your essay should explore the history of the concept, its development in psychoanalytic thinking and its wider implications. You should illustrate your understanding with detailed examples from your work setting and/or placement. Essay submitted for the award of MSc in Psychodynamic Counselling and Psychotherapy at Birkbeck College, University of London Introduction This paper attempts to trace the evolution of Freud’s concept of the Oedipus complex within psychoanalysis. One case will be considered in detail to illustrate relevant theoretical formulations focussing mainly on the work of Freud, Klein, Britton and Bion. The wider implications of the Oedipus complex will also be considered before I conclude. History of the Oedipus complex and Freud’s theoretic evolution Freud used the term Oedipus complex to describe a boy's feelings of desire for his mother and hostility towards his father which he regarded as a universal event of early childhood. The term is derived from the Greek myth of Oedipus who unwittingly killed his father and married his mother. In a letter to Fliess dating back to 1897, Freud (1954) attributes these ideas to his self analysis and indirectly to the Greek myth as presented in the popular Sophocles’ play ‘Oedipus Rex’. During the late 1890s Freud, theorised that fantasies built around the Oedipus complex were the primary cause of hysteria and other neurotic......

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...Complex Analysis Complex Numbers: A complex number z is an ordered pair (x,y) of real numbers x and y. z = (x,y) = x + iy The Real Part of z ie.Re(z) = x and the Imaginery part of z ie. Im(z) = y. Moreover,i2 = -1 which is an imaginery unit. a. The two imaginery numbers x + iy and a + ib are equal iff x = a and y =b, b. For z = x + iy, if x = 0,then z = iy (A pure imaginery number) and if y = 0 then z = x ( Pure real number). If z1 = x1 + iy1 and z2 = x2 + iy2, then the Addition, Multiplication, Subtraction and Division of two complex numbers respectively is defined as follows: z1 + z2 = (x1 + x2) + i(y1 + y2) z1 z2 = (x1 x2 - y1 y2) + i(x1 y2 + x2 y1) z1 − z2 = (x1 - x2) + i(y1 - y2) z = x/y = x + iy,where x = , y = ,z2 ≠ 0. Complex Conjugate Number The complex conjugate of the number z = x +iy is = x-iy Re(z) = x = (z + ) and Im(z) = (z - ) When z is real, z = x then z = Polar Form of Complex Numbers Let (x,y) be the Cartesian coordinates and (r,Ө) be the polar coordinates,then x = r cos Ө , y = r sin Ө Therefore, z = x+iy = r (cos Ө+ isin Ө) r = which is the absolute value or the modulus of z. Ө = arg z = tan which is the argument of z. Important Properties Generalized Triangle Inequality : Let Then, De Moivre’s formula : Nth Root of z : Limit, Continuity and Derivatives of Function of Complex variable: Limit......

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...Inferiority complex An abnormal or pathological state which, due to the tendency of the complex to draw unrelated ideas into itself, leads the individual to depreciate himself, to become unduly sensitive, to be too eager for praise and flattery, and to adopt a derogatory attitude toward others. ers. every human being is born with inferiority feelings. For physical weakness and defects in some parts of body will contribute to emotional problems. I have undergone two times of liver transplant. One is when I was 13 and the second time is at 19. Since my first transplant, I had a big scar on my abdomen. I used to think that I was a weaker person because I had less energy and easier to get tired compared to other normal people. In this connection, I always think that I am weaker than others. I feel inferior and have low self-confidence. These are some feelings and personality arose from psychological inferiority. http://www.davidcox.com.mx/library/L/Lin,%20Timothy%20-%20Inferiority%20Complex.pdf The My life is always connected to these previous unpleasant experiences. The parental attitudes are also important in shaping my personality. Since I was My parents seldom praise me when I behave well or get good academic results. They tend to think that it is my responsibility to be a good child. Their attitude makes me never feel proud of my achievements even I was viewed as a good student by others. My mum disapprove me when my behavior didn’t meet their......

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...“Military Industrial Complex”. What does this term mean, who coined the term and when? The Military Industrial Complex is generally defined as a coalition consisting of the military and industrialists who profit by manufacturing arms and selling them to the government. The Military Industrial Complex as coined by President Dwight D. Eisenhower in his farewell address on 1961. Great and sustained spending for defense and war he warned created power groups that could disastrously harm the nation future. How is this concept related to Third World military regimes? Military Industrial Complex is very influential and resourceful and related to financial crisis at the time primarily affected developing nations negatively. The first modern MIC arose in Britain, France and Germany in the 1800s and 1890s as part of increasing need to defend their respective empires both on the ground and at sea. What roles do developed nations play in the arms business in Third World countries? Please think of present day examples. Developing countries are the main recipients of the arms sales. Developing nations continue to be the primary focus of foreign arms sales activity by weapons suppliers though most are arms are supplied by just 2 or 3 major supplier. In spite of our global economic climate major purchase has continue to be made by a select few developing nations in these regions such as India in Asia, and Saudi Arabia in the Middle East. In the increasing of......

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