Rtionl Numbers vs Rel Numbers 1. Wht is? Answer. is rel number such tht ( ) =. R [ ( ) = ].. Prove tht (i) 1; (ii). Proof. (i) For ny rel numbers x, y, we hve x = y. This is necessry condition, but not sufficient one s ( 1) = 1 = (1), while 1 1. Note tht if ( ) =, nd (1) = 1, so 1. (ii) We first show tht 0. This follows from proving the squres of, nd 0 re different s in (i). Then, we wnt to prove tht nd re different s rel numbers, s they hve different signs. By the Lw of Trichotomy of rel number system, we know tht given ny rel number x R, we hve the following 3 possibilities: (i) x = 0; (ii) x > 0; (ii) x < 0. And only 1 of these possibilities holds. As 0, we know is positive or negtive. It follows from properties of rel numbers tht for ny x R, we hve x > 0 if nd only if x < 0. If > 0, then < 0, nd hence s they re of different signs. Similrly, the sme conclusion holds if < 0. 3. Prove tht is not n integer. Proof. One cn esily prove tht is differen from 0 nd ±1 s in question bove. Define f(x) = x for ll x R. By writing f(x) f(y) = x y = (x y)(x + y), one cn esily prove tht f(x) = x is incresing on the set (0, ) of positive rels, nd f(x) = x is decresing on the set (, 0) of negtive rels. If > 0, then = f( ) f() = 4, which is contrdiction. If < 0, then = f( ) f() = 4, which is contrdiction. Hence, one knows tht is not n integer. Absolute vlues of rel numbers 1. Definition. Absolute vlue α of rel number α R is defined s follows: α if α 0; α = α if α < 0.. Proposition. Let α R, then () α α nd α α; (b) α = α. Proof of () nd (b). We consider the following different cses: (i) α 0: then α = α ( ), hence α α. As 0 = α + ( α) 0 + ( α) = α, i.e. 0 α. So, α = α 0 α. Moreover, it follows from 0 α tht α = ( α) = α = α. (ii) α < 0: then α = α ( ), hence α α. As 0 = α + ( α) < 0 + ( α) = α, i.e. 0 α. So, α = α > 0 > α. Moreover, it follows from 0 < α tht α = α = α. 3. Proposition. If α, β R, then one hs α β α β nd α β. Proof. ( ) Suppose tht α β. Then it follows from proposition tht α α β nd α α β. ( ) Suppose tht both inequlities α β nd α β hold. By definition of bsolute vlue, we know tht α = α β or α = α β. In ny cse, we hve α β. 4. Proposition. If α, β R, then α + β α + β. Proof. For ny α, β R, we consider the following 4 different cses: (i) α 0 nd β 0 : Then it follows from α 0 nd β 0 tht α = α, nd β = β. So α + β 0 + 0 = 0, nd hence α + β = α + β. In this cse, we hve α + β = α + β = α + β, so the inequlity holds. (ii) α 0 nd β < 0 : Then α = α nd β = β. So we hve the following α + β < α + 0 = α = α = α + β. (iii) α < 0 nd β 0 : By interchnge the role of α nd β, the result follows from the cse (ii) bove. (iv) α < 0 nd β < 0 : Then ( α) > 0, nd ( β) > 0. Applying the conclusion of (i) bove to α, β insted, it follows from proposition (b) tht α + β = (α + β) = ( α) + ( β) ( α) + ( β ) = α + β. 1
