+ b n sin 2πnt. dt. However, we have also seen that, using Euler s Formula, trigonometric functions can be written in a complex exponential form,

Transcription

1 4 omple Analsis He is not a true man of science who does not bring some smpath to his studies, and epect to learn something b behavior as well as b application. It is childish to rest in the discover of mere coincidences, or of partial and etraneous laws. The stud of geometr is a pett and idle eercise of the mind, if it is applied to no larger sstem than the starr one. Mathematics should be mied not onl with phsics but with ethics; that is mied mathematics. The fact which interests us most is the life of the naturalist. The purest science is still biographical. Henr David Thoreau (87-86) We have seen that we can seek the frequenc content of a signal f (t) defined on an interval [, T] b looking for the the Fourier coefficients in the Fourier series epansion f (t) a + a n cos πnt T + b n sin πnt T. n The coefficients can be written as integrals such as a n T T f (t) cos πnt T However, we have also seen that, using Euler s Formula, trigonometric functions can be written in a comple eponential form, dt. In this chapter we introduce comple numbers and comple functions. We will later see that the rich structure of comple functions will lead to a deeper understanding of analsis, interesting techniques for computing integrals, and a natural wa to epress analog and discrete signals. cos πnt T eπint/t + e πint/t. We can use these ideas to rewrite the trigonometric Fourier series as a sum over comple eponentials in the form f (t) c n e πint/t, n where the Fourier coefficients now take the form c n T f (t)e πint/t dt. This representation will be useful in the analsis of analog signals, which are ideal signals defined on an infinite interval and containing a continuum

2 8 fourier and comple analsis The Bernoulli s were a famil of Swiss mathematicians spanning three generations. It all started with Jacob Bernoulli (654-75) and his brother Johann Bernoulli ( ). Jacob had a son, Nicolaus Bernoulli ( ) and Johann ( ) had three sons, Nicolaus Bernoulli II (695-76), Daniel Bernoulli (7-87), and Johann Bernoulli II (7-79). The last generation consisted of Johann II s sons, Johann Bernoulli III (747-87) and Jacob Bernoulli II ( ). Johann, Jacob and Daniel Bernoulli were the most famous of the Bernoulli s. Jacob studied with Leibni, Johann studied under his older brother and later taught Leonhard Euler and Daniel Bernoulli, who is known for his work in hdrodnamics. r θ Figure 4.: The Argand diagram for plotting comple numbers in the comple - plane. The comple modulus, +. of frequencies. We will see the above sum become an integral and will naturall find ourselves needing to work with functions of comple variables and performing integrals of comple functions. With this ultimate goal in mind, we will now take a tour of comple analsis. We will begin with a review of some facts about comple numbers and then introduce comple functions. This will lead us to the calculus of functions of a comple variable, including differentiation and integration of comple functions. 4. omple Numbers omple numbers were first introduced in order to solve some simple problems. The histor of comple numbers onl etends about five hundred ears. In essence, it was found that we need to find the roots of equations such as +. The solution is ±. Due to the usefulness of this concept, which was not realied at first, a special smbol was introduced - the imaginar unit, i. In particular, Girolamo ardano (5 576) was one of the first to use square roots of negative numbers when providing solutions of cubic equations. However, comple numbers did not become an important part of mathematics or science until the late seventh centur after people like Abraham de Moivre ( ), the Bernoulli famil, and Leonhard Euler (77-783) took them seriousl. A comple number is a number of the form + i, where and are real numbers. is called the real part of and is the imaginar part of. Eamples of such numbers are 3 + 3i, i i, 4i and 5. Note that i and 4i + 4i. There is a geometric representation of comple numbers in a two-dimensional plane, known as the comple plane. This is given b the Argand diagram as shown in Figure 4.. Here we can think of the comple number + i as a point (, ) in the -comple plane or as a vector. The magnitude, or length, of this vector is called the comple modulus of, denoted b +. We can also use the geometric picture to develop a polar representation of comple numbers. From Figure 4. we can see that in terms of r and θ, we have that r cos θ, r sin θ. (4.) omple numbers can be represented in rectangular (artesian), + i, or polar form, re iθ. Here we define the argument, θ, and modulus, r, of comple numbers. Thus, using Euler s Formula (Eample.34), we have + i r(cos θ + i sin θ) re iθ. (4.) So, given r and θ we have re iθ. However, given the artesian form, + i, we can also determine the polar form, since r +, tan θ. (4.3)

3 comple analsis 9 Note that r. Locating + i in the comple plane, it is possible to immediatel determine the polar form from the angle and length of the comple vector. This is shown in Figure 4.. It is obvious that θ π 4 and r. Eample 4.. Write + i in polar form. If one did not see the polar form from the plot in the -plane, then one could sstematicall determine the results. First, write + i in polar form, re iθ, for some r and θ. Using the above relations between polar and artesian representations, we have r + and tan θ. This gives θ π 4. So, we have found that + i e iπ/4. We can also define binar operations of addition, subtraction, multiplication, and division of comple numbers to produce a new comple number. The addition of two comple numbers is simpl done b adding the real and imaginar parts of each number. So, i i i + i Figure 4.: Locating + i in the comple -plane. We can easil add, subtract, multipl, and divide comple numbers. (3 + i) + ( i) 4 + i. Subtraction is just as eas, (3 + i) ( i) + 3i. We can multipl two comple numbers just like we multipl an binomials, though we now can use the fact that i. For eample, we have (3 + i)( i) 3 + i 3i + i( i) 5 i. We can even divide one comple number into another one and get a comple number as the quotient. Before we do this, we need to introduce the comple conjugate,, of a comple number. The comple conjugate of + i, where and are real numbers, is given as i. The comple conjugate of + i is given as i. omple conjugates satisf the following relations for comple numbers and w and real number. + w + w. w w... (4.4) One consequence is that the comple conjugate of re iθ is re iθ cos θ + i sin θ cos θ i sin θ re iθ.

4 fourier and comple analsis Another consequence is that re iθ re iθ r. Thus, the product of a comple number with its comple conjugate is a real number. We can also prove this result using the artesian form ( + i)( i) +. Now we are in a position to write the quotient of two comple numbers in the standard form of a real plus an imaginar number. Eample 4.. Simplif the epression 3+i i. This simplification is accomplished b multipling the numerator and denominator of this epression b the comple conjugate of the denominator: 3 + i i 3 + i i + i + i + 5i. Therefore, the quotient is a comple number and in standard form is given b + 5 i. We can also consider powers of comple numbers. For eample, ( + i) i, ( + i) 3 ( + i)(i) i. But, what is ( + i) / + i? In general, we want to find the nth root of a comple number. Let t /n. To find t in this case is the same as asking for the solution of t n given. But, this is the root of an nth degree equation, for which we epect n roots. If we write in polar form, re iθ, then we would naivel compute /n (re iθ) /n r /n e [ iθ/n r /n cos θ n + i sin θ ]. (4.5) n For eample, ( ) ( + i) / / e iπ/4 /4 e iπ/8. The function f () /n is multivalued: /n r /n e i(θ+kπ)/n, k,,..., n. But this is onl one solution. We epected two solutions for n. The reason we onl found one solution is that the polar representation for is not unique. We note that e kπi, k, ±, ±,.... So, we can rewrite as re iθ e kπi re i(θ+kπ). Now we have that

5 comple analsis /n r /n e i(θ+kπ)/n, k,,..., n. Note that these are the onl distinct values for the roots. We can see this b considering the case k n. Then we find that e i(θ+πin)/n e iθ/n e πi e iθ/n. So, we have recovered the n value. Similar results can be shown for the other k values larger than n. Now we can finish the eample we started. Eample 4.3. Determine the square roots of + i, or + i. As we have seen, we first write + i in polar form: + i e iπ/4. Then we introduce e kπi and find the roots: ( + i) / ( e iπ/4 e kπi) /, k,, /4 e i(π/8+kπ), k,, /4 e iπ/8, /4 e 9πi/8. (4.6) Finall, what is n? Our first guess would be n. But we now know that there should be n roots. These roots are called the nth roots of unit. Using the above result with r and θ, we have that For eample, we have n [cos πk n 3 [cos πk 3 πk + i sin n These three roots can be written out as ], k,..., n. ] πk + i sin, k,,. 3 The nth roots of unit, i n. 3 3, + i, 3 i. We can locate these cube roots of unit in the comple plane. In Figure 4.3, we see that these points lie on the unit circle and are at the vertices of an equilateral triangle. In fact, all nth roots of unit lie on the unit circle and are the vertices of a regular n-gon with one verte at. 4. omple Valued Functions i Figure 4.3: Locating the cube roots of unit in the comple -plane. We would like to net eplore comple functions and the calculus of comple functions. We begin b defining a function that takes comple numbers into comple numbers, f :. It is difficult to visualie such functions. For real functions of one variable, f : R R, we graph these functions b first drawing two intersecting copies of R and then proceed to map the domain into the range of f.

