Jens Sebel (Unversty of Appled Scences Kserslutern) An Interctve Introducton to Complex Numbers 1. Introducton We know tht some polynoml equtons do not hve ny solutons on R/. Exmple 1.1: Solve x + 1= for x: x x x x + 1= = 1 = 1 = 1 Problem: The equton hs no soluton on R/, s 1 R/. An ext strtegy s to ntroduce n mgnry unt wth = 1 or = 1, lterntvely. Then, 1 x + = hs two solutons: x = nd x=. Rule 1.1 The mgnry unt s defned s = 1 or = 1. Exmple 1.: Solve x x+ 5= for x: x x+ 5= x= 5 x= + 5 ( ) ( ) x= 1 4 x= 1+ 4 x= 1 4 1 x= 1+ 4 1 x= 1 1 x= 1+ 1 x= 1 x= 1+ The solutons x= 1 nd x= 1+ re complex numbers. In crtesn form they re composed from rel number nd n mgnry number. Normlly, complex numbers re denoted z. Accordng to the fundmentl theorem of lgebr, ny polynoml equton of grde n N/ hs n complex solutons. 1
Crtesn form Rule 1. Jens Sebel: An Interctve Introducton to Complex Numbers 1. Introducton A complex number z n crtesn form s wrtten z= + b,, b R/ wth ( z) = Re : rel prt of z b ( z) = Im : mgnry prt ofz : mgnry unt. Exmple 1.3: z1 = 1 nd z = 1+ re the solutons of exmple 1.. We hve = Re( z ) = 1, b ( z ) 1 1 = Im = 1 1 = Re( z ) = 1, b ( z ) = Im = Rule 1.3 C/ s the set of complex numbers z. It comprses ll other sets of numbers. N/ Z/ Q/ R/ C/ Fgure 1.1: Sets of numbers Any rel number x s complex number z= x+. Rel numbers do not hve mgnry unts. Rule 1.4 Complex numbers re presented s ponts or poston vectors n crtesn dgrm ( complex plne ). Any rel number s locted on the rel xs.
Jens Sebel: An Interctve Introducton to Complex Numbers 1. Introducton Im(z) b z= + b Fgure 1.: Complex plne Re(z) Exmple 1.4: Open the Applet Bsc Clcultons nd enter z1 = 1+ nd z =.5 1.5 n crtesn form. Polr form In studyng the Applet Bsc Clcultons n exmple 1.4 you my hve notced tht the complex number z1 = 1+ hs lso been dsplyed s ( ) 63.4349 z1 =.361 exp 63.4349 =.361 e. Ths s the so-clled exponentl form - specl cse of the polr form. Vector lgebr llows us to turn ny complex number z from crtesn form nto polr form. 3
Jens Sebel: An Interctve Introducton to Complex Numbers 1. Introducton Im(z) b z= + b r ϕ Re(z) Fgure 1.3: Geometrc presentton of z n polr form In order to rewrte z= + b n polr form we hve to determne the vlue of ϕ nd the length r of the poston vector z= + b. Rule 1.5 z s the bsolute vlue of complex number. It corresponds the length vector z= + b. Applyng Pythgors equton we get z = r= + b. r R/ of Rule 1.6 ϕ s the ngle between the postve rel xs nd the poston vector z= + b. ϕ s lso clled the rgument of z : ϕ = rgz. It cn be gven both n degree or rdn. Due to vector lgebr we hve = + s z r cosϕ r snϕ r ( cosϕ snϕ) z b = r cosϕ nd b= r snϕ. Then, we cn rewrte = + = +. Rule 1.7 ( ) z= r cosϕ+ r snϕ = r cosϕ+ snϕ s the trgonometrc form of complex number z= + b wth = r cosϕ nd b= r snϕ. b Regrdng fgure 1.3 we see tht tnϕ = nd hence 4
Jens Sebel: An Interctve Introducton to Complex Numbers 1. Introducton b rctn f >, b ϕ [, 9 ) 9 f =, b> b ϕ = rgz= 18 + rctn f < ϕ ( 9, 7 ) 7 f =, b< b 36 + rctn f >, b ϕ 7, 36 or ( ] b π rctn f >, b ϕ, π f =, b> b π 3 ϕ = rgz= π + rctn f < ϕ, π 3 πf =, b< b 3 π + rctn f >, b ϕ π, π (degree) (rdn). Exmple 1.5: The trgonometrc form of z1 = 1+ n exmple 1.4 s z1 =.361 ( cos 63.4349 + sn 63.4349 ), s ϕ 1= rctn = 63.4349 nd 1 r 1= 1 + =.361. The Euler equton enbles us to convert z r ( cosϕ snϕ) = + nto z= r e ϕ. Rule 1.8 z= r e ϕ s the exponentl form of complex number z. In Generl The polr form of z s depcted by r nd ϕ. It cn be dsplyed n ether trgonometrc form or exponentl form. Note tht t requres the trgonometrc form to turn the exponentl form nto the crtesn form. 5
Jens Sebel: An Interctve Introducton to Complex Numbers 1. Introducton Throughout the followng nd n the pplets we wll prefer the exponentl form to the trgonometrc form, s the frst one hs shorter notton nd less demndng lgebr. Note tht both pplets dsply the exponentl form s z r exp( ϕ ) =. Exmple 1.6: Open the Applet Bsc Clcultons nd enter 45 z1 1 e = nd.5 z e = n exponentl form. Note tht ϕ 1= rgz1 = 45 s n degree, wheres ϕ = rgz =.5 s n rdn. When enterng ϕ n degree you hve to dd. Exmple 1.7: In order to rewrte 45 z1 1 e = nd.5 z e = from exmple 1.6 n crtesn form, one hs to clculte s follows: ( ) ( ) z = e = + = + = + 45 1 1 1 cos 45 sn 45 1.771.771.771.771 ( ) z = e = + = +.5 cos.5 sn.5 1.63 1.1969 Exercse 1.1: Rewrte z= 1+ 3, z= 1+ 3, z= 1 3 nd z= 1 3 n exponentl form. Use the Applet Bsc Clcultons to check your nswers. Exercse 1.: Rewrte z e z e z e 75 9 85 =, =, = nd z.5 π = e n crtesn form. Use the Applet Bsc Clcultons to check your nswers. You cn enter.5 π s.5*p or p/4. s sqrt() nd 6