Complex Numbers 1. Let C = { + bi = (, b) R, b R } be the set of ll complex numbers. Let = + bi = (, b), we cll nd b re clled rel nd imginry prts of respectively, denoted by Re = nd Im = b.. Let = + bi nd w = c + di be two complex numbers with, b, c, d R. One cn see tht = w the corresponding vectors (, b) nd (c, d) re equl in R = c nd b = d. Remrk. w c or b d. 3. Addition + : C C C is binry opertion in C, defined to be usul ddition in R, i.e. for ny v = + bi nd w = c + di, the sum + w of nd w, is defined to be v + w = ( + c) + (b + d)i. One cn immeditely check the following: (i) Commuttive lw: v + w = w + v for ll v, w C; (ii) Associtive lw: (v + w) + = v + (w + ) for ll v, w, C; (iii) Zero: 0 + 0i = (0, 0), denoted by 0, stisfies 0 + v = v + 0 = v for ll v C; (iv) Additive inverse: For ny = + bi = (, b) C, we hve w = ( ) + ( b)i C, denoted by stisfying + w = w + = 0. Remrk. The tuple (C, +, 0) is clled n belin group. 4. Multipliction : C C C is binry opertion in C defined s follows: for ny v = + bi nd w = c + di, the product v w of v nd w, is defined to be v w = ( + bi) (c + di) = (c bd) + (d + bc)i. One cn immeditely check the following: (v) Commuttive lw: v w = w v for ll v, w C; (vi) Associtive lw: (v w) = v (w ) for ll v, w, C; (vii) Unit: 1 + 0i = (1, 0), denoted by 1, stisfies 1 v = v 1 = v for ll v C; (viii) Distributive lw: (v + w) = v + w nd (v + w) = v + w, for ll v, w, C. (ix) Multiplictive Inverse: For ny = + bi = (, b) C \ {0}, we hve w = +b + b +b i C, denoted by 1 or 1 stisfying w = w = 1. Remrk. The tuple (C, +,, 0, 1) stisfying (i)-(viii) is clled commuttive ring; The tuple (C, +,, 0, 1) stisfying (i)-(viii) is clled field. 5. Exmple. Prove (v)-(iv) for multipliction in C. Proof. Let v = + bi, w = c + di, = x + iy. (v) Following the definition of multipliction, we hve v w = ( + bi)(c + di) = (c bd) + (d + bc)i, nd w v = (c + di)( + bi) = (c db) + (cb + d)i = (c bd) + (bc + d)i = w v. (vi) (v w) = ((c bd) + (d + bc)i) (x + iy) = ((c bd)x (d + bc)y) + ((c bd)y + (d + bc)x) i = ((cx dy) b(cy + dx)) + ((cy + dx) + b(cx dy)) i = ( + bi) ((cx dy) + (cy + dx)i) = v (w ). (vii) 1 v = (1+0i) (+bi) = (1 0 b)+(1 b+0 )i = +bi = (+bi) (1+0i), in which the lst equlity follows from commuttive lw (v). (viii) (v + w) = ( ( + bi) + (c + di) ) (x + yi) = ( ( + c) + (b + d)i ) (x + yi) = ( ( + c)x (b + d)y ) + ( ( + c)y + (b + d)x )i = ( (x by) + (cx dy) ) + ( (y + bx) + (cy + dx) )i = ( (x by) + (y + bx)i ) + ( (cx dy) + (cy + dx)i ) = v + w. (ix) For ny = + bi C \ {0}, we know tht 0, then 0 or b 0, i.e. +b + b = > 0 or b > 0, hence + b > 0. In prticulr, ( the complex number +b i is well-defined, nd 1 = ( + bi) + b + b ) + b i ) ( ) + b b b b + b + + b + b + b i ( = + b b + b + + b + b i = 1. 6. Lw of Multiplictive Cnceltion. Let v, w, C. If v = w nd 0, then v = w. Proof. v (vii) = v 1 (ix) = v ( 1 ) (v) = (v ) 1 = (w ) (ix) = w 1 (vii) = w. 1 (v) = w ( 1 ) Remrk. The condition 0 is essentil, s 1 0 = 0 = 0 but 1. 7. Proposition. Let w, C. If w = 0, then = 0 or w = 0. Proof. We divide two possible cses: (i) 0, then the condition cn be rewritten s w = 0 = 0, it follows from Lw of Multiplictive Cnceltion tht w = 0. (ii) = 0, so the result follows. 8. Definition. Let = +bi C with, b R, define complex conjugte of, denoted by, to be = bi. Define modulus of, denoted by, to be + b.