6 fourier and comple analsis It would be more difficult to do this for comple functions. Imagine placing together two orthogonal copies of the comple plane,. One would need a four-dimensional space in order to complete the visualiation. Instead, one tpicall uses two copies of the comple plane side b side in order to indicate how such functions behave. Over the ears there have been several was to visualie comple functions. We will describe a few of these in this chapter. We will assume that the domain lies in the -plane and the image lies in the w-plane. We will then write the comple function as w f (). We show these planes in Figure 4.4 as well as the mapping between the planes. Figure 4.4: Defining a comple valued function, w f (), on for + i and w u + iv. v w f () w u Letting + i and w u + iv, we can write the real and imaginar parts of f () : w f () f ( + i) u(, ) + iv(, ). We see that one can view this function as a function of or a function of and. Often, we have an interest in writing out the real and imaginar parts of the function, u(, ) and v(, ), which are functions of two real variables, and. We will look at several functions to determine the real and imaginar parts. Eample 4.4. Find the real and imaginar parts of f (). For eample, we can look at the simple function f (). It is a simple matter to determine the real and imaginar parts of this function. Namel, we have Therefore, we have that ( + i) + i. u(, ), v(, ). In Figure 4.5 we show how a grid in the -plane is mapped b f () into the w-plane. For eample, the horiontal line is mapped to u(, ) and v(, ). Eliminating the parameter between these two equations, we have u v /4. This is a parabolic curve. Similarl, the horiontal line results in the curve u v /4. If we look at several curves, const and const, then we get a famil of intersecting parabolae, as shown in Figure 4.5.

7 comple analsis Figure 4.5: D plot showing how the function f () maps the lines and in the -plane into parabolae in the w-plane. v u Figure 4.6: D plot showing how the function f () maps a grid in the -plane into the w-plane..5 v u Eample 4.5. Find the real and imaginar parts of f () e. For this case, we make use of Euler s Formula (from Eample.34): e e +i e e i e (cos + i sin ). (4.7) Thus, u(, ) e cos and v(, ) e sin. In Figure 4.7 we show how a grid in the -plane is mapped b f () e into the w-plane Figure 4.7: D plot showing how the function f () e maps a grid in the -plane into the w-plane..5 v u Eample 4.6. Find the real and imaginar parts of f () /.

8 4 fourier and comple analsis v u Figure 4.8: D plot showing how the function f () maps a grid in the -plane into the w-plane. Thus, We have that / + (cos (θ + kπ) + i sin (θ + kπ)), k,. (4.8) u cos (θ + kπ), u cos (θ + kπ), for + and θ tan (/). For each k-value, one has a different surface and curves of constant θ give u/v c ; and, curves of constant nonero comple modulus give concentric circles, u + v c, for c and c constants. Eample 4.7. Find the real and imaginar parts of f () ln. In this case, we make use of the polar form of a comple number, re iθ. Our first thought would be to simpl compute ln ln r + iθ. However, the natural logarithm is multivalued, just like the square root function. Recalling that e πik for k an integer, we have re i(θ+πk). Therefore, ln ln r + i(θ + πk), k integer. The natural logarithm is a multivalued function. In fact, there are an infinite number of values for a given. Of course, this contradicts the definition of a function that ou were first taught. Thus, one tpicall will onl report the principal value, Log ln r + iθ, for θ restricted to some interval of length π, such as [, π). In order to account for the multivaluedness, one introduces a wa to etend the comple plane so as to include all of the branches. This is done b assigning a plane to each branch, using (branch) cuts along lines, and then gluing the planes together at the branch cuts to form what is called a Riemann surface. We will not elaborate upon this an further here and refer the interested reader to more advanced tets. omparing the multivalued logarithm to the principal value logarithm, we have ln Log + nπi. Figure 4.9: Domain coloring of the comple -plane assigning colors to arg(). We should note that some books use log instead of ln. It should not be confused with the common logarithm. 4.. omple Domain oloring Another method for visualiing comple functions is domain coloring. The idea was described b Frank A. Farris. There are a few approaches to this method. The main idea is that one colors each point of the -plane (the domain) according to arg() as shown in Figure 4.9. The modulus, f () is then plotted as a surface. Eamples are shown for f () in Figure 4. and f () /( ) in Figure 4.. We would like to put all of this information in one plot. We can do this b adjusting the brightness of the colored domain using the modulus of the

9 comple analsis 5 Figure 4.: Domain coloring for f (). The left figure shows the phase coloring. The right figure shows the colored surface with height f (). Figure 4.: Domain coloring for f () /( ). The left figure shows the phase coloring. The right figure shows the colored surface with height f (). function. In the plots that follow we use the fractional part of ln. In Figure 4. we show the effect for the -plane using f (). In the figures that follow, we look at several other functions. In these plots, we have chosen to view the functions in a circular window. One can see the rich behavior hidden in these figures. As ou progress in our reading, especiall after the net chapter, ou should return to these figures and locate the eros, poles, branch points, and branch cuts. A search online will lead ou to other colorings and superposition of the uv grid on these figures. As a final picture, we look at iteration in the comple plane. onsider the function f ().75.i. Interesting figures result when studing the iteration in the comple plane. In Figure 4.5 we show f () and f (), which is the iteration of f twent times. It leads to an interesting coloring. What happens when one keeps iterating? Such iterations lead to the stud of Julia and Mandelbrot sets. In Figure 4.6 we show si iterations of f () ( i/) sin. The following code was used in MATLAB to produce these figures. Figure 4.: Domain coloring for the function f () showing a coloring for arg() and brightness based on f (). Figure 4.3: Domain coloring for the function f (). (-i/)*sin(); min-; ma; min-; ma; N5; N5;

10 6 fourier and comple analsis Figure 4.4: Domain coloring for several functions. On the top row, the domain coloring is shown for f () 4 and f () sin. On the second row, plots for f () + and f () are shown. In the last (/ )( i)( i+) row, domain colorings for f () ln and f () sin(/) are shown. Figure 4.5: Domain coloring for f ().75.i. The left figure shows the phase coloring. On the right is the plot for f (). Figure 4.6: Domain coloring for si iterations of f () ( i/) sin.

11 comple analsis 7 linspace(min,ma,n); linspace(min,ma,n); [X,Y] meshgrid(,); comple(x,y); tmp; for n:6 tmp fn(tmp); end Ztmp; XXreal(Z); YYimag(Z); Rma(ma(X.^)); Rma(ma(XX.^+YY.^)); circle(:,:,) X.^+Y.^ < R; circle(:,:,)circle(:,:,); circle(:,:,3)circle(:,:,); addcirc(:,:,)circle(:,:,); addcirc(:,:,)circle(:,:,); addcirc(:,:,3)circle(:,:,); warning off MATLAB:divideBZero; hsvircleones(n,n,3); hsvircle(:,:,)atan(yy,xx)*8/pi+(atan(yy,xx)*8/pi<)*36; hsvircle(:,:,)hsvircle(:,:,)/36; lglog(xx.^+yy.^)/; hsvircle(:,:,).75; hsvircle(:,:,3)-(lg-floor(lg))/; hsvircle(:,:,) flipud((hsvircle(:,:,))); hsvircle(:,:,) flipud((hsvircle(:,:,))); hsvircle(:,:,3) flipud((hsvircle(:,:,3))); rgbirclehsvrgb(hsvircle); rgbirclergbircle.*circle+addcirc; image(rgbircle) ais square set(gca, XTickLabel,{}) set(gca, YTickLabel,{}) 4.3 omple Differentiation Net we want to differentiate comple functions. We generalie the definition from single variable calculus, f f ( + ) f () () lim, (4.9)

12 8 fourier and comple analsis provided this limit eists. The computation of this limit is similar to what one sees in multivariable calculus for limits of real functions of two variables. Letting + i and δ δ + iδ, then Figure 4.7: There are man paths that approach as. Figure 4.8: A path that approaches with constant. + δ ( + δ) + i( + δ). Letting means that we get closer to. There are man paths that one can take that will approach. [See Figure 4.7.] It is sufficient to look at two paths in particular. We first consider the path constant. This horiontal path is shown in Figure 4.8. For this path, + i, since does not change along the path. The derivative, if it eists, is then computed as f f ( + ) f () () lim lim lim u( +, ) + iv( +, ) (u(, ) + iv(, )) u( +, ) u(, ) v( +, ) v(, ) + lim i. (4.) The last two limits are easil identified as partial derivatives of real valued functions of two variables. Thus, we have shown that when f () eists, f () u + i v. (4.) A similar computation can be made if, instead, we take the vertical path, constant, in Figure 4.7). In this case, i and f f ( + ) f () () lim Therefore, lim u(, + ) + iv(, + ) (u(, ) + iv(, )) i u(, + ) u(, ) lim + lim i v(, + ) v(, ). f () v i u. (4.3) (4.) We have found two different epressions for f () b following two different paths to. If the derivative eists, then these two epressions must be the same. Equating the real and imaginar parts of these epressions, we have

