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1 Sum, Product, Modulus, onjugate, Definition.. Given (, y) R 2, a comple number z is an epression of the form z = + iy. (.) Given a comple number of the form z = + iy we define Re z =, the real part of z, (.2) Im z = y, the imaginary part of z. (.3) Eample.2. Let z = 2 3i, then Re z = 2 and Im z = 3. note that Im z is a real number. 2 3 z = 2 3i Definition.3. The set of all comple numbers is denoted by. If y = 0 in (.), then z = is a real number. In this sense R, If = 0, then z = iy is called a purely imaginary number. Eample.4. y = i, y = i, y = 3i, are all purely imaginary 3i i i The y ais is called the imaginary ais, as all the purely imaginary numbers are on the y ais.

2 . The sum of two comple numbers If z = + iy and w = u + iv, then z + w = ( + u) + i(y + v). (.4) Therefore, we add the real parts of z and w to get the real part of the sum. We also add the imaginary parts of z an w to get the imaginary part of the sum. Eample.5. If z = 5 + i and w = 2 3i, then z + w = (5 + 2) + i( 3) = 7 2i. Geometrically, the sum is obtained from parallelogram law for the sum of vectors: z = 5 + i z = 2 3i z + w = 7 i The modulus of a comple number z = + iy, denoted by z is the length of the vector, y. Therefore z = + iy = 2 + y 2. (.5) Eample.6. If z = i, then z is the length of the vector 0,, therefore z = =. Eample.7. The set of all comple numbers z such that z = 2 form a circle centered at the origin with radius 2. z = 2 2 Triangle inequality: If z and w are two comple numbers, then z + w z + w. (.6) 2

3 .2 The product of two comple numbers The product of two comple numbers is a comple number. If z = + iy and w = u + iv, then zw = (u yv) + i(v + yu) (.7) Therefore, Re (zw) = u yv, Im (zw) = v + yu. Eample.8. Let z = + 2i and w = + 3i, then zw = ( 6) + i(3 2); zw = 7 + i Eample.9. if z = i, then i 2 = zz = + i(0 + 0) =, i 2 = (.8) Remark.0. The algebraic manipulation of comple epression follows the same for the real epression, ecept we replace i 2 by. Eample.. (2 + 3i)( + i) = i + 3i + 3i i = 2 + 2i + 3i 3 = + 5i i + i = (2 + 3i)( i) ( + i)( i) 2 2i + 3i 3i2 = = 5 + i = i 2 Every comple number z 0 has a multiplicative inverse w = /z such that wz =. If z = + iy then z = iy 2 + y 2 = 2 + y 2 + i y 2 + y 2. (.9) Therefore Re z = 2 + y 2, Im z = y 2 + y 2. (.0) Eample.2. If z = i, then z = i: i = i. The comple conjugate of a comple number z = +iy, denoted z, is defined by z = iy. Geometrically, z is the reflection of z in the ais. If we reflect twice, we return to z: z = z 3

4 z = 2 + 3i z = 2 3i Some other properties of comple conjugate are. z + w = z + w. 2. zw = z w. 3. z = z, 4. z 2 = z z. (Therefore, z z is always a real number) Eample.3. Homework + 3i 2 i =. Show zw = z w. 2. Show /z = z/ z Show z 2 = z z. ( + 3i)(2 + i) (2 i)(2 + i) 4. If z = ± i, then z 2 2z + 2 = 0. = 5 + 5i 5 + 5i = = + i. 2 i Epress the following comple numbers in the form a + ib (a) 6. Show that 2+3i 4+i, i + 3 +i, (8 + 5i)2 ( ) i z 2i = z + 2i, i 2 = +2i 5, i3 = i i 4 = i 5 = i, i 27 = i 7. Identify the set of all comple numbers z such that < z Identify the set of all comple numbers z such that Im z. 9. Identify the set of all comple numbers z such that < Re z <. 0. Identify the set of all comple numbers z such that z i <. 4

5 2 Polar Representation (eponential form) A point (, y) (0, 0) in the plane can be described by polar coordinates r and θ, where r = z = 2 + y 2 > 0 and the real number θ represents the angle that the vector, y makes with the positive real ais. θ is measured in radian, lies in the same quadrant as (, y) does, and satisfies tan θ = y. θ is not unique. There are infinitely many θ s, with 2nπ distance from each other. Each value of θ is called an argument of z = + iy. arg z is the set of all arguments of z. There is one and only one θ in the interval ( π, π]. The principal value of arg z denoted by Arg z, is that unique value of Θ such that π < Θ π. arg z = {Arg z + 2nπ n = 0, ±, ±2, }. (2.) Eample 2.. Let Θ denote Arg z.. z =, then r = and Θ = 0, 2. z = 2, then r = 2 and Θ = π. 3. z = + i, then r = 2 and Θ = π 4, 4. z = + i, then r = 2 and Θ = 3π 4, 5. z = i, then r = 2, and Θ = 3π 4, 6. z = i, then r = 2, and Θ = π 4. z = 3 + 2i Θ z = 3 2i Let z = + iy. If r = z, and θ arg z, then = r cos θ, y = r cos θ. (2.2) Therefore, we can write z = r cos θ + ir sin θ = r(cos θ + i sin θ). (2.3) 5

6 Net we define the symbol e iθ or ep(iθ) via Euler s formula as Therefore, we can write z in its polar form as Eample z = in polar form is z = 2. z = in polar form is z = e iπ, therefore 3. z = 2 in polar form is z = 2e iπ. 4. i = e iπ/2 and i = e iπ/2. 5. z = + i in polar form is z = 2e iπ/4 6. z = + i in polar form is z = 2e i3π/4. 7. z = i in polar from is z = 2e iπ/4. e iθ = cos θ + i sin θ. (2.4) z = re iθ = z e iθ. (2.5) e iπ =. (2.6) Eample z = 2e iπ/3 in standard form is z = 2(cos π 3 + i sin π 3 ) = + i 3. Remark 2.4. It is clear from the Euler s formula (2.4) that e iθ is periodic of period 2π, i.e., e iθ = e i(θ+2kπ) e iθ+2kπ In general r e iθ = r 2 e iθ2 (r = r 2 and θ = θ 2 + 2kπ) Eample 2.5. ompute ( + i) 0. Solution. Since + i = 2e iπ/4, ( + i) 0 = [ 2e iπ/4 ] 0 = 2 0 (e iπ/4 ) 0 = 32e i5π/2. Using Euler s formula (2.4) you can convert this to the standard form to get ( + i) 0 = 32(cos 5π 2 + i sin 5π 2 ) = 32i. 6

7 2. Roots of omple Numbers We use the polar form to find the roots of a comple number. Eample 2.6. Find the 3rd roots of 5. Solution.. We have to find all comple numbers z such that z 3 = 5, i.e., we want to find all the solutions (roots) of p(z) = 0, where p(z) = z If p(z) is a polynomial of degree n, then the equation p(z) = 0 has n solutions. 3. Then five has three cube roots. To find them we write z in its polar form: z = re iθ, where r = z and θ = Arg z. Since 5 = 5e iπ in polar form, then Therefore (re iθ ) 3 = 5e iπ r 3 e i3θ = 5e iπ r 3 = 5, and 3θ = π + 2kπ, k = 0, ±, ±2, r = 5 /3, and θ = π 3 + 2kπ, k = 0, ±, ±2, 3 Observe that only k = 0,, 2 produce the three distinct roots: The first root corresponds to k = 0 : z 0 = 5 /3 e i π 3 Th second root corresponds to k = : The third root corresponds to k = 2 : z = 5 /3 e iπ = 5 /3 z 2 = 5 /3 e i 5π 3 = 5 /3 e i π 3 Notice that z 0 = z = z 2 = 5 /3. Therefore, all the roots are on a circled centered at zero with radius 5 /3 : z 0 z z 2 7

