Defiitios Complex Numbers Brief Notes A complex umber z is a expressio of the form: z = a + bi where a ad b are real umbers ad i is thought of as 1. We call a the real part of z, writte Re(z), ad b the imagiary part of z, writte Im(z). The complex umber z = a + bi is called real if b = 0 ad purely imagiary if a = 0. Two complex umbers are equal if they have both real ad imagiary parts the same. The complex cojugate of z = a + bi is defied to be z = a bi. Thus z = 3 + 6i is a complex umber with Re(z)= 3 ad Im(z)= 6, ad z = 3 6i. Complex arithmetic The golde rule for arithmetic of complex umbers is : the usual rules of algebra hold with the proviso that i = ii = 1, i.e wheever you ecouter a product of two is you replace it by 1. Thus we defie: (a + bi) + (c + di) = (a + c) + (b + d)i Examples: (a + bi) (c + di) = (a c) + (b d)i (a + bi)(c + di) = (ac bd) + (ad + bc)i. ( 3i) + (4 5i) = 6 8i, ( 3i) (4 5i) = + i, ( 3i)(4 5i) =.4 + ( 3)( 5)i + ( 5)i 3.4i = 8 15 10i 1i = 7 i. Note that if z = a + bi the z = a bi so zz = (a + bi)(a bi) = a + b which is real. This property give us the rule for dividig complex umbers: we multiply both umerator ad deomiator by the cojugate of the deomiator, e.g. 3i 1 i = ( 3i)(1 + i) (1 i)(1 + i) = + 4i 3i 6i (1 4i ) = 8 + i 5 = 8 5 + 1 5 i. The usual rules of arithmetic hold, e.g. for z 1, z, z 3 complex umbers: (z 1 + z )z 3 = z 1 z 3 + z z 3, z 1 z = z z 1, (z 1 z )z 3 = z 1 (z z 3 ). Thus we ca solve equatios, simultaeous equatios, quadratic equatios, etc., i the usual way, provided we always replace i by 1. 1
Example Solve the quadratic equatio z + (1 i)z 1 i = 0. By the formula for solutio of a quadratic equatio, z = (1 i) ± (1 i) 4( 1 i) = 1 + i ± 1 4i 4 + 4 + 4i = 1 + i ± 1 = i or i 1. The complex plae We represet z = a + bi by the poit (a, b) i the coordiate plae. This picture is called the complex plae or Argad diagram. The x-axis is called the real axis ad the y-axis is called the imagiary axis. We sometimes thik of z = a + bi as the vector from the origi (0, 0) to the poit with coordiates (a, b). The legth of this vector (i.e. the distace of (a, b) from the origi) is called the modulus or absolute value of z, writte z. The agle made by this vector with the real axis is the argumet of z, writte argz (ote that 0 has o argumet). By Pythagoras theorem z = a + b = zz so z = a + b = zz. Complex umbers are ofte very coveiet to describe geometric properties of the plae. Very useful is: To see this ote that (distace betwee z = a + bi ad w = c + di) = z w. z w = (a + bi) (c + di) = (a c) + (b d)i = (a c) + (b d) which, by Pythagoras theorem, is the distace betwee the poits (a, b) ad (c, d) i the coordiate plae. Certai lies ad curves i the plae have very simple equatios i terms of complex umbers. For example, z ( + i) = 3 is the equatio of the circle with cetre ( + i) ad radius 3, sice z will satisfy z ( + i) = 3 precisely whe the distace of z from the poit ( + i) equals 3.
Usig the same idea: z w = r is the circle with cetre w ad radius r z w = z u is the perpedicular bisector betwee the poits w ad u z w + z u = c is a ellipse with foci w ad u. Modulus-argumet form ad multiplicatio Let z = a + bi have modulus r = z ad argumet θ = argz. From the diagram ad also I particular a = r cos θ ad b = r si θ r = z = a + b ad ta θ = b/a. z = a + bi = r(cos θ + i si θ). We call r(cos θ + i si θ), where r = z ad θ = argz the modulus-argumet form or polar form of z. Thus, we have 1 + 3i = (cos(π/3) + i si(π/3)), + i = (cos(3π/4) + i si(3π/4)). [it is advisable to draw a diagram whe puttig a umber ito modulus-argumet form.] Note that a complex umber has may argumets: if argz = θ the..., θ 4π, θ π, θ, θ + π, θ + 4π,... are also valid argumets. The priciple argumet is the value with π < θ π. Modulus-argumet form is very useful whe multiplyig complex umbers, squarig, cubig, etc., sice: z 1 z = z 1 z arg(z 1 z ) = argz 1 argz. Thus: to multiply complex umbers, multiply the moduli ad add the argumets. Proof: Write z 1 ad z i modulus-argumet form, so that z 1 = r 1 (cos θ 1 + i si θ 1 ) ad z = r (cos θ + i si θ ). 3