13 comple analsis 9 u v v u. (4.4) These are known as the auch-riemann Equations. Theorem 4.. f () is holomorphic (differentiable) if and onl if the auch-riemann Equations are satisfied. Eample 4.8. f (). In this case we have alread seen that + i. Therefore, u(, ) and v(, ). We first check the auch-riemann Equations. u v v u. (4.5) Therefore, f () is differentiable. We can further compute the derivative using either Equation (4.) or Equation (4.3). Thus, f () u + i v + i(). This result is not surprising. Eample 4.9. f (). In this case we have f () i. Therefore, u(, ) and v(, ). But, u v and. Thus, the auch-riemann Equations are not satisfied and we conclude that f () is not differentiable. Another consequence of the auch-riemann Equations is that both u(, ) and v(, ) are harmonic functions. A real-valued function u(, ) is harmonic if it satisfies Laplace s Equation in D, u, or u + u. Theorem 4.. f () u(, ) + iv(, ) is differentiable if and onl if u and v are harmonic functions. This is easil proven using the auch-riemann Equations. u u v v u u. (4.6) The auch-riemann Equations. Augustin-Louis auch ( ) was a French mathematician well known for his work in analsis. Georg Friedrich Bernhard Riemann (86-866) was a German mathematician who made major contributions to geometr and analsis. Harmonic functions satisf Laplace s Equation.

14 3 fourier and comple analsis Eample 4.. Is u(, ) + harmonic? u + u +. No, it is not. Eample 4.. Is u(, ) harmonic? Yes, it is. u + u. The harmonic conjugate function. Given a harmonic function u(, ), can one find a function, v(, ), such f () u(, ) + iv(, ) is differentiable? In this case, v are called the harmonic conjugate of u. Eample 4.. Find the harmonic conjugate of u(, ) and determine f () u + iv such that u + iv is differentiable. The auch-riemann Equations tell us the following about the unknown function, v(, ) : v u, v u. We can integrate the first of these equations to obtain v(, ) d + c(). Here, c() is an arbitrar function of. One can check to see that this works b simpl differentiating the result with respect to. However, the second equation must also hold. So, we differentiate the result with respect to to find that v + c (). Since we were supposed to get, we have that c (). Thus, c() k is a constant. We have just shown that we get an infinite number of functions, v(, ) + k, such that f () + i( + k) is differentiable. In fact, for k, this is nothing other than f ().

15 comple analsis omple Integration We have introduced functions of a comple variable. We also established when functions are differentiable as comple functions, or holomorphic. In this chapter we will turn to integration in the comple plane. We will learn how to compute comple path integrals, or contour integrals. We will see that contour integral methods are also useful in the computation of some of the real integrals that we will face when eploring Fourier transforms in the net chapter omple Path Integrals In this section we will investigate the computation of comple path integrals. Given two points in the comple plane, connected b a path Γ as shown in Figure 4.9, we would like to define the integral of f () along Γ, f () d. Γ A natural procedure would be to work in real variables, b writing f () d [u(, ) + iv(, )] (d + id), Γ Γ since + i and d d + id. In order to carr out the integration, we then have to find a parametriation of the path and use methods from a multivariate calculus class. Namel, let u and v be continuous in domain D, and Γ a piecewise smooth curve in D. Let ((t), (t)) be a parametriation of Γ for t t t and f () u(, ) + iv(, ) for + i. Then Figure 4.9: We would like to integrate a comple function f () over the path Γ in the comple plane. Figure 4.: Eamples of (a) a connected set and (b) a disconnected set. Γ f () d t t [u((t), (t)) + iv((t), (t))] ( d dt + i d dt )dt. (4.7) Here we have used d d + id ( d dt + i d ) dt. dt Furthermore, a set D is called a domain if it is both open and connected. Before continuing, we first define open and connected. A set D is connected if and onl if for all, and in D, there eists a piecewise smooth curve connecting to and ling in D. Otherwise it is called disconnected. Eamples are shown in Figure 4. A set D is open if and onl if for all in D, there eists an open disk < ρ in D. In Figure 4. we show a region with two disks. For all points on the interior of the region, one can find at least one disk contained entirel in the region. The closer one is to the boundar, the Figure 4.: Locations of open disks inside and on the boundar of a region.

16 3 fourier and comple analsis i i i Figure 4.: ontour for Eample 4.3. smaller the radii of such disks. However, for a point on the boundar, ever such disk would contain points inside and outside the disk. Thus, an open set in the comple plane would not contain an of its boundar points. We now have a prescription for computing path integrals. Let s see how this works with a couple of eamples. Eample 4.3. Evaluate d, where the arc of the unit circle in the first quadrant as shown in Figure 4.. There are two was we could carr out the parametriation. First, we note that the standard parametriation of the unit circle is ((θ), (θ)) (cos θ, sin θ), θ π. For a quarter circle in the first quadrant, θ π, we let cos θ + i sin θ. Therefore, d ( sin θ + i cos θ) dθ and the path integral becomes d π (cos θ + i sin θ) ( sin θ + i cos θ) dθ. We can epand the integrand and integrate, having to perform some trigonometric integrations. π [sin 3 θ 3 cos θ sin θ + i(cos 3 θ 3 cos θ sin θ)] dθ. The reader should work out these trigonometric integrations and confirm the result. For eample, ou can use to write the real part of the integrand as The resulting antiderivative becomes sin 3 θ sin θ( cos θ)) sin θ 4 cos θ sin θ. cos θ cos3 θ. The imaginar integrand can be integrated in a similar fashion. While this integral is doable, there is a simpler procedure. We first note that e iθ on. So, d ie iθ dθ. The integration then becomes d π (e iθ ) ie iθ dθ i π e 3iθ dθ ie3iθ π/ 3i + i 3. (4.8)

17 comple analsis 33 Eample 4.4. Evaluate Γ d, for the path Γ γ γ shown in Figure 4.3. In this problem we have a path that is a piecewise smooth curve. We can compute the path integral b computing the values along the two segments of the path and adding the results. Let the two segments be called γ and γ as shown in Figure 4.3 and parametrie each path separatel. Over γ we note that. Thus, for [, ]. It is natural to take as the parameter. So, we let d d to find γ d d. For path γ, we have that + i for [, ] and d i d. Inserting this parametriation into the integral, the integral becomes i. γ d ( + i) id i. ombining the results for the paths γ and γ, we have Γ d + (i ) Eample 4.5. Evaluate γ d, where γ 3 3 is the path shown in Figure 4.4. In this case we take a path from to + i along a different path than in the last eample. Let γ 3 {(, ), [, ]} { + i, [, ]}. Then, d ( + i) d. The integral becomes γ 3 d ( + i )( + i) d ( + 3i 3 ) d [ + i 3 4 ] i. (4.9) In the last case we found the same answer as obtained in Eample 4.4. But we should not take this as a general rule for all comple path integrals. In fact, it is not true that integrating over different paths alwas ields the same results. However, when this is true, then we refer to this propert as path independence. In particular, the integral f () d is path independent if f () d f () d Γ Γ for all paths from to as shown in Figure 4.5. We can show that if f () d is path independent, then the integral of f () over all closed loops is ero: f () d. closed loops A common notation for integrating over closed loops is f () d. But first we have to define what we mean b a closed loop. A simple closed contour is a path satisfing i i γ γ i + i Figure 4.3: ontour for Eample 4.4 with Γ γ γ. i + i i γ γ 3 γ i Figure 4.4: ontour for Eample 4.5. Γ Figure 4.5: Γ f () d Γ f () d for all paths from to when the integral of f () is path independent. A simple closed contour. Γ

