Chapter 1. Complex Numbers. Dr. Pulak Sahoo

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1 Chapter 1 Complex Numbers BY Dr. Pulak Sahoo Assistat Professor Departmet of Mathematics Uiversity Of Kalyai West Begal, Idia sahoopulak1@gmail.com 1

2 Module-2: Stereographic Projectio 1 Euler s Formula By assumig that the ifiite series expasio e x = 1 + x + x2 2! + x3 3! +... of elemetary calculus holds whe x = iθ, we ca arrive at e iθ = cos θ + i si θ, which is called Euler s formula. I geeral, we defie The -th Root of Uity e z = e x+iy = e x.e iy = e x (cos y + i si y). The solutios of the equatio z = 1 where is a positive iteger are called the -th root of uity ad are give by z = cos 2kπ 2kπ + i si = e 2kπi, k = 0, 1, 2,..., 1. If we put ω = cos 2π + i si 2π = e 2πi, the roots are 1, ω, ω 2,..., ω. Geometrically they represet the vertices of a regular polygo of sides iscribed i a circle of radius oe havig ceter at the origi. The circle has the equatio z = 1 ad is ofte called the uit circle. 2 Poit at Ifiity The liear trasformatio z w = f(z), where f(z) = λz + µ, λ 0, 2

3 is a oe-oe mappig of the fiite complex plae oto itself. This is ot true of the iversio map z w = 1/z. Writig ito polar forms, we have z = re iθ ad w = ρe iφ, where ρ = 1/r. Therefore, the poits close to the origi i the z-plae, r 0, are mapped oto poits far away from the origi i the w-plae. All the poits iside a disk of small radius ε, i the z-plae, are mapped oto poits outside a disk of large radius 1/ε, i the w-plae. As ε 0, the disk i the z-plae shriks to the origi ad there is o image of z = 0 i the w-plae. Similarly, as the poit z moves farther ad farther away from the origi, its image i the w-plae moves closer ad closer to the origi i the w-plae, but there is o poit i the z-plae which ca be assiged w = 0 as the image uder iversio. It turs out to be useful to itroduce the cocept of a poit at ifiity, or z =, as a formal image of z = 0 uder the iversio map w = 1/z. The poit z = 0 ca the be regarded as the image of the poit at ifiity. The use of z = will always be uderstood i terms of a limitig process w 0, where w = 1/z. To examie the behavior of f(z) at z =, it suffices to let z = 1 w ad examie the behavior of f( 1 w ) at w = 0. For example, we say that the fuctio f(z) = 2z 1 z 3 f(1/w) = 2 w 1 3w teds to 2 as w 0. teds to 2 as z, because 3 Exteded Complex Plae By the exteded complex umber system, we shall mea the complex plae C together with a symbol which satisfies the followig properties : (a) If z C, the we have z + = z =, z/ = 0. (b) If z C, but z 0, the z. = ad z/0 =. (c) + =. = (d) /z = (z ). The set C { } is called the exteded complex plae ad is deoted by C. The ature of Argad plae at the poit at ifiity is made much clear by the use of Riema s spherical represetatio of complex umbers, which depeds o Stereographic Projectio. 3

4 Stereographic Projectio We cosider the Argad plae C ad a uit sphere Ŝ taget to C at z = 0. The diameter NS is perpedicular to C ad we call the poits N ad S the orth ad south poles of Ŝ respectively. Now we establish a oe-oe correspodece betwee the poits o the sphere ad the poits o the plae. To each poit A i the plae there correspods a uique poit A o the sphere. The poit A is the poit where the lie joiig A to the orth pole N itersects the sphere. Coversely, correspodig to each poit A o the sphere (except the orth pole) there exists a uique poit A i the plae. By defiig that the orth pole N correspods to the poit at ifiity, we ca say that there exists a oe-oe correspodece betwee the poits o the sphere ad the poits i the exteded complex plae. This sphere is kow as Riema sphere ad the correspodece is kow as stereographic projectio, (see Fig. 1.1). We cosider the sphere as Fig. 1.1: x x x 2 3 = 1, the plae of projectio as x 3 = 0 ad let (0, 0, 1) be the coordiate of N. For ay poit