8 Eample 2.7. Solve z 5 = i. Solution. Let z = re iθ, where r = z, then z 5 = r 5 e i5θ. Since i = 2e i3π/4, then z 5 = + i r 5 e i5θ = 2e i3π/4 Therefore, r 5 = 2 and 5θ = 3π 4 + 2kπ. r = 2 /0 and θ = 3π kπ 5, where k = 0,, 2, 3, 4. All the roots are on circle centered at the origin, with radius 2 /0 : The first root corresponds to k = 0 : z 0 = 2 0 e i 3π 20 Th second root corresponds to k = : z = 2 0 e iπ/4 The third root corresponds to k = 2 : z 2 = 2 0 e i 3π 20 The fourth root corresponds to k = 3 : z 3 = 2 0 e i 2π 20 The fifth root corresponds to k = 4 : z 4 = 2 0 e i 29π 20 z 2 z z 3 the radius is 2 /0 z 0 z 4 8

9 3 Maps 3. f(z) = z 2 Eample 3.. Polar Let f(z) = z 2 on an annulus domain. Let D f = {z : 2 < z 3, π/2 Arg z π},. Domain z plane D f f maps each point in D f to a point w = f(z). For eample 3i D f and f maps 3i to w = f(3i) which is w = 9. Similarly 3 f 9: f( 3) = To see how f moves points around, let z = re iθ, then f(z) = z 2 = (re iθ ) 2 = r 2 e i(2θ). i.e., f doubles the angle and squares the modulus of the points in D f. 4. Therefore, f takes D f to R f = {w : 4 < w 9, π Arg z 0}, D f f Rf 9

10 w plane 9 4 R f Eample 3.2. Let f(z) = z 2. If z = u+iv, then under f the image of z = u+iv to (u 2 v 2 ) + i2uv. onclusions:. All points in u-v plane on the curve u 2 v 2 = get mapped to the vertical line + si, where s = 2uv: v u2 v2 = u Remark 3.3. The graph of u 2 v 2 = is the 45 rotation of the wellknown graph of uv = (or u = /v). When z = u + iv moves on this graph, u 2 v 2 remains and you can check that 2uv varies between and. Therefore the image of the points on this graph, under f(z) = z 2 is the vertical line w = + is for s (, ): 0

11 w plane 2. All point in u-v plane on the curve 2uv = get mapped to the horizontal line w = r + i, where r = u 2 v 2 : v 2uv = u When z = u + iv moves on this graph, 2uv remains and you can check that u 2 v 2 varies between and. Therefore the image of the points on this graph, under f(z) = z 2 is the vertical line w = s + i for s (, ): w plane w = s + i Eample 3.4. Find the image of the region between the two curves in the following figure under the map f(z) = z 2

12 v u 2 v 2 = w = + is u u 2 v 2 = 3 w = 3 + is 3.2 f(z) = e z Recall that we defined e iy := cos y + i sin y. If z = + iy, we define e z by e z = e +iy := e e iy = e (cos y + i sin y). (3.) Remark 3.5. Let z = + iy. We can write e z in its polar form: e z = re iθ, where r = e = e z, θ = y. Since r = e > 0 for all R, then e z 0 for all z. When, then r 0, but it never touches zero.. If z moves on a horizontal line, then y remains constant. Therefore, e z moves on ray emanating from the origin: y z = + 5π 3 i z = + π 6 i π/6 e z e z 2. If z moves on a vertical line, then remains constant. Therefore, r = e = e z is a constant. Therefore, e z moves on a circle centered at zero and radius r = e : 2

13 z = 3 + iy z = + iy 0 + iy e 3 e 0 = e = e Eample 3.6. Find the image of the rectangle R = {z : 0. < Im z < 0.2, 0.2 < Re z 0.3} y R under the map w = e 2iz Solution. w = e 2i(+iy) = e 2y+2i = e 2y e 2i.. On blue: = 0.3, then w = e 2y e.6i. The angle is 0.6rad 35, and the modulus is e 2y where 0. < y < 0.2: On purple: y = 0.2, then w = e.4 e i2, where 0.2 <

14 35 3. On orange: = 0.2, then w = e 2y e 0.4i Then the angle is 0.4 rad 23, and the modulus is e 2y where 0. < y < 0.2: On black: y = 0., then w = e 0.2 e i2, where.2 < 0.3: Limits The limit theory for the comple functions follows closely the theory for the functions from R 2 to R 2. We start with the definition of the limit. We will not use the definition to find the limits, instead we use the theorems from the calculus. 4

15 Definition 4.. Let f :, and z D f. By lim z z0 f(z) = w 0 we mean the following: for every δ > 0, there is ɛ > 0 such that z z 0 < ɛ implies that f(z) w 0 < δ. (4.) Aleis claims that lim z +2i f(z) = i: y f z 0 = + 2i w 0 = i I want to check this claim. So I choose a neighborhood of w 0, i.e., a disk centered at w 0 with a radius that I choose, say δ: (Let B δ denote this disk) y z 0 = + 2i δ To prove her clime, Aleis must be able to find a neighborhood B ɛ of z 0, say of radius ɛ, such that the image of B ɛ under f is inside B δ y ɛ f δ 5

16 Simply for any δ in w-plane we have to find a ɛ in z-plane, such ɛ δ that f maps inside. ɛ Since is centered at z 0, we only need to know its radius ɛ which depends on δ. Eample 4.2. Aleis claims that lim z i iz =. The neighborhood B δ I choose for is a disk centered at with radius δ > 0: B δ = {w : w ( ) < δ} = {w : w + < δ}. Aleis must find ɛ > 0, such that an ɛ-neighborhood B ɛ of i gets mapped inside B δ, i.e., z i < ɛ iz + < δ. }{{}}{{} z is in B ɛ f(z) is in B δ Since iz + = i(z i) = i z i = z i, then she rewrites the above as z i < ɛ z i < δ Aleis can easily conclude that ɛ is any positive number smaller δ. 4. Useful theorems and facts Let z = + iy denote points in the z-comple plane. hoose a point z 0 in the z-plane and write Let f : be a map from z-plane to w-plane and where u, v : R 2 R are real-valued functions. Then is equivalent to the following z 0 = 0 + iy 0 (4.2) f(z) = u(, y) + iv(, y), (4.3) lim z z 0 f(z) = a + ib lim u(, y) = a, lim (,y) ( 0,y 0) v(, y) = b (4.4) (,y) ( 0,y 0) 6