18 34 fourier and comple analsis a. The end point is the same as the beginning point. (This makes the loop closed.) b. The are no self-intersections. (This makes the loop simple.) f () d if the integral is path in- dependent. Figure 4.6: The integral f () d around is ero if the integral Γ f () d is path independent. A loop in the shape of a figure eight is closed, but it is not simple. Now consider an integral over the closed loop shown in Figure 4.6. We pick two points on the loop, breaking it into two contours, and. Then we make use of the path independence b defining to be the path along but in the opposite direction. Then, f () d f () d + f () d f () d f () d. (4.) Assuming that the integrals from point to point are path independent, then the integrals over and are equal. Therefore, we have f () d. Eample 4.6. onsider the integral d for the closed contour shown in Figure 4.4 starting at following path γ, then γ and returning to. Based on the earlier eamples and the fact that going backward on γ 3 introduces a negative sign, we have d d + d d ( γ γ γ 3 + i ) i auch s Theorem A curve with parametriation ((t), (t)) has a normal (n, n ) ( d dt, d dt ). Net we want to investigate if we can determine that integrals over simple closed contours vanish without doing all the work of parametriing the contour. First, we need to establish the direction about which we traverse the contour. We can define the orientation of a curve b referring to the normal of the curve. Recall that the normal is a perpendicular to the curve. There are two such perpendiculars. The above normal points outward and the other normal points toward the interior of a closed curve. We will define a positivel oriented contour as one that is traversed with the outward normal pointing to the right. As one follows loops, the interior would then be on the left. We now consider (u + iv) d over a simple closed contour. This can be written in terms of two real integrals in the -plane. (u + iv) d (u + iv)(d + i d) u d v d + i v d + u d. (4.) These integrals in the plane can be evaluated using Green s Theorem in the Plane. Recall this theorem from our last semester of calculus:

19 comple analsis 35 Green s Theorem in the Plane. Theorem 4.3. Let P(, ) and Q(, ) be continuousl differentiable functions on and inside the simple closed curve as shown in Figure 4.7. Denoting the enclosed region S, we have ( Q P d + Q d P ) dd. (4.) S S Using Green s Theorem to rewrite the first integral in Equation (4.), we have ( v u d v d u ) dd. If u and v satisf the auch-riemann Equations (4.4), then the integrand in the double integral vanishes. Therefore, S u d v d. Figure 4.7: Region used in Green s Theorem. Green s Theorem in the Plane is one of the major integral theorems of vector calculus. It was discovered b George Green (793-84) and published in 88, about four ears before he entered ambridge as an undergraduate. In a similar fashion, one can show that v d + u d. We have thus proven the following theorem: auch s Theorem Theorem 4.4. If u and v satisf the auch-riemann Equations (4.4) inside and on the simple closed contour, then (u + iv) d. (4.3) orollar f () d when f is differentiable in domain D with D. Either one of these is referred to as auch s Theorem. Eample 4.7. Evaluate 3 4 d. Since f () 4 is differentiable inside the circle 3, this integral vanishes. We can use auch s Theorem to show that we can deform one contour into another, perhaps simpler, contour. Theorem 4.5. If f () is holomorphic between two simple closed contours, and, then f () d f () d. One can deform contours into simpler ones. Proof. We consider the two curves and as shown in Figure 4.8. onnecting the two contours with contours Γ and Γ (as shown in the figure), is seen to split into contours and and into contours and. Note that f () is differentiable inside the newl formed regions between the curves. Also, the boundaries of these regions are now simple closed curves.

20 36 fourier and comple analsis Γ Γ Figure 4.8: The contours needed to prove that f () d f () d when f () is holomorphic between the contours and. R γ 3 i i i i γ γ 4 γ Figure 4.9: The contours used to compute d R. Note that to compute the integral around R we can deform the contour to the circle since f () is differentiable in the region between the contours. R γ 3 i i i i γ γ 4 γ Figure 4.3: The contours used to compute d R. The added diagonals are for the reader to easil see the arguments used in the evaluation of the limits when integrating over the segments of the square R. Therefore, auch s Theorem tells us that the integrals of f () over these regions are ero. Noting that integrations over contours opposite the positive orientation are the negative of integrals that are positivel oriented, we have from auch s Theorem that and f () d + f () d Γ f () d f () d Γ f () d + f () d Γ f () d f () d. Γ In the first integral, we have traversed the contours in the following order:, Γ, backward, and Γ. The second integral denotes the integration over the lower region, but going backward over all contours ecept for. ombining these results b adding the two equations above, we have f () d + f () d Noting that + and +, we have f () d f () d, as was to be proven. d f () d f () d. Eample 4.8. ompute R for R the rectangle [, ] [ i, i]. We can compute this integral b looking at four separate integrals over the sides of the rectangle in the comple plane. One simpl parametries each line segment, perform the integration and sum the four separate results. From the deformation theorem, we can instead integrate over a simpler contour b deforming the rectangle into a circle as long as f () is differentiable in the region bounded b the rectangle and the circle. So, using the unit circle, as shown in Figure 4.9, the integration might be easier to perform. More specificall, the the deformation theorem tells us that R d d The latter integral can be computed using the parametriation e iθ for θ [, π]. Thus, d d i π π ie iθ dθ e iθ dθ πi. (4.4) πi b deforming the original simple closed Therefore, we have found that R contour. For fun, let s do this the long wa to see how much effort was saved. We will label the contour as shown in Figure 4.3. The lower segment, γ 4, of the square can be

21 comple analsis 37 simple parametried b noting that along this segment i for [, ]. Then, we have d γ 4 d i ln i ( ln( ) πi 4 ) ( ln( ) 3πi ) 4 πi. (4.5) We note that the arguments of the logarithms are determined from the angles made b the diagonals provided in Figure 4.3. Similarl, the integral along the top segment, + i, [, ], is computed as d γ d + i ln + i ( ln( ) + 3πi 4 ) ( ln( ) + πi ) 4 πi. (4.6) The integral over the right side, + i, [, ], is d γ id + i ln + i ( ln( ) + πi 4 ) ( ln( ) πi ) 4 πi. (4.7) Finall, the integral over the left side, + i, [, ], is d γ 3 id + i ln + i Therefore, we have that R d ( ln( ) + 5πi 4 ) ( ln( ) + 3πi ) 4 πi. (4.8) γ d + γ d + γ 3 d + d γ 4

22 38 fourier and comple analsis πi + πi + πi + πi ( ) πi 4 πi. (4.9) This gives the same answer as we found using a simple contour deformation. Morera s Theorem. The converse of auch s Theorem is not true, namel f () d does not alwas impl that f () is differentiable. What we do have is Morera s Theorem (Giacinto Morera, ): Theorem 4.6. Let f be continuous in a domain D. Suppose that for ever simple closed contour in D, f () d. Then f is differentiable in D. The proof is a bit more detailed than we need to go into here. However, this theorem is useful in the net section Analtic Functions and auch s Integral Formula In the previous section we saw that auch s Theorem was useful for computing particular integrals without having to parametrie the contours or for deforming contours into simpler contours. The integrand needs to possess certain differentiabilit properties. In this section, we will generalie the functions that we can integrate slightl so that we can integrate a larger famil of comple functions. This will lead us to the auch s Integral Formula, which etends auch s Theorem to functions analtic in an annulus. However, first we need to eplore the concept of analtic functions. A function f () is analtic in domain D if for ever open disk < ρ ling in D, f () can be represented as a power series in. Namel, f () c n ( ) n. n There are various tpes of complevalued functions. A holomorphic function is (comple) differentiable in a neighborhood of ever point in its domain. An analtic function has a convergent Talor series epansion in a neighborhood of each point in its domain. We see here that analtic functions are holomorphic and vice versa. If a function is holomorphic throughout the comple plane, then it is called an entire function. Finall, a function which is holomorphic on all of its domain ecept at a set of isolated poles (to be defined later), then it is called a meromorphic function. This series converges uniforml and absolutel inside the circle of convergence, < R, with radius of convergence R. [See the Appendi for a review of convergence.] Since f () can be written as a uniforml convergent power series, we can integrate it term b term over an simple closed contour in D containing. In particular, we have to compute integrals like ( ) n d. As we will see in the homework problems, these integrals evaluate to ero for most n. Thus, we can show that for f () analtic in D and on an closed contour ling in D, f () d. Also, f is a uniforml convergent sum of continuous functions, so f () is also continuous. Thus, b Morera s Theorem, we have that f () is differentiable if it is analtic. Often terms like analtic, differentiable, and holomorphic are used interchangeabl, though there is a subtle distinction due to their definitions. As eamples of series epansions about a given point, we will consider series epansions and regions of convergence for f () +.

