INTRODUCTION TO COMPLEX NUMBERS
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1 INTRODUCTION TO COMPLEX NUMBERS The numers -4, -3, -, -1, 0, 1,, 3, 4 represent the negtve nd postve rel numers termed ntegers. As one frst lerns n mddle school they cn e thought of s unt dstnce spced ponts long strght lne s shown- The sc mthemtcl opertons wth these numers re: ddton : A + B sutrcton: A-B multplcton: AxB nd dvson: A/B There re numers whch le etween these whole numers. They re the rtonls such s 5/ whose dgtl expnson repets tself fter n unts nd the rrtonls such s sqrt( where the dgts never repet themselves. Although the product of two ntegers wll lwys e n nteger, the sme s not true for the roots of ntegers. Tke, for exmple, the squre root of whch s nontermntng rrtonl numer. If one tkes thngs one step further sks wht s sqrt(-, one fnds- ( ( 1 where sqrt( ssgvenove ut wht s the menng of sqrt(-1? One clls ths quntty the mgnry numer. Wth ths defnton t s possle to extend the set of ll numers nto n even lrger set of complex numers - z+ wth representng the rel prt nd the mgnry prt of the numer z. The stndrd desgnton s Re(z nd Im(z. We lso hve tht the solute vlue of z equls A(z z sqrt( +. On tkng the squre of z we hve- z ++((( - + snce -1. Tkng more powers of we fnd the followng propertes-
2 So tht we cn wrte ny nteger power p of s- p (4n+r r wth n nteger nd the remnder r0, 1, or 3 Tht s 359 ( The sc ddton, sutrcton, multplcton nd dvson lws for complex numers remn s they were for rel numers. Therefore- ( (-1+ nd (3-+(-+1+ A convenent wy to plot complex numer z s y mens of n Argnd Dgrm n whch the rel prt of complex numer s mesured long the x xs nd the mgnry porton mesured long the y xs. We cn represent z n ether ts Crtesn form or ts polr form. They red respectvely- z+sqrt( + exp[ rctn(/]r exp(θ Here z Rsqrt( + s the mpltude(or modulus nd θrg(zrctn(/ the rgument of z. By replcng y n complex numer one produces ts complex conjugte desgnted y z. One lwys hs tht z z z s rel numer. Here s grph of the complex numer z+ nd ts conjugte n the Argnd dgrm-
3 Let us next look t the numer exp(zexp(+ n more detl. Expndng ths functon s Tylor seres we hve- e + e e e (1! 4 + 4! ( 3! 5 + 5!... But the two nfnte seres n the curly rcket re recognzed s cos( nd sn(. Hence one hs the fmous formul frst derved y Leonrd Euler, nmely,- exp( + e [cos( ] On settng 0 nd π/ nd π we hve exp(π/ nd exp(π-1, respectvely. Thus one cn conclude tht- If n we hve tht- n exp( nπ/ cos( nπ/ nπ/ exp( π/ whch s rel ut rrtonl numer. Also we hve tht z hs the soluton- 1/3 /3 z ( 1 cos( π / 3 π/ 3 (1+ 3 / Ths, however, s not the only soluton snce there re two more whch cn e gotten y rottng wy from the frst soluton y θ±π/3 rdns. The other two solutons re- cos(π+sn(π-1 nd cos(-π/3+sn(-π/3(1-3/ The root of complex numer cn e wrtten s- z 1/ N R 1/ N ( θ + πk exp[ ] N where k 0, ± 1, ±, etc Ths equlty s known n the lterture s the demovre Formul. It follows from the form of the polr representton of the complex numer z nd clerly shows the presence of N multple roots. A queston I used to sk my undergrdute nlyss clss t the Unversty of Flord ws Are there ny rel solutons to the pth root of when p s n nteger? The nswer s no nd here s the proof-
4 (1 4k exp[ π ] p 1/ p + To hve rel soluton requres tht sn[π(1+4k/(p]0 whch mples tht p(1+4k/n wth n0,±1,± etc. We see tht the numertor n ths expresson s n odd numer nd the denomntor s lwys n even numer. Hence p cn never e n nteger nd so no pure rel solutons cn exst! Note tht pure rel roots re possle f p equls certn rtonl numers such s /3. One cn derve numerous trgonometrc denttes usng the Euler Formul s strtng pont. Frst settng 0 we hve the results- exp(cos(+sn( nd exp(-cos(-sn( Addng nd sutrctng these together, we rrve t the denttes- cos( + e nd sn( e whch upon replcng y c produces the hyperolc functons- cos( c c + e c cosh( c nd sn( c c e c snh( c Also, on lettng A+B, we otn the well known trgonometrc denttes- 1 cos( A+ cos( cos( sn( sn( nd- {[cos( ][cos( ] + [cos( ][cos( ] } 1 sn( A+ sn( cos( + sn( cos( {[cos( ][cos( ] [cos( ][cos( ] } The doule ngle formuls- cos(a cos A sn A 1 sn A nd sn( sn( cos( follow on settng AB. We cn lso use the complex numer representtons for sn( nd cos( to develop the qudruple ngle formuls-
5 4 3 5 cos(4a 1 8cos ( + 8cos ( nd sn(4 8sn( 4sn ( + 16sn ( Certn defnte ntegrls cn lso e ncely solved usng complex numers. Consder the ntegrl- K sn( xexp( x dx Im x 0 x 0 exp( + xdx wth nd > 0 Here Im stnds for the mgnry prt of the functon just lke Re would stnd for the rel prt. After smple ntegrton of the exponentl functon we hve- K 1 ( + Im Im ( ( + + I rememer how we derved ths result mny ters go n my frst college clculus clss y much longer route nvolvng severl ntegrton y prts. A eneft of the complex numer pproch s not only ts ese compred to other methods ut lso the fct tht t wll sometmes yeld ddtonl nformton such s, n ths cse, tht- L cos( xexp( x Re exp( + xdx + x 0 x 0 Fnlly let us show how one cn plot functon such s Fz n R n exp(nθ n the complex plne. Specfclly let n e ny postve power greter thn one ncludng non nteger vlues. Also set z+. On susttutng these vlue nto F we fnd r Modulus F( + n/ nd Θ Argument Fn rctn(/ On elmntng the n, one fnds- rexp(αθ wth α[ln( + ]/[ rctn(/]constnt Ths fgure represents the logrthmc Sprl of Bernoull nd t looks s follows for 1
6 It ws ths fgure whch I used s demonstrton n our undergrdute complex nlyss clss whch led to the dscovery of the nteger sprl- rn nd θnπ/4 for ll postve ntegers n. It produces the nterestng pcture-
7 n whch ll even ntegers le long the x or y xs whle ll odd ntegers fll long the dgonls y±x.
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