17 Eample 4.3. Find lim z (+i) iz + log z. Solution. f(z) = iz + log z = i( + iy) + log 2 + y 2 = i + [log 2 + y 2 y] Therefore u(, y) = log 2 + y 2 y and v(, y) =. We also have ( 0, y 0 ) = (, ). Therefore lim u(, y) = log 2, (,y) (,) lim v(, y) =. (,y) (,) Therefore lim z (+i) f(z) = (log 2 ) + i Sometimes we want to show that the limit does not eist. One way to show this, is to let (, y) approach ( 0, y 0 ) from two different directions, and find two different limits. This contradicts the uniqueness of limit. Therefore, the limit does not eist. Eample 4.4. Show that u(, y) = 2 y 2 2 +y 2 does not have limit as (, y) (0, 0). Proof. On the -ais the value of the function is : u(, y) =, for (, y) on the ais. Therefore lim u(, y) =, when (, y) approaches (0, 0) on the -ais. On the y-ais the value of the function is : u(, y) =, for (, y) on the y ais. Therefore lim u(, y) =, when (, y) approaches (0, 0) on the y-ais. We found two different (directional) limits, then the limit does not eist. To show that a comple-valued function does not have a limit, we show either the real part u(, y) or the imaginary part v(, y) does not have a limit, using the above technique. Eample 4.5. Show the f(z) = z z does not have a limit as z 0. Proof. Since f(z) = z z = z2 z 2 = (2 y 2 )+i2y 2 +y, then 2 u(, y) = 2 y 2 2, v(, y) = 2y + y2 2 + y 2 7

18 Since u(, y) does not have a limit, as (, y) (0, 0), then f(z) does not have a limit as z lim z z0 f(z) = In comple plane we do not have ± ; we have. For real valued functions, if f() moves unboundedly toward right on the real ais, as 0, we say lim f() = +, 0 lim 0 + = If f() moves unboundedly toward left on the real ais, as 0, we say lim f() =, 0 lim 0 = In comple plane, if f(z) moves further and further from the origin, as z z 0, we say lim f(z) = lim z z 0 z 0 z = lim z 0 z = means that as z goes closer and closer to the origin, z moves further and further from the origin. It could move in any direction. All we can tell about its location is that z is outside a large disk centered at the origin. Let D(0, a) = {z : 0 < z 0 < a} (4.5). If z D(0, 20 ) then f(z) = z is outside D(0, 20). 8

19 2. If z D(0,.00) then f(z) is outside D(0, 000). y v f(z) = z u 3. If z D(0,.00000) then f(z) is outside D(0, ). y v f(z) = z u The red region is called a neighborhood of infinity. An ɛ-neighborhood of infinity is a region R define by R = {w : w > /ɛ} 9

20 4.2.2 lim z f(z) = w 0 Roughly speaking, lim z z = 0 means if we choose z in an ɛ- neighborhood of, then z is in anɛ-neighborhood of zero ɛ nbhd of y f(z) = z v ɛ nbhd of 0 u Theorem 4.6. If z 0 and w 0 are points in the z and w planes, respectively, then. lim z z0 f(z) = if and only if lim z z0 f(z) = 0 2. lim z f(z) = w 0 if and only if lim z 0 f( z ) = w 0 3. lim z f(z) = if and only if lim z 0 = f( z ) = 0. Eample lim z iz+3 z+ = since lim z z+ iz+3 = 0 2z+i 2. lim z z+ = 2 since lim z 0 (2/z)+i /z+ = lim z 0 2+iz +z = 2 3. lim z 2z 3 z 2 + = since lim z 0 (/z2 )+ (2/z 3 ) = lim z 0 z+z3 2 z 3 = ontinuous Functions Definition 4.8. A comple-valued function f : is continuous at z 0 if. f(z) has limit as z z 0, 2. f(z 0 ) is defined, 3. lim z z0 f(z) = f(z 0 ) 20

21 A function f(z) = u(, y) + iv(, y) is continuous at z = + iy if and only if u(, y) and v(, y) are continuous at ( 0, y 0 ). 5 Derivatives Let f : be function. Let z 0 be a point in the domain D f of f. The derivative f of f is a comple valued function, whose value at z 0 is f (z 0 ) = lim z z 0 f(z) f(z 0 ) z z 0, (5.) provided the limit eists. z is a number close to z 0 which approaches to z 0. We can denote z by z 0 + z: z = z 0 + z. (5.2) Then the fact that z z 0 is equivalent to z 0. This notation allows us to rewrite (5.) as f f(z 0 + z) f(z 0 ) (z 0 ) = lim. (5.3) z 0 z If we replace z 0 by z we have f (z) = lim z 0 f(z + z) f(z). (5.4) z All these formulas are equivalent. However, (5.3) is usually used when we want to find f at a given point z 0 D f, while (5.4) is used to find f (z) for a typical z D f. Eample 5.. Let f(z) = z 2. At any point z f f(z + z) f(z) (z) = lim z 0 z (z + z) 2 z 2 = lim z 0 z ( z) 2 + 2z z = lim z 0 z = lim z + 2z = 2z. z 0 To carry the computation correctly, we need to bear in mind that z is a fied comple number and z is another comple number that converges to zero. Net eample shows that the limit in (5.), might not eist at some points (or at any z). In this we say f is not differentiable at that z. 2

22 Eample 5.2. Let f(z) = z. Then, since (z + z) = z + z, f(z + z) f(z) z = (z + z) z z = z z In Eample 4.5 we showed that z z does not have a limit as 0. (To see the equivalence, remember that z is a comple number just like z. ) Eample 5.3. Let f(z) = z 2. ompute f(z + z) f(z). Solutioin. f(z + z) f(z) = z + z 2 z 2. Since z 2 = z z, then z + z 2 = (z + z)(z + z) = (z + z)(z + z) = zz + z z + zz + z z = z 2 + z z + zz + z z. Therefore, f(z + z) f(z) = z z + zz + z z. (5.5) It is possible that the real part u(, y) and the imaginary part v(, y) of a comple function f have partial derivatives, and yet the function is not (comple) differentiable. See the net eample. Eample 5.4. Let f(z) = z 2. Then f(z) = ( 2 + y 2 ) + 0i. Therefore, u(, y) = 2 + y 2, and v(, y) = 0. Therefore, u and v both have continuous partial derivatives with respect to and y: u(, y) = 2, y u = 2y, v(, y) = y v(, y) = 0 But, from (5.5), f(z + z) f(z) = z z + z + z. z z Does this have a limit as z 0? The answer is negative. The last term 0. The middle term z does not depend on z and remains unchanged. The first term on the R-H side does not have a limit, for z 0.(See homework#4). Therefore, f(z) is differentiable only at z = 0. Remark 5.5. Instead of z we could use any other letter say q, z, etc. But from (5.2) we see that z denotes the difference z := z z 0 There is a convention in mathematics literature to for denoting differences of values. Similarly, we use w to denote the difference in the w-plane, i.e., w := f(z + z) f(z) 22