23 comple analsis 39 Eample 4.9. Find the series epansion of f () + about. This case is simple. From hapter we recall that f () is the sum of a geometric series for <. We have f () + ( ) n. n Thus, this series epansion converges inside the unit circle ( < ) in the comple plane. Eample 4.. Find the series epansion of f () + about. We now look into an epansion about a different point. We could compute the epansion coefficients using Talor s formula for the coefficients. However, we can also make use of the formula for geometric series after rearranging the function. We seek an epansion in powers of. So, we rewrite the function in a form that is a function of. Thus, f () + + ( + ) 3 + ( ). This is not quite in the form we need. It would be nice if the denominator were of the form of one plus something. [Note: This is similar to what we had seen in Eample.33.] We can get the denominator into such a form b factoring out the 3. Then we would have f () ( ). The second factor now has the form r, which would be the sum of a geometric series with first term a and ratio r 3 ( ) provided that r <. Therefore, we have found that f () 3 [ 3 ( ] n ) n for 3 ( ) <. This convergence interval can be rewritten as < 3, which is a circle centered at with radius 3. In Figure 4.3 we show the regions of convergence for the power series epansions of f () + about and. We note that the first epansion gives that f () is at least analtic inside the region <. The second epansion shows that f () is analtic in a larger region, < 3. We will see later that there are epansions which converge outside these regions and that some ield epansions involving negative powers of. We now present the main theorem of this section: i i i i Figure 4.3: Regions of convergence for epansions of f () + about and.

24 4 fourier and comple analsis auch Integral Formula Theorem 4.7. Let f () be analtic in < ρ and let be the boundar (circle) of this disk. Then, f ( ) f () d. (4.3) πi Proof. In order to prove this, we first make use of the analticit of f (). We insert the power series epansion of f () about into the integrand. Then we have f () [ n c n ( ) n ] [ ] c + c ( ) + c ( ) +... c + c + c ( ) (4.3) }{{} analtic function As noted, the integrand can be written as f () c + h(), where h() is an analtic function, since h() is representable as a series epansion about. We have alread shown that analtic functions are differentiable, so b auch s Theorem h() d. Noting also that c f ( ) is the first term of a Talor series epansion about, we have [ ] f () c d + h() d f ( ) d. We need onl compute the integral d to finish the proof of auch s Integral Formula. This is done b parametriing the circle, ρ, as shown in Figure 4.3. This is simpl done b letting ρe iθ. (Note that this has the right comple modulus since e iθ. Then d iρe iθ dθ. Using this parametriation, we have ρ d π iρe iθ dθ ρe iθ π i dθ πi. Figure 4.3: ircular contour used in proving the auch Integral Formula. Therefore, as was to be shown. f () d f ( ) d πi f ( ),

25 comple analsis 4 cos 6+5 d. Eample 4.. ompute 4 In order to appl the auch Integral Formula, we need to factor the denominator, ( )( 5). We net locate the eros of the denominator. In Figure 4.33 we show the contour and the points and 5. The onl point inside the region bounded b the contour is. Therefore, we can appl the auch Integral Formula for f () cos 5 4 Therefore, we have cos ( )( 5) d 4 4 cos ( )( 5) to the integral f () d πi f (). ( ) πi cos() d. We have shown that f ( ) has an integral representation for f () analtic in < ρ. In fact, all derivatives of an analtic function have an integral representation. This is given b f (n) ( ) n! f () d. (4.3) πi ( ) n+ This can be proven following a derivation similar to that for the auch Integral Formula. Inserting the Talor series epansion for f () into the integral on the right hand side, we have f () ( ) n+ d c m m c m m ( ) m d ( ) n+ d. (4.33) ( ) n m+ Picking k n m, the integrals in the sum can be computed using the following result: { d ( ) k+, k, (4.34) πi, k. ρ 4 3i i i i i 3i 4 Figure 4.33: ircular contour used in computing 4 cos 6+5 d. The proof is left for the eercises. The onl nonvanishing integrals, d, occur when k n m ( ) n m+, or m n. Therefore, the series of integrals collapses to one term and we have f () ( ) n+ d πic n. We finish the proof b recalling that the coefficients of the Talor series epansion for f () are given b Then, and the result follows. c n f (n) ( ). n! f () πi d f (n) ( ( ) n+ ) n!

26 4 fourier and comple analsis Laurent Series 3 Re-indeing a series is often useful in series manipulations. In this case, we have the series ( ) n n n The inde is n. You can see that the inde does not appear when the sum is epanded showing the terms. The summation inde is sometimes referred to as a dumm inde for this reason. Reindeing allows one to rewrite the shorthand summation notation while capturing the same terms. In this eample, the eponents are n. We can simplif the notation b letting n j, or j n +. Noting that j when n, we get the sum j ( )j j. Until this point we have onl talked about series whose terms have nonnegative powers of. It is possible to have series representations in which there are negative powers. In the last section we investigated epansions of f () + about and. The regions of convergence for each series was shown in Figure 4.3. Let us reconsider each of these epansions, but for values of outside the region of convergence previousl found. Eample 4.. f () + for >. As before, we make use of the geometric series. Since >, we instead rewrite the function as f () + +. We now have the function in a form of the sum of a geometric series with first term a and ratio r. We note that > implies that r <. Thus, we have the geometric series f () ( ) n. n This can be re-indeed 3 as f () n ( ) n n j ( ) j j. Note that this series, which converges outside the unit circle, >, has negative powers of. Eample 4.3. f () + for > 3. As before, we epress this in a form in which we can use a geometric series epansion. We seek powers of. So, we add and subtract to the to obtain: f () + + ( + ) 3 + ( ). Instead of factoring out the 3 as we had done in Eample 4., we factor out the ( ) term. Then, we obtain f () + ( ) [ + 3 ( ) ]. Now we identif a and r 3 ( ). This leads to the series f () n n ( 3 ( 3 ( ) ) n ) n ( ) n. (4.35) This converges for > 3 and can also be re-indeed to verif that this series involves negative powers of.

27 comple analsis 43 This leads to the following theorem: Theorem 4.8. Let f () be analtic in an annulus, R < < R, with a positivel oriented simple closed curve around and inside the annulus as shown in Figure Then, f () j a j ( ) j + j b j ( ) j, with and a j f () d, πi ( ) j+ b j f () d. πi ( ) j+ The above series can be written in the more compact form f () c j ( ) j. j Such a series epansion is called a Laurent series epansion, named after its discoverer Pierre Alphonse Laurent (83-854). R R Eample 4.4. Epand f () in the annulus < <. ( )(+) Using partial fractions, we can write this as f () 3 [ + ]. + We can epand the first fraction,, as an analtic function in the region > and the second fraction, +, as an analtic function in <. This is done as follows. First, we write Figure 4.34: This figure shows an annulus, R < < R, with a positivel oriented simple closed curve around and inside the annulus. Then, we write + [ ( )] ( ) n. n [ ] n n. Therefore, in the common region, < <, we have that ( )( + ) 3 [ ( ) n n n ( ) n 6( n n ) n + n ] n+ ( ) 3 n. (4.36) We note that this is not a Talor series epansion due to the eistence of terms with negative powers in the second sum.

28 44 fourier and comple analsis Eample 4.5. Find series representations of f () throughout the ( )(+) comple plane. In the last eample we found series representations of f () in the ( )(+) annulus < <. However, we can also find epansions which converge for other regions. We first write f () [ 3 + ]. + We then epand each term separatel. The first fraction,, can be written as the sum of the geometric series n, <. n 3 3 i i i i Figure 4.35: Regions of convergence for Laurent epansions of f () ( )(+). 4 This series converges inside the unit circle. We indicate this b region in Figure In the last eample, we showed that the second fraction, +, has the series epansion + [ ( )] ( ) n. n which converges in the circle <. This is labeled as region in Figure Regions and intersect for <, so, we can combine these two series representations to obtain [ ] ( )( + ) n + ( 3 ) n, <. n n In the annulus, < <, we had alread seen in the last eample that we needed a different epansion for the fraction. We looked for an epansion in powers of / which would converge for large values of. We had found that ( ) n, >. n This series converges in region 3 in Figure ombining this series with the one for the second fraction, we obtain a series representation that converges in the overlap of regions and 3. Thus, in the annulus < <, we have ( )( + ) 3 [ ( ) n n n n+ So far, we have series representations for <. The onl region not covered et is outside this disk, >. In Figure 4.35 we see that series 3, which converges in region 3, will converge in the last section of the comple plane. We just need one more series epansion for /( + ) for large. Factoring out a in the denominator, we can write this as a geometric series with r / : + [ + ] ( ) n. n ].

