Presentation of complex number in Cartesian and polar coordinate system

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1 1 a + bi, aεr, bεr i = 1 z = a + bi a = Re(z), b = Im(z) give z = a + bi & w = c + di, a + bi = c + di a = c & b = d The complex cojugate of z = a + bi is z = a bi The sum of complex cojugates is real: z + z = 2a The product of complex cojugates is real: z z = a 2 + b 2 Properties of complex cojugates: Re(z) = 1 2 (z + z ) Im(z) = 1 2i (z z ) (z 1 ± z 2 ) = z 1 ± z 2 (z ) = z (z 1 z 2 ) = z 1 z 2 & ( z 1 ) = z 1 z 2 z, z (z ) = (z ) Presetatio of complex umber i Cartesia ad polar coordiate system Modulus or Absolute value: z = r = x 2 + y 2 Argumet: arg z = θ = arc ta y x be careful x = rcos θ, y = rsi θ z = x + yi Cartesia form = r(cos θ + i si θ) polar form modulus argumet form = re iθ = r cis θ Euler form Recall that the argumet should be measured i radias Arga plae is the complex plae Very useful for fast coversios from Cartesia ito Euler form it will give you visually positio of the poit, ad therefore quadrat for the agle z = i z = e i3π 2 2 i = 2e i3π 2 i = e iπ 2 1 = e i0

2 2 Practicality of Euler s form z 1 = x 1 + iy 1 = r 1 (cos θ 1 + i si θ 1 ) & z 2 = x 2 + iy 2 = r 2 (cos θ 2 + i si θ 2 ) Product is: z 1 z 2 = (x 1 + iy 1 )(x 2 + iy 2 ) = r 1 r 2 (cos θ 1 + i si θ 1 )(cos θ 2 + i si θ 2 ) = r 1 r 2 (cos θ 1 cos θ 2 si θ 1 si θ 2 + i (cos θ 1 si θ 2 + si θ 1 cos θ 2 ) = r 1 r 2 [cos(θ 1 + θ 2 ) + i si(θ 1 + θ 2 )] z 1 z 2 = z 1 z 2 modulus of product is product of modulus & arg (z 1 z 2 ) = arg (z 1 ) + arg (z 2 ) argmet of product is sum of argumets z 1 z 2 = [ z 1 e iθ 1] [ z 2 e iθ 2] = z 1 z 2 e i(θ 1+θ 2 ) Quotiet is: z 1 = (x 1 + iy 1 ) z 2 (x 2 + iy 2 ) = r 1 cos θ 1 + i si θ 1 = r 1 cos θ 1 + i si θ 1 cos θ 2 i si θ 2 r 2 cos θ 2 + i si θ 2 r 2 cos θ 2 + i si θ 2 cos θ 2 i si θ 2 = r 1 r 2 cos θ 1 cos θ 2 + si θ 1 si θ 2 + i (si θ 1 cos θ 2 cos θ 1 si θ 2 ) cos 2 θ 2 + si 2 θ 2 = r 1 r 2 [cos( θ 1 θ 2 ) + i si( θ 1 θ 2 )] z 1 = z 1 z 2 z 2 modulus of quotiet is quotiet of modulus & arg ( z 1 z 2 ) = arg (z 1 ) arg (z 2 ) z 1 = [ z 1 e iθ1] z 2 [ z 2 e iθ 2] = z 1 z 2 ei(θ 1 θ 2 ) argmet of quotiet is differece of argumets Coclusio: Euler form of complex umbers follows ordiary algebra: a a m = a +m It is easy to multiply two complex umber i ay form, but what if you have 10 factors or z or z Properties of modulus ad argumet z = z & arg (z ) = arg z zz = z 2 e iθ = e i(θ+k2π) kεz z 1 = z 2 z 1 = z 2 & θ 2 = θ 1 + 2kπ kεz

3 3 De Moivre s Theorem z = [r(cosθ + i siθ)] = r (cosθ + i siθ) z = [re iθ ] = r e iθ = r (cos θ + i si θ) (cos θ + i si θ) = (cosθ + i siθ) Applicatio of DeMoivre s Theorem (a + b) = ( r ) a r b r = a + ( 1 ) a 1 b + + ( r ) a r b r + + b r=0 ( ( 1)( 2) ( r + 1) ) = r r! =! r! ( r)! =! ( r)! r! = ( r ) ( 0 ) 1 Certai trigoometric idetities ca be derived usig DeMoivre s theorem. We ca for istace express cos, si ad ta i terms of cos, si ad ta. Example: We ca fid a expressio for cos5 Re(cos5 isi5 ) The: cos 5θ = Re (cosθ + i siθ) 5 Re cos isi = 5 (usig DeMoivre s theorem) = Re(cos 5 θ + 5i cos 4 θ siθ + 10i 2 cos 3 θ si 2 θ + 10i 3 cos 2 θ si 3 θ + 5i 4 cos θ si 4 θ + i 5 si 5 θ) = Re(cos 5 θ + 5i cos 4 θ siθ + 10i 2 cos 3 θ si 2 θ + 10i 3 cos 2 θ si 3 θ + 5i 4 cos θ si 4 θ + i 5 si 5 θ) = cos 5 θ 10 cos 3 θ si 2 θ + 5 cos θ si 4 θ cos5 cos 10cos si 5cos si. (*) If required, the right had side ca be expressed etirely i terms of cos oly. We get: cos5 cos 10cos (1 cos ) 5cos (1 cos ) = cos 10cos 10cos 5cos 10cos 5cos cos cos 20cos 5cos Note 1: We ca also get a idetity for si5 : si 5θ = Im (cosθ + i siθ) 5