23 Therefore we define the notation to write more compactly. w z := f(z + z) f(z) z 5. Differentiation Formulas If f is differentiable at z, then we denote its derivative at z by either df dz (z), d dz f(z) or f (z). The following basic formulas can be derived from the definition of the derivative.. If c is a comple constant then d dz c = Linearity: If f(z) and g(z) are differentiable and c is a comple constant, then d dz (f(z) + cg(z)) = d dz f(z) + c d g(z) (5.6) dz 3. For any positive integer n: d dz zn = nz n (5.7) 4. Product rule: If f and g are differentiable at z, then so is their product and d dz (fg)(z) = f (z)g(z) + f(z)g (z). (5.8) 5. Quotient rule: If f and g are differentiable at z, and g(z) 0, then so is their quotient f g (z) and d f dz g (z) = f (z)g(z) f(z)g (z) g 2. (5.9) (z) 6. hain rule: If f is differential at z and g is differentiable at w = f(z), then so is their composition gof(z) and d dz gof(z) = g of(z)f (z). (5.0) Eample 5.6. If f(z) = (i+z 2 ) 7, then by the chain rule, f (z) = 7(i+z 2 ) 6 (2z) 23

24 5.2 auchy-riemann Formula onsider the definition of the derivative f of f: f f(z + z) f(z) (z) = lim z 0 z If f (z) eists, i.e. if the limit on the R-H-S is defined, then the limit would be the same when choose a specific manner in which z 0. We choose the ais and the y ais and compute the limit separately: On the ais: z is a real number, we call it : z =. Furthermore, if z = + iy, then z + z = + + iy. In this case we can write f(z) = u(, y) + iv(, y), f(z + z) = u( +, y) + iv( +, y). Therefore, after rearrangement, f(z + z) f(z) = u( +, y) u(, y) + i { v( +, y) iv(, y) }. Then f(z + z) f(z) z = u( +, y) u(, y) Then, by the definition of partial derivative + i v( +, y) iv(, y). f f(z + z) f(z) (z) = lim = u (, y) + iv (, y) z 0 z We just proved that, if f (z) eists, then f (z) = u (, y) + iv (, y). (5.) On the y ais: z is purely imaginary, we call it i y, where y R: z = i y. Furthermore, if z = + iy, then z + z = + i(y + y). In this case, we can write, And, similarly, f(z + z) f(z) z f(z) = u(, y) + iv(, y), f(z + z) = u(, y + y) + iv(, y + y). = u(, y + y) u(, y) i y v(, y + y) v(, y) + i. i y (5.2) 24

25 Therefore, after taking the limit as z 0, f (z) = v y (, y) iu y (, y) (5.3) Since the limit is unique, then comparing (5.) and (5.3) yields that f (z) eists { u (, y) = v y (, y) u y (, y) = v (, y). (5.4) But, as we saw in Eample 5.4, eistence of f (z) is not trivial. The second part of the following theorem gives the sufficient condition for eistence of f (z). Theorem If f (z) eists, then u (, y) = v y (, y) u y (, y) = v (, y). 2. Furthermore, if u, u y, v and v y eist and are continuous, then f (z) eists. We also have Theorem 5.8. If f (z) eists, then f (z) = u (, y) + iv (, y) or f (z) = v y (, y) iv y (, y) Eample 5.9. We are revisiting Eample 5.4. Let f(z) = z 2. Therefore u(, y) = 2 + y 2 and u(, y) = 0. Then u = 2, u y = 2y, v = 0 and v y = 0 eist and are continuous. But the -R eq. requires that 2 = 0, and 2y = 0, which holds only at = y = 0. Therefore, f(z) is differentiable only at z = 0. Eample 5.0. Let f(z) = z. This is a function from \{0}. Since u(, y) = 2 +y and v() = y 2 2 +y on R 2 \{(0, 0)}, then 2 and u (, y) = y2 2 ( 2 + y 2 ) 2 = v y(, y) (, y) R 2 \{(0, 0)} u y (, y) = Therefore, f(z) = z 2y ( 2 + y 2 ) 2 = v (, y) (, y) R 2 \{(0, 0)} is differentiable at all z 0, and f (z) = u (, y) + iv (, y) = z 2. 25

26 5.3 Polar coordinates Sometimes a function f is given in terms of the polar coordinates of z: f(z) = u(r, θ) + iv(r, θ). Assume, that partial derivatives u r, u θ, v r and v θ eist and are continuous at z = re iθ. The following theorem indicates the sufficient condition for eistence of f (z). Theorem 5.. If ru r = v θ and u θ = rv r, then f is differentiable at z and f (z) = e iθ (u r + iv r ) Remember that the principal value Arg z of arg z is the value θ satisfying π < θ π. We define the principal value of log z to be Log z = log z + iarg z, z 0. (5.5) Eample 5.2. Log 2i = log 2i + iarg 2i = log 2 + i π 2. Eample 5.3. Let Log z = u(r, θ) + iv(r, θ), where z \(, 0]. Then u(r, θ) = r and v(r, θ) = θ. y Log z v \(, 0] π π u Then ru r (r, θ) = r = v θ and u θ = 0 = rv r. Therefore, Log z is differentiable at all z \(, 0]. Of course we could write Log z in terms of (, y) and use the Theorem Analytic Functions and Singularities Definition 5.4. f : is called analytic at z 0 if f is differentiable in a neighborhood of z 0. Being differentiable at z 0 is necessary, but not sufficient for being analytic at z 0 : 26

27 Eample 5.5. f(z) = z 2 is not analytic at z 0 = 0, though it is differentiable at 0. Because there is no neighborhood of z 0 = 0, at which f(z) is differentiable. (f(z) is differentiable only at z 0 = 0.) Eample 5.6. f(z) = z 2, is not analytic anywhere. Eample 5.7. f(z) = z is not analytic at z = 0, because it is not differentiable at z = 0, though it is differentiable in any deleted neighborhood of zero. Eample 5.8. f(z) = z is analytic at every z 0. Because for any non-zero z, there is a neighborhood of z that does not include z = 0. Therefore, f(z) will be differentiable at all points in the neighborhood. A point like z = 0 is called a singularity for f(z) = z : Definition 5.9. A point z 0 is a singularity for f(z), if f is not differentiable at z 0, but in any neighborhood of z 0 there is a point at which f is analytic. Eample z = 0 is not a singularity for f(z) = z 2. This is because, f(z) is not analytic anywhere Eample 5.2. f(z) = z2 + z 2 +2 is analytic in ecept for the singular points z = ±i 2 Definition A function that is analytic everywhere in is called entire. Eample f(z) = e z and all polynomials p(z) = a 0 + a z + + a n z n are entire. 6 Elementary Functions 6. Logarithm y y = e y 0 This figure shows how to find a := log y 0, for y 0 > 0 a In calculus logarithm is defined to be the inverse of the function y = e. By this we mean, for a given number y 0 > 0, log y is a number a R such that e a = y 0. (6.) 27

28 Definition of log y has to do with solving equation e a = y, i.e., log y = a e a = y We know that: if y 0 0, then the equation e a = y 0 does not have a solution. In comple variables, for a given number w, if we find a such that e a = w, we say: a = log w But there is an issue here to be addressed. Unlike real case e z is periodic in Im z, i.e., If a solve e a = w, then so does a + i2kπ for k = 0, ±, ±2,. This is because e a+i2kπ = e a e 2ikpi = w = w Eample 6.. e +i = ee i, which is a number of modulus e and argument radian. Therefore log ee i = + i, + i ± i2π, + i ± i4π : Since e z takes + i to w 0 = ee i y w 0 = ee i z 0 = + i It also takes all the these points to w 0 too y z 2 = + i( + 4π) z = + i( + 2π) w 0 = ee i z 0 = + i Therefore, z = + i( 2π) z 2 = + i( 4π) log ee i = { + i( + 2kπ), k Z} This defines log z as a multi-valued map, which is not a function. The multi-valued map log z is not a function. 28