29 comple analsis 45 This series converges for >. Therefore, it converges in region 4 and the final series representation is [ ( )( + ) ( 3 ) ] n n n n Singularities and The Residue Theorem In the last section we found, that we could integrate functions satisfing some analticit properties along contours without using detailed parametriations around the contours. We can deform contours if the function is analtic in the region between the original and new contour. In this section we will etend our tools for performing contour integrals. The integrand in the auch Integral Formula was of the form g() f (), where f () is well behaved at. The point is called a singu- Singularities of comple functions. larit of g(), as g() is not defined there. More specificall, a singularit of f () is a point at which f () fails to be analtic. We can also classif these singularities. Tpicall, these are isolated singularities. As we saw from the proof of the auch Integral Formula, g() f () has a Laurent series epansion about, given b g() f ( ) + f ( ) + f ( )( ) It is the nature of the first term that gives information about the tpe of singularit that g() has. Namel, in order to classif the singularities of f (), we look at the principal part of the Laurent series of f () about, j b j( ) j, which consists of the negative powers of. There are three tpes of singularities: removable, poles, and essential singularities. The are defined as follows: lassification of singularities.. If f () is bounded near, then is a removable singularit.. If there are a finite number of terms in the principal part of the Laurent series of f () about, then is called a pole. 3. If there are an infinite number of terms in the principal part of the Laurent series of f () about, then is called an essential singularit. Eample 4.6. f () sin has a removable singularit at. At first it looks like there is a possible singularit at, since the denominator is ero at. However, we know from the first semester of calculus that. Furthermore, we can epand sin about and see that lim sin sin 3 ( +...) 3! 3! Thus, there are onl nonnegative powers in the series epansion. So, is a removable singularit.

30 46 fourier and comple analsis Simple pole. Double pole. Eample 4.7. f () e ( ) n has poles at for n a positive integer. For n, we have f () e. This function has a singularit at. The series epansion is found b epanding e about : f () e e e + e + e ( ) +....! Note that the principal part of the Laurent series epansion about onl has e one term,. Therefore, is a pole. Since the leading term has an eponent of, is called a pole of order one, or a simple pole. For n we have f () e. The series epansion is again found b ( ) epanding e about : f () e ( ) e e ( ) + e + e! + e ( ) ! Note that the principal part of the Laurent series has two terms involving ( ) and ( ). Since the leading term has an eponent of, is called a pole of order, or a double pole. Eample 4.8. f () e has an essential singularit at. In this case, we have the series epansion about given b ( ) n n! n! n. f () e n n We see that there are an infinite number of terms in the principal part of the Laurent series. So, this function has an essential singularit at. Poles of order k. In the above eamples we have seen poles of order one (a simple pole) and two (a double pole). In general, we can sa that f () has a pole of order k at if and onl if ( ) k f () has a removable singularit at, but ( ) k f () for k > does not. Eample 4.9. Determine the order of the pole of f () cot csc at. First we rewrite f () in terms of sines and cosines. f () cot csc cos sin. We note that the denominator vanishes at. How do we know that the pole is not a simple pole? Well, we check to see if ( ) f () has a removable singularit at : lim cos ( ) f () lim ( sin lim sin ) ( lim ) cos sin lim cos sin. (4.37) We see that this limit is undefined. So now we check to see if ( ) f () has a removable singularit at : lim ( ) f () lim cos sin

31 comple analsis 47 ( lim lim ) ( lim ) cos sin sin cos(). (4.38) sin In this case, we have obtained a finite, nonero, result. So, is a pole of order. We could have also relied on series epansions. Epanding both the sine and cosine functions in a Talor series epansion, we have f () cos sin! +... ( 3! ). Factoring a from the epansion in the denominator, f ()! +... ( 3! +.. ( ).) + O( ), we can see that the leading term will be a /, indicating a pole of order. We will see how knowledge of the poles of a function can aid in the computation of contour integrals. We now show that if a function, f (), has a pole of order k, then f () d πi Res[ f (); ], Integral of a function with a simple pole inside. Residues of a function with poles of order k. where we have defined Res[ f (); ] as the residue of f () at. In particular, for a pole of order k the residue is given b Residues for Poles of order k d Res[ f (); ] k [ ] lim (k )! d k ( ) k f (). (4.39) Proof. Let φ() ( ) k f () be an analtic function. Then φ() has a Talor series epansion about. As we had seen in the last section, we can write the integral representation of an derivative of φ as φ (k ) ( ) (k )! πi Inserting the definition of φ(), we then have φ (k ) ( ) (k )! πi φ() ( ) k d. f () d. Solving for the integral, we have the following result: πi d f () d k [ ] (k )! d k ( ) k f () πi Res[ f (); ] (4.4)

32 48 fourier and comple analsis Note: If is a simple pole, the residue is easil computed as Res[ f (); ] lim ( ) f (). The residue for a simple pole. In fact, one can show (Problem 8) that for g and h analtic functions at, with g( ), h( ), and h ( ), [ ] g() Res h() ; g( ) h ( ). Eample 4.3. Find the residues of f () (+) ( +4). f () has poles at, i, and i. The pole at is a double pole (pole of order ). The other poles are simple poles. We compute those residues first: Res[ f (); i] lim ( i) i ( + ) ( + i)( i) lim i ( + ) ( + i) i (i + ) (4i) 5 i. (4.4) Res[ f (); i] lim i ( + ) ( + i)( i) lim i ( + ) ( i) i ( i + ) ( 4i) 5 + i. (4.4) The residue of f () at is the coefficient of the ( ) term, c b, of the Laurent series epansion about. For the double pole, we have to do a little more work. [ ] d Res[ f (); ] lim ( + ) d ( + ) ( + 4) [ ] d lim d + 4 d lim d lim d d [ ] + 4 ( ) ( + 4) [ ] ( + 4) 5. (4.43) Eample 4.3. Find the residue of f () cot at. We write f () cot cos sin and note that is a simple pole. Thus, cos Res[cot ; ] lim cos(). sin Another wa to find the residue of a function f () at a singularit is to look at the Laurent series epansion about the singularit. This is because the residue of f () at is the coefficient of the ( ) term, or c b.

33 comple analsis 49 Eample 4.3. Find the residue of f () at using a Laurent series (3 ) epansion. First, we need the Laurent series epansion about of the form c n n. A partial fraction epansion gives f () (3 ) ( 3 + ). 3 The first term is a power of. The second term needs to be written as a convergent series for small. This is given b Thus, we have found 3 f () 3 3( /3) ( ) n 3. (4.44) 3 n ( ) + ( ) n 3. 3 n The coefficient of can be read off to give Res[ f (); ] 3. Eample Find the residue of f () cos at using a Laurent series epansion. In this case, is an essential singularit. The onl wa to find residues at essential singularities is to use Laurent series. Since Finding the residue at an essential singularit. cos! + 4! 4 6! , then we have ( f ()! + 4! 4 ) 6! ! + 4! 3 6! (4.45) From the second term we have that Res[ f (); ]. We are now read to use residues in order to evaluate integrals. d sin. Eample Evaluate We begin b looking for the singularities of the integrand. These are located at values of for which sin. Thus,, ±π, ±π,..., are the singularities. However, onl lies inside the contour, as shown in Figure We note further that is a simple pole, since Figure 4.36: d sin. i i ontour for computing Therefore, the residue is one and we have lim ( ) sin. d sin πi.

34 5 fourier and comple analsis The Residue Theorem. In general, we could have several poles of different orders. For eample, we will be computing d. The integrand has singularities at, or ±. Both poles are inside the contour, as seen in Figure One could do a partial fraction decomposition and have two integrals with one pole each integral. Then, the result could be found b adding the residues from each pole. In general, when there are several poles, we can use the Residue Theorem: The Residue Theorem Theorem 4.9. Let f () be a function which has poles j, j,..., N inside a simple closed contour and no other singularities in this region. Then, f () d πi N j Res[ f (); j ], (4.46) where the residues are computed using Equation (4.39), d Res[ f (); ] k [ ] lim (k )! d k ( ) k f (). Figure 4.37: A depiction of how one cuts out poles to prove that the integral around is the sum of the integrals around circles with the poles at the center of each. The proof of this theorem is based upon the contours shown in Figure One constructs a new contour b encircling each pole, as show in the figure. Then one connects a path from to each circle. In the figure two separated paths along the cut are shown onl to indicate the direction followed on the cut. The new contour is then obtained b following and crossing each cut as it is encountered. Then one goes around a circle in the negative sense and returns along the cut to proceed around. The sum of the contributions to the contour integration involve two integrals for each cut, which will cancel due to the opposing directions. Thus, we are left with f () d f () d f () d f () d f () d. 3 Of course, the sum is ero because f () is analtic in the enclosed region, since all singularities have been cut out. Solving for f () d, one has that this integral is the sum of the integrals around the separate poles, which can be evaluated with single residue computations. Thus, the result is that f () d is πi times the sum of the residues. d. Eample Evaluate We first note that there are two poles in this integral since ( )( + ). In Figure 4.38 we plot the contour and the two poles, denoted b an. Since both poles are inside the contour, we need to compute the residues for each one. The are each simple poles, so we have