The z 1 z = r 1 (cos θ 1 + i si θ 1 )r (cos θ + i si θ ) = r 1 r (cos θ 1 + i si θ 1 )(cos θ + i si θ ) = r 1 r ( (cos θ1 cos θ si θ 1 si θ ) + (si θ 1 cos θ + cos θ 1 si θ ) ) = r 1 r (cos(θ 1 + θ ) + i si(θ 1 + θ )), usig the additio formulae from trigoometry. But this is just the complex umber with modulus r 1 r ad argumet θ 1 +θ, that is z 1 z = r 1 r ad arg(z 1 z ) = θ 1 +θ, as required. Example: From above, (1 + 3i)( + i) = (cos(π/3) + i si(π/3)) (cos(3π/4) + i si(3π/4)) =. (cos(π/3 + 3π/4) + i si(π/3 + 3π/4)) = 4 (cos(13π/1) + i si(13π/1)). Powers ad de Moivre s theorem From above, for a positive iteger power : z = z, arg(z ) = argz. ( ) For each positive iteger m, we have that 1 = z m z m ; thus ad 1 = z m z m = z m z m so z m = z m 0 = arg 1 = arg(z m z m ) = m argz + arg(z m ) so arg(z m ) = m argz. Thus ( ) is also true for egative itegers = m. I particular, takig z = cos θ + i si θ, this gives (cos θ + i si θ) = cos θ + i si θ (de Moivre s theorem), which is valid for ay positive or egative iteger. de Moivre s theorem has may importat applicatios. (i) Powers of umbers. e.g. to fid (1 + 3i) 8. I modulus-argumet form (1 + 3i) 8 = (cos(π/3) + i si(π/3)) so (1 + 3i) 8 = 8 (cos(π/3) + i si(π/3)) 8 = 8 (cos(8π/3) + i si(8π/3)) = 8 (cos(π/3) + i si(π/3)) = 56( 1/ + i 3/) = 18 + 18 3i. (ii) Trigoometric fuctios. e.g. to expad cos 3θ, usig de Moivre s theorem ad multiplyig out gives: cos 3θ + i si 3θ = (cos θ + i si θ) 3 = cos 3 θ + 3 cos θ i si θ + 3 cos θ i si θ + i 3 si 3 θ = cos 3 θ + 3i cos θ si θ 3 cos θ si θ i si 3 θ. 4
Equatig real ad imagiary parts gives cos 3θ = cos 3 θ 3 cos θ si θ = cos 3 θ 3 cos θ(1 cos θ) = 4 cos 3 θ 3 cos θ Similarly for cos 4θ, si 4θ, etc. si 3θ = 3 cos θ si θ si 3 θ = 3(1 si θ) si θ si 3 θ = 3 si θ 4 si 3 θ. (iii) th roots of complex umbers. Care is required sice complex umbers have square roots, 3 cube roots, etc. Let be a positive iteger. Note that by de Moivre s theorem, for every iteger k ( cos θ + πk + i si θ + πk ) = cos(θ + πk) + i si(θ + πk) = cos θ + i si θ. Hece ( ) ( ) θ + πk θ + πk cos + i si is a th root for every iteger k. Takig k = 0, 1,,..., 1 gives the differet values. Example: To fid the cube roots of i: Note that i = cos(π/) + i si(π/) = cos(π/ + πk) + i si(π/ + πk) for all k. By de Moivre s theorem ( cos((π/ + πk)/3) + i si((π/ + πk)/3) ) 3 = cos(π/ + πk) + i si(π/ + πk) = i, so takig k = 0, 1, gives the three cube roots as cos(π/6) + i si(π/6) = 3/ + i/, cos(5π/6) + i si(5π/6) = 3/ + i/, cos(9π/6) + i si(9π/6) = i. Note that the th roots of ay complex umber are symmetrically arraged at the vertices of a regular -sided polygo cetered at the origi. Euler s formula The expoetial series is e x = 1 + x + x! + x3 3! + x4 4! + x5 5! +. Assume this series is valid for complex x as well as real x. Substitute x = iθ to get e iθ = 1 + iθ + i θ! + i3 θ 3 3! + i4 θ 4 4! + i5 θ 5 5! + = 1 + iθ θ! iθ3 + θ4 3! 4! + iθ5 + 5! ) ) = (1 θ! + θ4 4! + + i (θ θ3 3! + θ5 5! + = cos θ + i si θ 5
recallig the series for cos θ ad si θ. Thus e iθ = cos θ + i si θ. (1) This is Euler s formula ad is the reaso why the expressio cos θ +i si θ is so importat. There are may cosequeces of this formula: 1. Puttig θ = π i (1) gives e iπ = cos π + i si π = 1. Thus we get Euler s idetity e iπ + 1 = 0.. We have e iθ = cos( θ) + i si( θ) = cos θ i si θ o replacig θ by θ i (1). Addig ad subtractig this to (1) e iθ + e iθ = cos θ ad e iθ e iθ = i si θ, so cos θ = eiθ + e iθ ad si θ = eiθ e iθ. i This is remiiscet of the defiitios of hyperbolic fuctios. 3. Usig (1) ad the rules of expoets (cos θ + i si θ) = (e iθ ) = e iθ = cos θ + i si θ which is a quick derivatio of de Moivre s Theorem. 4. Agai usig the rules of expoets e i(θ+φ) = e iθ+iφ = e iθ e iφ so applyig (1) to both sides cos(θ + φ) + i si(θ + φ) = (cos θ + i si θ)(cos φ + i si φ) = (cos θ cos φ si θ si φ) + i(si θ cos φ + cos θ si φ) so equatig real ad imagiary parts gives the additio formulae for trigoometry. 6