29 We have two following tasks:. to make log z single valued: This is done by choosing one of the values of log z. We then use this single-valued function, 2. to make a continuous function: This is done by finding the discontinuities and eliminating those points from the domain. This new function, with smaller domain is called Log z. Step : To make the function log z single valued, we notice that the distance between any two z k is at least 2π. Therefore, we can choose a horizontal strip of width 2π, to only have one solution in this stip. For now, we specifically choose the strip Γ = {z : π < Im z π} (6.2) In the figure below the purple infinite strip Γ tells us to choose the value z 0 = + i for log ee i : y z 2 = + i( + 4π) e z z = + i( + 2π) e z e z w 0 = ee i z 0 = + i e z e z z = + i( 2π) z 2 = + i( 4π) 29

30 y z 2 = + i( + 4π) z = + i( + 2π) log w 0 w 0 = ee i z 0 = + i z = + i( 2π) z 2 = + i( 4π) Log z is the value of log z which is in the purple strip Γ, i.e., z 0 = + i: y z 2 = + i( + 4π) z = + i( + 2π) Log w 0 w 0 = ee i Γ z 0 = + i z = + i( 2π) z 2 = + i( 4π) 30

31 Eample 6.2. Find log 4e 2i : Solution. If w = log 4e 2i, then w must solve e w = 4e 2i. Let assume w = + iy. Then e e iy = 4e 2i { e = 4 = log 4 e iy = e 2i y = 2 + 2kπ (6.3) Then. y w = {log 4 + i(2 + 2kπ), k Z} w 2 = log 4 + i(2 + 4π) z=4e 2i v w = log 4 + i(2 + 2π) w = log z = log 2e 2i w 0 = log 4 + 2i u w = log 4 + i(2 2π) w 2 = log 4 + i(2 4π) Net, as usual we let z = re iθ move appropriately, and study the trace of its image under w = log z. If we let z rotate, i.e., we let θ change. If, as in Eample 6.2, we let w = +iy, then from Eq. (6.3), we see that is independent of θ and y = θ + 2kπ. This means that w only moves vertically. Similarly, if z moves radially, then the imaginary part of log z remains unchanged, and its real part moves horizontally; moving outward and inward, correspond to Re log z increase to infinity and decrease to. Remember that z 0. Therefore we have the following figure: 3

32 y v w = log z u Step 2: We now have a single-valued function w = Log z whose domain is and it s range is a strip of π < π π. (as of now Γ only includes its boundary y = π) y Regardless of how close z 3 is to z 2, their images w 2 and w 3 are far from each other w = Log z v z z 2 z 3 w 2 w w 3 y = π u y = π Eample 6.3. Evaluate Log z at z = e (π.00)i, z 2 = i, and z 3 = e (π+.00)i. 32

33 Solution. Log z = Log e (π.00)i = (π.00)i, Log i = Log e πi = π and Log z 3 = Log e (π+.00)i = ( π +.00)i Therefore, Log z is not continuous at z 2. Similar observation can be made for every z (, 0]. Therefore, we remove the discontinuities from the domain of Log z. This is equivalent to ecluding y = π from the image: y Log z v \(, 0] π π u Log z = log z + iarg z 6.. Branches of Logarithm (6.2) is only one of the many choices we have for making log z single valued. For eample, any strip of width 2π is an option. In general, we can define the range to be Γ = {z : α < Im z < 2π + α}. (6.4) This version of log z on the slit comple plane \{re iθ : θ = α} (6.5) is analytic and is called a branch of logarithm. The real parts of different branch of log are equal: log z = log z + iθ, θ arg z (α, α + 2π), where the log on the L-H-S is the branch of comple logarithm corresponding to Γ and log on the R-H-S is the familiar logarithm for the real numbers. 33

34 \{re iθ : θ = α} y branch of log z v α y = α + 2π y = α u If z = re iθ, where θ (α, α + 2π), then log z = log z + iθ defines a branch log whose values are in the strip Γ defined in (6.4), and its domain is the slit plane defined in (6.5). Furthermore, branches of log z are analytic in their domain and d dz log z = re iθ 6..2 Properties of Logarithm For n Z we have For n =, 2, log(z z 2 ) = log z + log z 2 (6.6) log(z /z 2 ) = log z log z 2 (6.7) z n = e n log z (6.8) z /n = e n log z (6.9) 6.2 omple Eponents For α, we define z α to be multi valued function It is multi-valued, because log z is. Therefore z α = e α log z, z 0. (6.0) z α α[log z +i(arg z+2kπ)] = e = e αlog z e α2kπi (6.) 34

35 Note that, by (6.), if α is an integer, then z n is single-valued for n Z. We can define a single-valuded function z α, by specifying a branch of log in (6.). Then the corresponding values of z α is called a branch of z α. In particular, if we use Log z in (6.) the resulting branch of z α is called the principal value of z α. P.V. z α = e αlog z (6.2) When the brach is specified, the single valued function z α is analytic in the slit comple plane: Assume a branch of log is specified. Then z α is analytic in the slit comple plane, and d dz zα = αz α (6.3) Eample 6.4. onsider the branch of log that corresponds to < α < + 2π. Let f(z) = z i. Find f(i + ) and f (i + ). Solution. Note that π/4 / (, + 2π), therefore we choose θ = 2π + π 4 = 9π 4. Since + i = 2e i9π/4, log(i + ) = log i + + i 9π 4 = log 2 + i9π 4. f(i + ) = (i + ) i = e (i )(log(i+)) = e (i )(log 2+ i9π 4 ) = e 9π 4 log 2+i(log 2 9π 4 ). f (z) = (i )z i 2 = (i )e (i 2) log z = (i )e (i 2)[log z +iα ], where α is an argument of z which is in interval (, + 2π). As we saw, α = 9π 4 for z = + i: f ( + i) = (i )e (i 2)[log 2+ 9πi 4 ] (6.4) Eample 6.5. Find P.V. (i + ) i. P.V. (i + ) i = e (i )Log (i+) = e (i )[log 2+i π 4 ]. (6.5) You can rearrange the power, to make it look nicer. 35

36 7 Integrals - auchy s Theorem This week we discuss the following topics:. ontour integrals: concept and computation 2. Upper bounds for the moduli of the ontour integrals 3. Anti derivatives Definition 7.. A continuous curve or contour in is a continuous map γ : [a, b]. Eample 7.2. Let be the upper-half unit circle traversed counter clockwise. can be formulated by y γ : [0, π] : γ(t) = e it (7.) γ(t) = e it = cos t + i sin t 0 π t Eample 7.3. Let f(z) = z, and γ as above. We can restrict f to γ and define f : [0, π] (Step ) f(t) := f(γ(t)) = e it We then observe that z = cos t + i sin t dz = sin t dt + i cos t dt = ( sin t + i cos t) dt (Step 2) Then from (Step ) and (Step 2) we define π f(z) dz = f(t)( sin t + i cos t) dt (7.2) γ We can easily compute the right-hand-side as follows π f(z) dz = (cos t i sin t)( sin t + i cos t) dt γ = 0 π 0 0 ( π ) ( cos t sin t + sin t cos t) dt + i sin 2 t + cos 2 tdt = iπ 0 36