35 comple analsis 5 and Then, [ ] Res ; [ ] Res ; lim( ) lim +, (4.47) lim lim. (4.48) d πi( ). i i i i Eample Evaluate + 3 ( ) (+) d. In this eample, there are two poles, inside the contour. [See Figure 4.39.] is a second-order pole and is a simple pole. Therefore, we need to compute the residues at each pole of f () + ( ) (+) : Res[ f (); ] lim! lim [ ( ) + d d ( + 4 ( + ) ( ) ( + ) ) 4 9. (4.49) ] Figure 4.38: d. ontour for computing 3i Res[ f (); ] lim ( + ) + ( ) ( + ) + lim ( ) 5 9. (4.5) The evaluation of the integral is found b computing πi times the sum of the residues: ( + 4 ( ) d πi ( + ) ) πi. 9 3 Eample ompute 3 e / d. In this case, is an essential singularit and is inside the contour. Laurent series epansion about gives A i i 3 3 i i 3i 3 Figure 4.39: ontour for computing + 3 ( ) (+) d. 3 e / 3 n n! n n! 3 n n ( ) n 3 +! + 4 3! + 8 4! + 6 5! (4.5)

36 5 fourier and comple analsis The residue is the coefficient of, or Res[ 3 e / ; ] 5. Therefore, 3 e / d 4 5 πi. omputation of integrals of functions of sines and cosines, f (cos θ, sin θ). Eample Evaluate π dθ +cos θ. Here we have a real integral in which there are no signs of comple functions. In fact, we could appl simpler methods from a calculus course to do this integral, attempting to write + cos θ cos θ. However, we do not get ver far. One trick, useful in computing integrals whose integrand is in the form f (cos θ, sin θ), is to transform the integration to the comple plane through the transformation e iθ. Then, cos θ eiθ + e iθ sin θ eiθ e iθ i i ( + ), ( ). Under this transformation, e iθ, the integration now takes place around the unit circle in the comple plane. Noting that d ie iθ dθ i dθ, we have 4 3 Figure 4.4: π dθ +cos θ. i i ontour for computing π dθ + cos θ i i + d i ( + d ) + ( + ) d (4.5) We can appl the Residue Theorem to the resulting integral. The singularities occur at the roots of Using the quadratic formula, we have the roots ± 3. The location of these poles are shown in Figure 4.4. Onl + 3 lies inside the integration contour. We will therefore need the residue of f () at this simple pole: i +4+ Res[ f (); + 3] lim + ( ( + i 3)) i lim + 3 i lim + 3 i ( + 3) ( ( + 3))( ( 3)) ( 3) + 3 ( 3) i 3 i 3. (4.53) 3

37 comple analsis 53 Therefore, we have π ( ) dθ + cos θ i d i πi 3 π 3. (4.54) 3 Before moving on to further applications, we note that there is another wa to compute the integral in the last eample. Karl Theodor Wilhelm Weierstraß (85-897) introduced a substitution method for computing integrals involving rational functions of sine and cosine. One makes the substitution t tan θ and converts the integrand into a rational function of t. One can show that this substitution implies that The Weierstraß substitution method. and sin θ t t, cos θ + t + t, dθ dt + t. The details are left for Problem 8. In order to see how it works, we will appl the Weierstraß substitution method to the last eample. Eample Appl the Weierstraß substitution method to compute π dθ +cos θ. π dθ + cos θ + t +t dt + t dt t + 3 ( )] 3 3 [tan 3 3 t π 3. (4.55) Infinite Integrals Infinite integrals of the form f () d occur often in phsics. The can represent wave packets, wave diffraction, Fourier transforms, and arise in other applications. In this section, we will see that such integrals ma be computed b etending the integration to a contour in the comple plane. Recall from our calculus eperience that these integrals are improper integrals. The wa that one determines if improper integrals eist, or converge, is to carefull compute these integrals using limits such as R f () d lim f () d. R R For eample, we evaluate the integral of f () as R ( R ) d lim d lim R R R ( R).

38 54 fourier and comple analsis One might also be tempted to carr out this integration b splitting the integration interval, (, ] [, ). However, the integrals d and d do not eist. A simple computation confirms this. R ( R ) d lim d lim. R R The auch principal value integral. Therefore, f () d f () d + f () d R does not eist while lim R R f () d does eist. We will be interested in computing the latter tpe of integral. Such an integral is called the auch Principal Value Integral and is denoted with either a P, PV, or a bar through the integral: P f () d PV R f () d f () d lim f () d. (4.56) R R If there is a discontinuit in the integral, one can further modif this definition of principal value integral to bpass the singularit. For eample, if f () is continuous on a b and not defined at [a, b], then b ( ɛ b ) f () d lim f () d + f () d. a ɛ a +ɛ In our discussions we will be computing integrals over the real line in the auch principal value sense. Eample 4.4. ompute d in the auch Principal Value sense. 3 In this case, f () is not defined at. So, we have 3 ( d ɛ 3 lim ɛ lim ɛ d 3 + ɛ ) d 3 ) ( ɛ ɛ. (4.57) omputation of real integrals b embedding the problem in the comple plane. R R Γ R Figure 4.4: ontours for computing P f () d. R We now proceed to the evaluation of principal value integrals using comple integration methods. We want to evaluate the integral f () d. We will etend this into an integration in the comple plane. We etend f () to f () and assume that f () is analtic in the upper half plane (Im() > ) ecept at isolated poles. We then consider the integral R R f () d as an integral over the interval ( R, R). We view this interval as a piece of a larger contour R obtained b completing the contour with a semicircle Γ R of radius R etending into the upper half plane as shown in Figure 4.4. Note that a similar construction is sometimes needed etending the integration into the lower half plane (Im() < ) as we will later see. The integral around the entire contour R can be computed using the Residue Theorem and is related to integrations over the pieces of the contour b R f () d f () d + f () d. (4.58) R Γ R R

39 comple analsis 55 Taking the limit R and noting that the integral over ( R, R) is the desired integral, we have P f () d f () d lim f () d, (4.59) R Γ R where we have identified as the limiting contour as R gets large. Now the ke to carring out the integration is that the second integral vanishes in the limit. This is true if R f () along Γ R as R. This can be seen b the following argument. We parametrie the contour Γ R using Re iθ. Then, when f () < M(R), π f () d f (Re iθ )Re iθ dθ Γ R π R f (Re iθ ) dθ < RM(R) π dθ πrm(r). (4.6) So, if lim R RM(R), then lim R Γ R f () d. We now demonstrate how to use comple integration methods in evaluating integrals over real valued functions. Eample 4.4. Evaluate d. + We alread know how to do this integral using calculus without comple analsis. We have that d ( ) ( π ) + lim tan R π. R We will appl the methods of this section and confirm this result. The needed contours are shown in Figure 4.4 and the poles of the integrand are at ±i. We first write the integral over the bounded contour R as the sum of an integral from R to R along the real ais plus the integral over the semicircular arc in the upper half comple plane, R d R + d R + + Γ R d +. Net, we let R get large. We first note that f () goes to ero fast enough on Γ + R as R gets large. R f () R + R e iθ Thus, as R, R f () and R. So, R + R cos θ + R 4. d + d +. We need onl compute the residue at the enclosed pole, i. Res[ f (); i] lim( i) i + lim i + i i. R Figure 4.4: d +. Γ R i R i ontour for computing

40 56 fourier and comple analsis Then, using the Residue Theorem, we have ( ) d + πi π. i Eample 4.4. Evaluate P sin d. For this eample, the integral is unbounded at. onstructing the contours as before we are faced for the first time with a pole ling on the contour. We cannot ignore this fact. We can proceed with the computation b carefull going around the pole with a small semicircle of radius ɛ, as shown in Figure Then the principal value integral computation becomes ( sin ɛ sin R ) sin P d lim d + d. (4.6) ɛ,r R ɛ We will also need to rewrite the sine function in terms of eponentials in this integral. There are two approaches that we could take. First, we could emplo the definition of the sine function in terms of comple eponentials. This gives two integrals to compute: sin P d ( e i P i d P e i ) d. (4.6) ε Γ R The other approach would be to realie that the sine function is the imaginar part of an eponential, Im e i sin. Then, we would have ( sin e P d i ) Im P d. (4.63) R ε ε R Figure 4.43: ontour for computing P sin d. We first consider P e i d, which is common to both approaches. We use the contour in Figure Then we have R e i d e i ɛ Γ R d + R e i d + e i R ɛ d + ɛ e i d. The integral e i R d vanishes since there are no poles enclosed in the contour! The sum of the second and fourth integrals gives the integral we seek as ɛ and R. The integral over Γ R will vanish as R gets large according to Jordan s Lemma. Jordan s Lemma gives conditions for the vanishing of integrals over Γ R as R gets large. We state a version of Jordan s Lemma here for reference and give a proof at the end of this chapter. Jordan s Lemma If f () converges uniforml to ero as, then lim f ()e ik d, R R where k > and R is the upper half of the circle R.