4 4 = Im(cos 5 θ + 5i cos 4 θ siθ + 10i 2 cos 3 θ si 2 θ + 10i 3 cos 2 θ si 3 θ + 5i 4 cos θ si 4 θ + i 5 si 5 θ) = Im(cos 5 θ + 5i cos 4 θ siθ + 10i 2 cos 3 θ si 2 θ + 10i 3 cos 2 θ si 3 θ + 5i 4 cos θ si 4 θ + i 5 si 5 θ) = 5 cos 4 θ siθ 10 cos 2 θ si 3 θ + si 5 θ si 5θ = 5 cos 4 θ siθ 10 cos 2 θ si 3 θ + si 5 θ (**) If required, the right had side ca be expressed etirely i terms of si si (1 si ) si 10(1 si ) si si = si 10si 5si 10si 10si si si si 20si 5si Note 2: We ca also get a expressio for ta5 by dividig equatio (*) by equatio (**): si 5 5cos si 10cos si si ta5 cos cos 10cos si 5cos si Dividig every term o the top ad bottom by 5 cos gives: ta cos si 10 cos si si cos cos cos cos 10 cos si 5cos si cos cos cos = 3 5 ta 10 ta ta ta 5ta Questio: a) Fid a expressio for cos4 i terms of cos oly. b) Fid a expressio for si 4 i terms of si oly. c) Show that 3 4t 4t ta 4, where t = taθ t t Questios: 1. Fid a expressio for cos6 i terms of c cos ad s si 2. Fid a expressio for si 7 i terms of si oly. 3. Fid a expressio for ta 7 i terms of t ta.

5 5 Fidig -th root of a complex umber z z = z e iθ = z e i(θ+k2π) z (θ+k2π) i θ + k2π = z e = z {cos ( There are exactly th roots of z. Oly the values k = 0, 1,, - 1 give differet values of z. θ + k2π ) + i si ( ) } k = 0, 1, 2, 3, 1 Geometrically, the th roots are the vertices of a regular polygo with sides i Arga plae. 7 1 Example: Fid 5 th root of 1. Or show that if ω = e 2 5 πi, the the 5 th roots of uity ca be expressed as ,,,, 1 = e i(0+k2π) kεn 5 1 = e i(0+k2π) 5 k = 0, 1, 2, 3, 4 1 So the 5 th roots of uity are e 0, e i 2π/5, e i 4π/5, e i 6π/5, e i 8π/5 (r = 1) Example: Fid the cube roots of 8 8i, i.e. fid z such that z 3 = 8 8i. 8 8i = [ 128, π ] = [ 128, π + 2kπ] (from Arga plae straight forward) 4 4 The cube roots of 8 8i are: 6 i (cos( ) si( )) 6 7 i (cos( ) si( )) i (cos( ) si( )) 2 The cube roots (to 3 sf) are: i i i Note: The cube roots of 8 8i ca be show o a Argad diagram:

6 6 REAL POLYNOMIALS are polyomials with real coefficiets. Two polyomials are equal if ad oly if they have the same degree (order) ad correspodig terms have equal coefficiets. if 2x 3 + 3x 2 4x + 6 = ax 3 + bx 2 cx + d the a = 2, b = 3, c = 4, d = 6 Remaider If P(x) is divided by D(x) util a remaider R is obtai: P(x) R(x) = Q(x) + D(x) D(x) D(x) is the divisor Q(x) is the quotiet P(x) = D(x)Q(x) + R(x) R(x) is the remaider Whe we divide by a polyomial of degree 1 ("ax+b") the remaider will have degree 0 (a costat) The Remaider Theorem Whe a polyomial P(x) is divided by (x k) util a remaider R is obtai, the R = P(x): proof: P(x) x k = Q(x) + R x k P(x) = Q(x)(x k) + R P(k) = R The Factor Theorem

7 7 k is a zero of P(x) (x k) is a factor of P(x) proof: k is a zero of P(x) P(k) = 0 R = 0 as P(x) = Q(x)(x k) + R = Q(x)(x k) = (x k) is a factor of P(x) Corollar (x k) is a factor of P(x) there exists a polyomial Q(x) such that P(x) = (x k) Q(x)

8 8 P(x) is real ad 3 + i is zero. 3 i is zero [ax 3 + 9x 2 + ax 30] = [x ( 3 + i)][x ( 3 i)](ax + b) [ax 3 + 9x 2 + ax 30] = (x 2 + 6x + 10)(ax + b) = ax 3 + (6a + b)x 2 + (10a + 6b)x + 10b 6a + b = 9 & 10b = 30 10a + 6b = a 10b = 30 b = 3 10a + 6b = a 9a + 6b = 0 a = 2 or 6a + b = 9 a = 2 liear factor (ax+b) = 2x 3 so zeroes are: 3 ± i ad 3 2 Polyomials: Sums ad Products of Roots ax 2 + bx + c = 0 p, q roots the p + q = b a (x p)(x q) = 0 x 2 (p + q)x + pq = 0 pq = c a x 2 + b a x + c a = 0 a x 3 + b x 2 + cx + d = 0 p, q, r roots p + q + r = b a (x p)(x q)(x r) = 0 (x p)(x q)(x r) = 0 [ x 2 (p + q)x + pq](x r) = 0 x 3 (p + q + r) x 2 + (pq + pr + qr)x pqr(x r) = 0 pq + pr + qr = c a pqr = d a x 3 + b a x2 + c a x + d a = 0 i geeral: β 1, β 2, β 3,, β are roots of the equatio a x + a 1 x a 1 x 1 + a 0 = 0 (a 0 0) secod coefficiet sum of the roots = first coefficiet = a 1 a product of the roots = ( 1) first coefficiet free term = ( 1) a a 0