37 Eample 7.4. Let f(z) = ire z, and γ the straight line from 0 to + 2i. ompute f(z)dz (7.3) Solution. First observe that γ(t) = t + i2t for t [0, ]. Therefore Step : f on γ is: Step 2: dz is computed as follows: γ f(γ(t)) = ire (γ(t)) = ire (t + i2t) = it (7.4) dz = dγ(t) = γ (t) dt = ( + 2i) dt Therefore, ire z dz = γ = 0 f(γ(t))γ (t) dt = 2t dt + i it( + 2i)dt t dt = + i Properties of the contour integrals Let α : [, ] be defined by α(t) = t+ 2 + i(t + ) is again a straight line from z = 0 to z = + 2i. y γ(t) +2i 0 t α(t) t Using either α or γ to compute f(z)dz results the same number: If α and γ are as above, then f(γ(t))γ (t)dt = f(z)dz = 0 f(α(t))α (t) dt (7.5) 37

38 y α(t) = e (π t)i 0 π t We denote by, a curve whose image lies on the image of the curve, but it is traversed in opposite direction. Since in Eample 7.2 we denoted the upper half unit circle traveled from to by, then we let denote the upper half unit circle traveled from to. If α is a parameterization of, then γ is a parameterization of. Since α (t) = ie i(π t), then Therefore, we have z dz = π 0 π = = 0 f(α(t))α (t)dt e i(t π) ie i(t π) dt = iπ f(z)dz. If and are two opposite curves, then f(z) dz = f(z) dz (7.6) Remark 7.5. If γ : [a, b] is a parameterization of the curve, then γ is NOT in general a parameterization of. 7.2 An Estimate If ((t), y(t)) determines a curve in R 2, for t [a, b], then l = b a 2 (t) + y 2 (t) dt (7.7) is the length of the curve. If we think of ((t), y(t)) as the real part and the imaginary part of γ(t), then 2 (t) + y 2 (t) = α (t) (7.8) 38

39 Assume γ : [a, b] is a parameterization of a curve. Assume f(α(t)) M, for all t [a, b]. Then b f(z) dz f(α(t)) α (t) dt M where l is the length of the curve. a b a α (t) dt = Ml, (7.9) Eample 7.6. Show that e z z dz πe, where is the upper half of the unit circle, described counter clockwise. Proof. Let f(z) = ez z, and let γ(t) = eit be a parameterization for. Then e eit f(γ(t)) = e it = ecos t+i sin t e it = e cos t e. Since the length of the curve is π, then the result follows from (7.9). 7.3 Path dependence Eample 7.7. onsider f(z) = z and two points and. We choose two curves from to. ompare i f(z)dz for i =, 2. y : γ(t) = e it 2 : α(t) = e it 0 t π 0 t π 2 Solution. f(z) dz = πi by Eample 7.3. f(z) dz = 2 π f(α(t))α (t) dt = i π 0 0 e it e it dt = iπ 39

40 Therefore z dz 2 z dz (7.0) Eample 7.8. Let be the whole circle described counter clock wise starting from. ompute dz. (7.) z Solution. Of course, we can use the parameterization and write : γ(t) = e it, 0 t 2π (7.2) z dz = 2π 0 e iθ γ (t)dt = 2πi (7.3) Or we can use the results from Eample 7.7 to find the answer. First note that = 2, (7.4) where and 2 are the red and blue curves in Eample 7.7 respectively. Then z dz = z dz dz (7.5) 2 z And we evaluated both integrals in Eample 7.7. y y : γ(t) = e it 2 : α(t) = e it 0 t π 0 t π : γ(t) = e it 0 t 2π 2 = 2 (7.6) 40

41 Note that f(z) dz = f(z) dz 2 f(z) dz = 0 (7.7) Remark 7.9. We will learn later that failing to be path independent is closely connected to the fact that f(z) = z has a singular point inside. By this we mean that for any two curves and 2 such that their union does not have 0 inside we have f(z) dz = f(z) dz, and f(z) dz = 0 2 hold 7.4 Path independence, Anti derivatives The fundamental theorem of calculus states that if F () = f() then b a f() d = F (b) F (a) Therefore, eistence of the antiderivative F of f allows us to use only the last points to compute the definite integral. This is a simple version of path independence. There is an etension of this fact to the comple plane. Let D be an open domain in and the continuous function f : D has an anti-derivative F in D, i.e., F is an analytic in D and F (z) = f(z) for all z D. (7.8) Then the integral of f is independent of the path, i.e., f(z) dz is the same for all curve whose end points are the same. Therefore, we replace the complicated curve with the one which makes the computation easier, as long as our suggested curve stays in D. There is an inverse for this fact too. Below we state the precise theorem. Theorem 7.0. Suppose the f is continuous on a domain D. The the following are equivalent. f has an antiderivative F in D 2. f(z) dz = 0 for all closed curves lying entirely in D 3. f(z) dz has the same value for any curve in D starting from z and ending at z 2, and f(z) dz = F (z 2 ) F (z ), (7.9) where z and z 2 are the starting point and the end points of respectively. 4

42 Definition 7.. A set of points z = + iy in the comple plane is said to be an arc if = (t), y = y(t), a t b, (7.20) where (t) and y(t) are continuous functions. Jordan arc if it does not cross itself: t t 2 γ(t ) γ(t 2 ), An arc is a simple arc or where γ(t) = (t) + iy(t) defines. When the arc is simple ecept for γ(a) = γ(b), we say that is a simple closed curve. Such a curve is called positively oriented when it is in the counter clockwise direction. Suppose now that (t) and y (t) are defined and continuous on [a, b]. Then is called differentiable and we define γ (t) = (t) + iy (t) a t b. (7.2) γ (t) is a number in the comple plane for each t and γ (t) = [ (t)] 2 + [y (t)] 2. An arc : γ(t) = (t) + iy(t) where a t b is called smooth if γ (t) 0 for all a < t < b and continuous on a t b. A contour, or piecewise smooth arc is an arc consisting of a finite number of smooth arcs joined end to end. When only the initial and final values of γ(t) are the same, a contour is called a simple closed contour. Eample 7.2. Eample,2,3,4 on page auchy-goursat Theorem Theorem 7.3 (auchy-goursat Theorem I). If a function f is analytic at all points interior to and on a simple closed contour, then f(z) dz = 0. Eample 7.4. Find z dz, where is:. 42

43 2. Eample 7.5. Evaluate ( + esin z2 ) 2, where is a triangle with vertices 2e i+sin i, i + and i. Before we prove the -G theorem, we state a theorem from calculus. Theorem 7.6 (Green s Theorem). Let D be a bounded domain in the plane whose boundary = D consists of a finite number of disjoint piecewise smooth closed curves. Let P (, y) and Q(, y) be continuously differentiable functions on D. Then P d + Q dy = (Q P y ) d dy (7.22) Eample 7.7. Let us use the Green s Theorem to evaluate yd, where D is the quarter-disk in the first quadrant. Solution. Set P (, y) = y and Q(, y) = 0. Then /2 yd = ddy = r cos θrdrdθ = /3. D 0 D 0 Proof of -G Theorem. We prove this theorem under an etra assumption that f is continuous. Fact: The derivative f of an analytic function f is always continuous. Therefore, we do not need to add this assumption. But, right now we are not able to prove this fact. Let γ : [a, b] be the parameterization of. Then by definition b f(z) dz = f(γ(t))γ (t) dt Let f(z) = P (, y) + iq(, y), and γ(t) = (t) + iy(t). Then a f(γ(t)) = P ((t)), y(t)) + iq((t), y(t)) Since γ (t) = (t) + iy (t). Therefore, we suppress t, we have f(γ)γ = P (, y) Q(, y)y + i { P (, y)y + Q(, y) } 43