41 comple analsis 57 A similar result applies for k <, but one closes the contour in the lower half plane. [See Section for the proof of Jordan s Lemma.] The remaining integral around the small semicircular arc must be done separatel. We have ɛ e i d π ep(iɛe iθ ) π ɛe iθ iɛe iθ dθ i ep(iɛe iθ ) dθ. Taking the limit as ɛ goes to ero, the integrand goes to i and we have So far, we have that e i P ɛ e i d πi. Note e d lim ɛ ɛ i d πi. At this point, we can get the answer using the second approach in Equation (4.63). Namel, ( sin e P d i ) Im P d Im(πi) π. (4.64) that we have not previousl done integrals in which a singularit lies on the contour. One can show, as in this eample, that points on the contour can be accounted for using half of a residue (times πi). For the semicircle ɛ, the reader can verif this. The negative sign comes from going clockwise around the semicircle. It is instructive to carr out the first approach in Equation (4.6). We will need to compute P e i d. This is done in a similar maaner to the above computation, being careful with the sign changes due to the orientations of the contours as shown in Figure We note that the contour is closed in the lower half plane. This is because k < in the application of Jordan s Lemma. One can understand wh this is the case from the following observation. onsider the eponential in Jordan s Lemma. Let R + i I. Then, e ik e ik( R+i I ) e k I e ik R. As gets large, the second factor just oscillates. The first factor would go to ero if k I >. So, if k >, we would close the contour in the upper half plane. If k <, then we would close the contour in the lower half plane. In the current computation, k, so we use the lower half plane. Working out the details, we find the same value for e i P d πi. Finall, we can compute the original integral as sin P d ( e i P i d P e i ) d (πi + πi) i π. (4.65) R ε ε ε Γ R Figure 4.44: ontour in the lower half plane for computing P e i d. R This is the same result as we obtained using Equation(4.63).

42 58 fourier and comple analsis i i Figure 4.45: Eample with poles on contour. d +. Eample Evaluate In this eample, there are two simple poles, ±i ling on the contour, as seen in Figure This problem is similar to Problem c, ecept we will do it using contour integration instead of a parametriation. We bpass the two poles b drawing small semicircles around them. Since the poles are not included in the closed contour, the Residue Theorem tells us that the integral over the path vanishes. We can write the full integration as a sum over three paths: ± for the semicircles and for the original contour with the poles cut out. Then we take the limit as the semicircle radii go to ero. So, d d + + d +. The integral over the semicircle around i can be done using the parametriation i + ɛe iθ. Then + iɛe iθ + ɛ e iθ. This gives + d π + lim iɛe iθ ɛ iɛe iθ + ɛ e iθ dθ π dθ π. [ ] As in the last eample, we note that this is just πi times the residue, Res + ; i i. Since the path is traced clockwise, we find that the contribution is πires π, which is what we obtained above. A Similar computation will give the contribution from i as π. Adding these values gives the total contribution from ± as ero. So, the final result is that d +. Eample Evaluate e a +e d, for < a <. In dealing with integrals involving eponentials or hperbolic functions it is sometimes useful to use different tpes of contours. This eample is one such case. We will replace with and integrate over the contour in Figure Letting R, the integral along the real ais is the integral that we desire. The integral along the path for π leads to a multiple of this integral since + πi along this path. Integration along the vertical paths vanishes as R. This is captured in the following integrals: R + πi R R + πi Figure 4.46: Eample using a rectangular contour. R R e a + e d R e a π R + e d + R + R e a(+πi) + e +πi d + e a(r+i) d + er+i π e a( R+i) d (4.66) + e R+i We can now let R. For large R, the second integral decas as e (a )R and the fourth integral decas as e ar. Thus, we are left with e a ( R d lim + e R ( e πia ) e a d eπia + e R R R e a ) + e d e a d. (4.67) + e

43 comple analsis 59 We need onl evaluate the left contour integral using the Residue Theorem. The poles are found from + e. Within the contour, this is satisfied b iπ. So, [ e a ] Res + e ; iπ e lim ( iπ) a iπ + e eiπa. Appling the Residue Theorem, we have Therefore, we have found that ( e πia e ) a + e d πieiπa. e a πieiπa d + e e πia π sin πa, < a < Integration over Multivalued Functions We have seen that some comple functions inherentl possess multivaluedness; that is, such functions do not evaluate to a single value, but have man values. The ke eamples were f () /n and f () ln. The nth roots have n distinct values and logarithms have an infinite number of values as determined b the range of the resulting arguments. We mentioned that the wa to handle multivaluedness is to assign different branches to these functions, introduce a branch cut, and glue them together at the branch cuts to form Riemann surfaces. In this wa we can draw continuous paths along the Riemann surfaces as we move from one Riemann sheet to another. Before we do eamples of contour integration involving multivalued functions, let s first tr to get a handle on multivaluedness in a simple case. We will consider the square root function, w / r / e i( θ +kπ), k,. There are two branches, corresponding to each k value. If we follow a path not containing the origin, then we sta in the same branch, so the final argument (θ) will be equal to the initial argument. However, if we follow a path that encloses the origin, this will not be true. In particular, for an initial point on the unit circle, e iθ, we have its image as w e iθ /. However, if we go around a full revolution, θ θ + π, then e iθ +πi e iθ, but w e (iθ +πi)/ e iθ / e πi w. Here we obtain a final argument (θ) that is not equal to the initial argument! Somewhere, we have crossed from one branch to another. Points, such as

44 6 fourier and comple analsis the origin in this eample, are called branch points. Actuall, there are two branch points, because we can view the closed path around the origin as a closed path around comple infinit in the compactified comple plane. However, we will not go into that at this time. We can demonstrate this in the following figures. In Figure 4.47 we show how the points A through E are mapped from the -plane into the w-plane under the square root function for the principal branch, k. As we trace out the unit circle in the -plane, we onl trace out a semicircle in the w- plane. If we consider the branch k, we then trace out a semicircle in the lower half plane, as shown in Figure 4.48 following the points from F to J. Figure 4.47: In this figure we show how points on the unit circle in the -plane are mapped to points in the w-plane under the principal square root function. B A E E D B A D Figure 4.48: In this figure we show how points on the unit circle in the -plane are mapped to points in the w-plane under the square root function for the second branch, k. H G F J F J I G H I Figure 4.49: In this figure we show the combined mapping using two branches of the square root function. B A E D D B E F A J G G H I H F J I We can combine these into one mapping depicting how the two comple

45 comple analsis 6 planes corresponding to each branch provide a mapping to the w-plane. This is shown in Figure A common wa to draw this domain, which looks like two separate comple planes, would be to glue them together. Imagine cutting each plane along the positive -ais, etending between the two branch points, and. As one approaches the cut on the principal branch, one can move onto the glued second branch. Then one continues around the origin on this branch until one once again reaches the cut. This cut is glued to the principal branch in such a wa that the path returns to its starting point. The resulting surface we obtain is the Riemann surface shown in Figure 4.5. Note that there is nothing that forces us to place the branch cut at a particular place. For eample, the branch cut could be along the positive real ais, the negative real ais, or an path connecting the origin and comple infinit. We now look at eamples involving integrals of multivalued functions. Eample Evaluate + d. + d. We consider the contour integral The first thing we can see in this problem is the square root function in the integrand. Being that there is a multivalued function, we locate the branch point and determine where to draw the branch cut. In Figure 4.5 we show the contour that we will use in this problem. Note that we picked the branch cut along the positive -ais. We take the contour to be positivel oriented, being careful to enclose the two poles and to hug the branch cut. It consists of two circles. The outer circle R is a circle of radius R and the inner circle ɛ will have a radius of ɛ. The sought-after answer will be obtained b letting R and ɛ. On the large circle we have that the integrand goes to ero fast enough as R. The integral around the small circle vanishes as ɛ. We can see this b parametriing the circle as ɛe iθ for θ [, π] : π ɛe + d iθ + (ɛe iθ ) iɛeiθ dθ ɛ iɛ 3/ π e 3iθ/ + (ɛ e iθ dθ. (4.68) ) It should now be eas to see that as ɛ, this integral vanishes. The integral above the branch cut is the one we are seeking, d. The + integral under the branch cut, where re πi, is re + d πi + r dr e4πi r dr. (4.69) + r We note that this is the same as that above the cut. Up to this point, we have that the contour integral, as R and ɛ, is + d + d. Figure 4.5: Riemann surface for f () /. R i ɛ i Figure 4.5: An eample of a contour which accounts for a branch cut.