5 A = (x 1, x 2, x 3 ) o the sphere we have poit A = (x, y, 0) i the x 3 -plae where the lie NA meets the plae of projectio. Obviously, the poits (0, 0, 1), (x 1, x 2, x 3 ) ad (x, y, 0) are colliear ad the equatio of the lie is From this we get x 1 x = x 2 y = x x = x 1 1 x 3, y = x 2 1 x 3. Therefore z = x + iy = x 1+ix 2 1 x 3. From this we get ad hece z 2 = x2 1 + x 2 2 (1 x 3 ) 2 = 1 + x 3 1 x 3, x 3 = z 2 1 z Also we see that z + z z 2 +1 = x 1 ad z z i( z 2 +1) = x 2. I this way we ca establish a oe-oe correspodece betwee the poits i the exteded complex plae ad poits o the sphere. Example 1.1. Fid all the roots of the equatio z (1 z) = 0. Solutio. Let w = z. The the give equatio becomes 1 z w = 1 = cos 2kπ + i si 2kπ, where k is a iteger. Therefore w = cos 2kπ 2kπ +i si we get z = w w+1 2kπ cos. Hece z = cos 2kπ 2kπ +i si 2kπ +i si = e 2kπi +1 e 2kπi +1, k = 0, 1, 2, 3. Agai from w = z 1 z, k = 0, 1, 2, 3. Example 1.2. Fid all the values of z for which z 5 = 32 ad locate these values i the Argad plae. Solutio. z 5 = 32 = 32(cos 2kπ + i si 2kπ), k = 0, ±1, ±2,... This gives [ ( ) ( )] 2kπ 2kπ z = 2 cos + i si, k = 0, 1, 2, 3,

6 If k = 0, the z = z 1 = 2[cos 0 + i si 0] = 2. If k = 1, the z = z 2 = 2[cos 2π + i si 2π]. 5 5 If k = 2, the z = z 3 = 2[cos π + i si π]. 5 5 If k = 3, the z = z = 2[cos 6π + i si 6π]. 5 5 If k =, the z = z 5 = 2[cos 8π + i si 8π]. 5 5 These are the oly roots of the give equatio. The values of z are idicated i the figure (see Fig. 1.2). Note that they are equally spaced alog the circumferece of a circle havig ceter at the origi ad the radius 2. Aother way of sayig this is that the roots are represeted by the vertices of a regular polygo. Fig. 1.2: Example 1.3. Fid all the roots of ( 8 8 3i) 1 ad exhibit them geometrically. Solutio. ( 8 8 3i) 1 = [ ( ( 16 cos 2kπ + π 3 ) ( ( 2kπ + π ) 3 = 2 cos + i si k = 0, 1, 2, 3. Therefore all the four roots are z 1 = 2(cos π/3 + i si π/3); z 3 = 2(cos π/3 + i si π/3); or 1 + i 3, 3 + i, 1 i 3, 3 i. ( + i si 2kπ + π 3 ( 2kπ + π 3 )), z 2 = 2(cos 5π/6 + i si 5π/6); 6 z = 2(cos π/6 i si π/6); ))] 1

7 The roots lie at the vertices of a square iscribed i a circle of radius 2 cetered at the origi, also are equally spaced with differece of agle π/2 (see Fig. 1.3). Fig. 1.3: Example 1.. Establish the relatio : 2 = ( ) kπ si, 2. Solutio. Let 1, ρ 1, ρ 2,..., ρ be the roots of uity, where ρ k = e 2kπi, k = 1, 2,..., 1. The z 1 = (z 1)(z ρ 1 )(z ρ 2 )...(z ρ ). Dividig both sides by z 1 ad lettig z 1, we obtai = (1 ρ 1 )(1 ρ 2 )...(1 ρ ). Takig cojugate of both sides, we obtai = (1 ρ 1 )(1 ρ 2 )...(1 ρ ). 7

8 Therefore 2 = = = = (1 ρ k )(1 ρ k ) (1 e 2kπi ( 2 1 cos 2kπ ( ) kπ si 2 ( kπ = 2 2() si 2 )(1 e 2kπi ) Takig the oegative square root of both sides we obtai the required result. Example 1.5. For ay two ozero complex umbers z 1 ad z 2 prove that Solutio. We have z 1 + z 2 z 1 + z 2 ) ). z 1 z 1 + z 2 z 2 2( z 1 + z 2 ). z 1 z 1 + z 2 z 2 = z 1 + z 2 z 1 z 2 +z 2 z 1 z 1 z 2 = z 1 + z 2 (z 1 z 2 +z 2 z 1 ) z 1 z 2 z 1 + z 2 ( z 1 z 2 + z 2 z 1 ) z 1 z 2 = 2 z 1 + z 2 z 1 z 2 z 1 z 2 = 2 z 1 + z 2 2( z 1 + z 2 ). 8

Complex Analysis Spring 2001 Homework I Solution

Complex Analysis Spring 2001 Homework I Solution Complex Aalysis Sprig 2001 Homework I Solutio 1. Coway, Chapter 1, sectio 3, problem 3. Describe the set of poits satisfyig the equatio z a z + a = 2c, where c > 0 ad a R. To begi, we see from the triagle