44 Therefore, remembering that P and Q are functions of (t) and y(t), b f(z) dz = (P Qy )dt + i(p y + Q )dt a = { P d Qdy } + i { P d + Qdy } Then, by the Green s Theorem f(z) dz = (Q + P y )ddy + i R R (Q P y )ddy Since, by the hypothesis, f is analytic, then it satisfies the R equation: { P = Q y P y = Q (7.23) The proof is complete. 7.6 Simply onnected Domains How if crosses itself? For eample where is the interior of the curve: To address this issue, we change our perspective. Instead of choosing the curve () first, and investigating the analyticity of f in the interior of, we consider a domain D. A Simply onnected Domain D is a domain such that every simple closed contour within it encloses only points of D. D D SD not SD The following theorem for integrating on a not simple curves as above as long as they stay in a simply connected domain. 44

45 Theorem 7.8. If a function is analytic throughout a simply connected domain D, then for every closed contour in D. f(z) dz = 0, Proof. We only proof the theorem for the case that intersects itself finitely many times. Assume we are given the following curve which starts from γ(a) = z 0 : z 0 = D We can decompose this curve to two simple contours as follows z 0 = 2 = + 2 D Then f(z)dz = f(z)dz + f(z)dz = = 0 2 Eample 7.9. If is any closed contour lying in the open disk z < 2, then ze z (z 2 + 9) 5 dz = 0 45

46 Assume f is analytic throughout a simply connected domain D. By Theorem (7.8), integral over every closed contour is zero. Then 2 of Theorem 7.0 holds. Therefore of Theorem 7.0 must hold. We just proved the following orollary A function f that is analytic throughout a simply connected domain D must have an antiderivative everywhere in D. 7.7 Multiply onnected Domain A domain that is not simply connected is called multiply connected. The auchy-goursat theorem for simply connected domain The following theorem is the auchy-gousat theorem (7.8) for multiply connected domain does not hold any longer. Eample 7.2. Let D denote the comple plane with z = 0 deleted. Then D is not a simply connected domain, as the unit circle encloses a point not in D. We know that f(z) = z is analytic in D, but if we let denote the unit circle, f(z)dz 0. The following version of auchy Goursat s theorem is suitable for the multiply connected domains. Theorem Suppose that. is a simple closed contour, described in the counterclockwise direction, 2. k s are simple closed contours for k =,, n, lying inside, with clockwise orientation. We assume that the interior of k s are also disjoint. If a function f is analytic on all of those contours and throughout the multiply connected domain consisting of the points inside and eterior to each k, then n f(z) dz = f(z)dz. (7.24) k k= 2 D Proof. 46

47 The theorem states that if f is analytic on,, 2 and the area that is eterior to, 2 and interior of, then (7.24) holds. To see this, on each curve choose two points: D By connecting these points through curve and 2 and 3 we have : 3 2 D Define two simple contours A and B as follows ontour A ontour A = D 47

48 ontour B D ontour B = Then we can write: ( ) = ( ) + ( ) }{{}}{{} contour:a contour:b Since A and B are simple, and f is analytic on and inside them, therefore f(z)dz = 0, and f(z)dz = 0. By theorem 7.3. Therefore f(z)d(z) = A A B f(z)dz + f(z)dz = 0. B If we omit from the above figures, we have the following: 2 D Then we have f(z)dz = f(z)dz = 2 f(z)dz 2 48

49 Since 2 is negatively oriented, then 2 is positively oriented. This observation proves the following result. To get rid of the negative signs, we let 2 be positively oriented. orollary 7.23 (Principal of deformation of paths). Let and 2 denote positively oriented simple closed curves, where is interior to 2 as the figure below. If a function f is analytic in the closed region consisting of those contours and all points between them, then f(z) = 2 f(z)dz (7.25) 2 D Eample When is any positively oriented simple closed contour surrounding the origin, the corollary can be used to show that dz z = 2πi 49

50 7.8 auchy s Integrals Let be the unit circle, oriented counterclockwise. We used a parameterization technique to show that dz 2πi z = (7.26) y : γ(t) = e it 0 t 2π Then we can use the principal of deformation of paths to show that dz =, (7.27) 2πi z for every simple closed curve : D 50

51 Theorem 7.25 (auchy Integral Formula). Let f be analytic everywhere inside and on a simple closed contour, taken in the positive sense. If z 0 is any point interior to, then f(z)dz f(z 0 )2πi = (7.28) z z 0 z 0 D Proof. We draw a very small circle r centered at z 0 of radius r, such that r fits entirely inside : r z 0 D dz In (7.28) we replace 2πi by r z z 0. Therefore we have to show that dz f(z)dz f(z 0 ) = (7.29) r z z 0 z z 0 Since f(z 0 ) is a number, we can take it inside the integral to have f(z 0 )dz f(z)dz = (7.30) z z 0 z z 0 On the right-hand-side, since f(z) z z 0 r is analytic between and r, then by the 5

52 Principle of Path Deformation, we have f(z 0 )dz = z z 0 r r r f(z)dz z z 0 (7.3) So, we need to show difference of the two integrals is zero. Because the contours are the same, then f(z 0 )dz f(z)dz = 0 (7.32) z z 0 z z 0 Then after taking the common denominator, we need to show that f(z 0 ) f(z)dz = 0 (7.33) z z 0 r To show this, we estimate the modulus of the integral. We know that f(z 0 ) f(z)dz r z z 0 M r2πr, (7.34) where M r is the maimum of on r. Because the function f is f(z0) f(z)dz z z 0 analytic at z 0 then f(z0) f(z)dz z z 0 f (z 0 ) and therefore, f(z 0 ) f(z)dz z z 0 f (z 0 ) < 2 f (z 0 ) Therefore, we can replace M r by 2 f (z 0 ), when r is very small. Therefore, f(z 0 ) f(z)dz z z 0 4 f (z 0 ) πr, (7.35) r 52

53 8 Series 8. Sequences, Series and onvergence Definition 8. (auchy sequence). Let {S n } n= be a sequence of comple numbers. s n is called a auchy sequence if ɛ > 0 n 0 such that n, m n 0 implies that S n S m < ɛ. (8.) Intuitively, this means, whatever small ɛ > 0 is, passed some n 0 the distance between any two terms is smaller than ɛ. Eample 8.2. Let S n = i+ 8 ni is auchy. Because ɛ = 0 : Passed n 0 = 30 the distance between any two terms is less than /0 ɛ = 00 : Passed n 0 = 300 the distance between any two terms is less than /00 In general passed n 0 = 3/ɛ, the distance between any two terms is less than ɛ > 0. Definition 8. is very similar to definition of limits: Definition 8.3 (auchy sequence). Let {S n } n= be a sequence of comple numbers. We say when the following holds: lim S n = l n ɛ > 0 n 0 such that n n 0 implies that S n l < ɛ. (8.2) Eample 8.4. Let S n = +ni n+. Then S n auchy, and l = i. To prove this claim, we must show that for any ɛ > 0 we have to find n 0 such that n n 0 then + ni i < ɛ (8.3) n + We simplify (8.3), as follows + i 2 n + < ɛ < ɛ (8.4) n + 2 n >. (8.5) ɛ Now we rewrite our claim, to simply verify its truth: For any ɛ > 0 we have to find n 0 such that n n 0 then n > 2 ɛ. It is correct, because it is enough to let n 0 = 2 ɛ. 53