9. Proposition. (i) 0 for ll C, nd (ii) the equlity holds if nd only if = 0. Proof. (i) Let = + bi C, with = Re nd b = Im R. Then, b 0, so + b 0. = + b 0. (ii) = 0 + b = = 0 = b = 0 = b = 0 = + bi = 0. 10. Proposition. Let = + bi C, where, b R, then we hve (i) = Re() Im()i. (ii) Re = +, nd Im = i. (iii) = =, nd =. (iv) If 0, then 1 = Proof.. (i) If = + bi with, b R, then Re() =, nd Im() = b, so = bi = Re() Im()i. (ii) Re = bi, nd Im = i (iii) It follows from commuttive lw for multipliction tht = = (+bi)+( bi) = + = (+bi) ( bi) i = i. = ( + bi)( bi) = ( b( b) ) + (( b) + b)i = + b = ( + b ) =, nd hence =. (iv) The result follows from 1 = + b + b + b i = bi + b = 11. Theorem. Let = + bi, w = c + di C with, b, c, d R. Then. (i) Re(), nd Im(). (ii) w = w, nd w = w for non-ero w C. Proof. (i) Re() = = = + b =, nd similrly, Im() = b = b = b + b =. (ii) w = ( + bi)(c + di) = (c bd) + (d + bc)i = (c bd) + (d + bc) = ( c cbd + b d ) + ( d + dbc + b c ) = ( c + b d ) + ( d + b c ) = ( c + d ) + (b d + b c ) = (c + d ) + b (d + c ) = ( + b )(c + d ) = ( + b ) (c + d ) = w. If w 0, then 1 = 1 = w 1 w = w 1 w, so w = 1 w = 1 w = 1 w = w. 1. Theorem. (Tringle Inequlity). For ny, w C, + w + w. Proof. We consider two different cses: (1) + w = 0 : then + w = 0 = 0, nd it follows from, w 0 tht + w = 0 = 0 + w. () + w 0 : then 1 = +w +w Re( w +w ) + Re( +w ) +w + w + w + w., so 1 = Re1 = Re( +w +w ) = Re( +w = +w + +w ) + Re( w w +w = + w +w +w ), so we hve Remrks. (i) Equlity holds bove if nd only if there exist α, β 0 nd (α, β) (0, 0) such tht α = βw. The proof is left to the reder s n exercise. Moreover, one cn prove the following two corollries by using tringle inequlity. (ii) We will give nother proof of tringle inequlity with qudrtic function. 13. Corollry. For ny 1,,, n C, we hve 1 + + + n 1 + + n. 14. Corollry. For ny 1,, 3 C, we hve 1 3 1 + 3. Proof. It follows from tringle inequlity of complex numbers tht 1 3 = ( 1 ) + ( 3 ) 1 + 3. 15. Proposition. For ny, w C, we hve w ± w. Proof. = w + ( w) w + w, so w w, nd by interchnging the roles of nd w, we hve w w = ( w) = 1 w = w. Hence, w w. Moreover, replcing w by w, we hve w = w ( w) = + w. Complex Numbers in Polr form. 1. Euler Formul. exp(iθ) = cos θ + i sin θ for ll rel θ. Remrk. The proof is beyond the present knowledge for first yer student, s it demnd the understnding of power series.. Use Euler s formul, define complex-vlued exponentil function exp : C C s follows: exp() = e x+iy = e x e iy = e x (cos y + i sin y) = e x cos y + ie x sin y, where x = Re nd y = Im. 3. Proposition. exp( + w) = exp() exp(w) for ll, w C. Proof. Let = x + iy nd w = u + iv, where u, v, x, y R. Then we hve exp( + w) = exp((x + u) + i(y + v)) = e x+u cos(y + v) + ie x+u sin(y + v) = e x+u (cos y cos v sin y sin v) + ie x+u (sin y cos v + cos y sin v) = (e x cos y e u cos v e x sin y e u sin v) + i(e x sin y e u cos v + e x cos y e u sin v) = (e x cos y + ie x sin y) (e u cos v + ie u sin v) = e x+iy e u+iv = exp() exp(w). 3