54 The two definitions are very similar. Indeed, they are equivalent: Theorem 8.5 (onvergence of the auchy sequences). For a sequence {S n } n= of comple numbers the following are equivalent:. S n is a auchy sequence. 2. There is a comple number l such that lim S n = l (8.6) n Theorem 8.6. Suppose that S n = X n + iy n and S = X + iy. If then lim n S n = S. Eample 8.7. In Eample 8.4 we have S n = n 0 and lim n n+ =, then S = i and lim n = X n (8.7) lim n = X, n (8.8) n+ +i n n+. Since lim n n+ = + ni lim = i. (8.9) n n + Eample 8.8. Let S n = ( i 2 )n. Show that lim n S n = 0. Proof. It is easier to use the Definition (8.). Since S n l = ( i 2 )n = 2 n, then everything boils down to show that For any ɛ > 0 we have to find n 0 such that n n 0 then 2 n > ɛ. And this is doable by letting n 0 = ɛ. Net we review a lemma which is very helpful in computing the sum of numbers of a certain form. Lemma 8.9. For every c \{}: + c + c c n = cn+ c (8.0) Eample 8.0. Let c n = ( i 2 )n, for n = 0,,, and define S n = n k= c k = i 2 i i ( i n )n. Is S n auchy? What is lim S n? n 54

55 Soluiton. By (8.0) S n = ( i 2 )n+ i 2 (8.) We can write S n = 2 2 i [ + ( i 2 )n ]. Since lim n ( i 2 )n = 0, by Eample 8.8, then lim S n = lim n n n ( i 2 )k = 2 2 i k= Motivated by this eample, we introduce the following definition Definition 8.. Let c, c 2, be an infinite sequence of comple numbers. Define n S n := c k = c + + c n (8.2) k= If S n convergent and lim n S n = S, then we write c n = S (8.3) k=0 Furthermore, S n s are called the partial sums of S. We say k= c n is convergent, If lim n S n eists. If lim n S n does not eist, we say k= c n is divergent. orollary 8.2. If k= c n converges, then lim n c n = 0. Proof. We know that c n = S n S n. Since k= c n converges, then S n converges. Therefore, lim c n = lim S n lim S n = S S = 0 n n n Eample 8.3. Show that k= ik is divergent. Proof. c n = i n. Since lim n c n = 0 does not hold, then the sum is divergent. Eample 8.4. Does 4 k= + i denote a comple number? 3 k+4 2 k Solution. If we show that that the real part 4 k= is equal to R and 3 k+4 the imaginary part k= is equal to y R, then the infinite sum is equal to 2 k + iy. 55

56 In calculus we saw that ( ) n n= n is convergent, but n= ( )n n, the harmonic series, is divergent. And we said ( ) n n= n is convergent but not absolutely. This observation motivates the following definition Definition 8.5. n= c n is called absolutely convergent, when n= c n is a convergent series of real numbers. 8.2 Taylor Series Definition 8.6. Power series are series of the form a n (z z 0 ) n = a 0 + a (z z 0 )+a 2 (z z 0 ) a n (z z 0 ) n + n=0 Some functions have the power series representation for a given z 0 which is valid for z such that z z 0 < R 0. Eample 8.7. Let z 0 = 0 and R 0 =, then f(z) = power series representation z has the following z = z n whenever z <. (8.4) n=0 How about f(z) = e z or f(z) = sin z? Is there a power series representation for these functions? The following theorem no only answers this question affirmatively, but also produces the desired power series Theorem 8.8. Suppose that a function f is analytic throughout a disk z z 0 < R 0. Then f(z) has the power series representation f(z) = a n (z z 0 ) n z z 0 < R 0, (8.5) n=0 where a n = f (n) (z 0 ) n 0 n = 0,, 2,. (8.6) Remark 8.9 (The uniqueness of the Taylor series). Given z 0, the power series representation is unique. Therefore, if we let z 0 = 0 and f(z) = z, then Theorem 8.8 produces the same series that we obtained in (8.4). Eample Read carefully Eample,2,4,5 on page 92-5 of your tet book. We did all these eamples in class. 56

57 8.3 Laurent Series onsider f(z) = z. We know the power series representation z = holds only for z <. what about z >? If we work a bit harder, we can epand f(z) in different way around z = 0 for z >. Note that if z >, then z <, and n=0 z n f(z) = z = z z = z w, (8.7) where w = z, therefor w <. Therefore, f(z) = z ( + w + w2 + ) = z ( + z + + ), z >. z2 Therefore, we have the following result Lemma 8.2. If z >, then z = z n ( z > ). (8.8) n= We can use (8.8) to give such epansions for functions like f(z) = ( z)(2 ) in different domains, but before that we need to eplain what we are trying to find Definition A Laurent series is an epression of the form n= a n (z z 0 ) n. (8.9) Notice that, unlike Taylor series, n could be negative here. For eample in the Laurent series representation of f(z) = z, when z > 0, then { 0 n 0 a n = n (8.20) Eample Let f(z) = ( z)(2 z). Find the Laurent series representation centered at z 0 = 0. Solution. f(z) has two singularities z = and z = 2. Therefore, we have 3 areas:. O: is a disk 2. G and P : are annuli 57

58 y f(z) is defined at z 0 = 0 and 2 are bad points of f(z) Therefore, f(z) is analytic throughout the orange annulus : < z < 2. G O 2 P We also write Therefore:. On G: f (z) := z = f 2 (z) := 2 z = f(z) = z }{{} n=0 2 z }{{} f 2(z) f (z) n=0 zn n=0 z n n=0 n=0 2n z n+ in G in O P zn 2 in G O n+ in P (8.2) (8.22) (8.23) f(z) = z n + ( z 2 2 )n (8.24) = = z n + n=0 n=0 n n=0 z n 2 n+ (8.25) ( + 2 n+ )zn, z <. (8.26) n=0 58

59 Therefore, a n = f (n) (0) n! = ( + 2 n+ ) 2. On O: f(z) = z n z n +, < z < 2 (8.27) 2n+ n=0 3. On P : Therefore, f(z) = a n = n= n= ( + 2 n )z n (8.28) { ( + 2 n ) n 0 n 0 Eample Let f(z) = e /z. Then f(z) is analytic in z > 0. y f(z) is not defined at z 0 = 0 The center of the annulus is a bad point of f(z) the annulus is of the form : 0 < z < R 2. We use the Taylor epansion of e z to find the Laurent series of f(z). Since e z = n=0 z n! z < (8.29) Then We can also write it as e /z = n=0 n!z n z > 0 (8.30) e /z = n=0 n= z n ( n)! z > 0 (8.3) The following theorem is a result for a general function 59

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