4. Proposition. e = 1 if nd only if = nπi for some n Z. Proof. ( ) e nπi = cos(nπ) + i sin(nπ) = 1 for ll n Z. ( ) Write = x + iy where x, y R. Then 1 + 0i = 1 = e = e x+iy = e x e iy = e x (cos y + i sin y) = e x cos y + ie x sin y, so we hve 1 = e x cos y nd 0 = e x sin y. Then e 0 = 1 = 1 + 0 = (e x cos y) + (e x sin y) = e x (cos y + sin y) = e x. As exp is n incresing rel-vlued function defined on R, so 0 = x, i.e. x = 0. Hence, we hve cos y = 1 nd sin y = 0. By the lst eqution, we hve y = kπ for some k Z. Moreover, 1 = cos y = cos(kπ) = ( 1) k, thus k is even, i.e. n = k for some k Z. 5. Nottion. Let, b R. We cll nd b re congruent modulo π if there exists integer n Z such tht b = nπ. In this cse, we denote by b (mod π). For exmple, π π (mod π), nd α 6π + α (mod π) for ny α R. 6. Proposition. For ny complex numbers = re iθ, w = Re iα, where r, R R, then (i) w = rre i(θ+α) nd (ii) rg( w) = rg + rg w (mod π), i.e. rg( w) = rg + rg w + nπ for some n Z. (iii) rg( 1 ) = rg (mod π) for ll C \ {0}. Proof. (i) w = re iθ Re iα = rre iθ e iα = rre i(θ+α). (ii) By (i) rg( w) = (θ + α) + nπ = rg + rg w + nπ for some n Z. (iii) By (ii) rg(1/) + rg rg(1/ ) = rg(1) = 0 (mod π), so rg(1/) rg (mod π). Remrk. The integer n in (ii) my not be ero in generl. This mens the rgument turns the multipliction of complex numbers into ddition modulo π, insted of exct ddition in rel number R. For exmple 1 = e iπ, so rg( 1) = π, nd rg(1) = 0. However, 1 = ( 1) ( 1), nd rg( 1) + rg( 1) = π + π = π, while rg(( 1) ( 1)) = rg(1) = 0. Equtions of lines nd circles in complex plne 1. Let l be line in xy-plne, then l = { (x, y) R x + by + c = 0}, where c, b, c R nd (, b) (0, 0). In this cse, we write l : x + by + c = 0. Express point (x, y) l by = x + iy C, then x = Re() = +, nd y = Im =. Hence, rewrite the eqution of l in terms of s 0 = (x + by + c) = ( + + b i + c) = + ib + ib + c = ( ib) + ( + bi) + c. Hence, l = { C α + α + c = 0 }, where α = + ib 0.. Proposition. (i) Let l be line in the complex plne C, then l is defined by n eqution α + α + c = 0 for some complex number α 0 nd rel number c. (ii) Let α = + ib 0, nd c R, the set S = { C α + α + c = 0 } represents line in C. Proof. (i) We hd just proved bove. (ii) Let = x + iy, then S 0 = α + α + c = ( ib)(x + iy) + ( + bi)(x iy) + c = (x + by) + i(y bx) + (x + by) + i(bx y) + c = x + by + c, so the set S is line given by the eqution 0 = x + by + c. 3. Let u, v C be two distinct points, nd l u,v be the set of of points in the perpendiculr bisector of the segment joining u nd v. Then we hve (i) l u,v is given by the loci { C u = v }. (ii) l u,v : + = v u, where = v u 0. Solution. (i) is just resttement of the perpendiculr bisector. (ii) l u,v u = v u u + u = v v + v, u u + u = v v + v, (v u) + (v u) = v u. 4
Isometries of complex plne 1. Definition. Let l be line through fixed point O in R. Define mp Refl l : R R s follows. For ny vector OZ R, let M be point in l such tht ZM l. Mrk point W on the line through Z nd M such tht ZM = MW. Then Refl l ( OZ) = OM for ll OZ R.. Let C with = 1, nd rg = e iθ. Define f() = for ll C, Prove tht f is the reflection bout the line l : e iθ/ + e iθ/ = 0. Proof I. Let = re iα, where α R, then f() = f(re iα ) = = e iθ re iα = e iθ re iα = re i(θ α). So, rg f() = (θ α) (mod π) nd rg = α (mod π). Then we hve rg( f() ) rg(f()) θ e iθ/ (θ α) θ θ eiθ/ α rg rg (mod π). e iθ/ Proof II. This proof requires little more on liner lgebr. For ny = xe iθ/ + iye iθ/ with x, y R, then f(xe iθ/ +iye iθ/ ) = f() = = e iθ (xe iθ/ + iye iθ/ ) = e iθ (xe iθ/ iye iθ/ ) = xe iθ/ iye iθ/. 5