Complex Numbers. where x represents a root of Equation 1. Note that the ± sign tells us that quadratic equations will have

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1 Comple Numbers I spite of Calvi s discomfiture, imagiar umbers (a subset of the set of comple umbers) eist ad are ivaluable i mathematics, egieerig, ad sciece. I fact, i certai fields, such as electrical egieerig, aeroautical egieerig ad quatum mechaics, progress has bee criticall depedet o comple umbers ad their behavior. I the cotet of mathematical biolog, a umber of models we ll be workig with ca ield solutios that ivolve comple umbers, so we eed to refresh our memories about comple umbers ad how to work with them. The Quadratic Formula ad Comple Numbers Aside from allowig us to solve difficult problems such as 9 +??, probabl the most frequet situatio i which most of us have ecoutered comple umbers has bee whe fidig the roots of quadratic equatios of the form f ( ) A + B + C 0 Equatio 1 You ma recall from our high-school algebra course that the roots of Equatio 1 are the values of for which the equatio is eactl equal to zero, ad that we ca easil solve for the roots of Equatio 1 b meas of the quadratic formula, which usuall takes the form B ± B AC A where represets a root of Equatio 1. Note that the ± sig tells us that quadratic equatios will have B AC two roots that differ i value b a quatit equal to the value of (which ma equal zero). A Let s work with Equatio 1 ad set A C. This leads to the followig epressio for the quadratic formula B ± B 16 which we will ow solve for B K. The table o the et page gives the results of these calculatios.

2 B ± B 16 B B AC 9 0., 0 1, 1 0. ± 0. ± ± ± ± ± ± ± , , We thus see that a quadratic equatio has two roots (which ma, depedig o the values of A, B, ad C, be equal to each other). However, the roots for B,, K, probabl look a little strage, ivolvig that 1. We re about to eter the strage ad fabulous world of comple umbers Imagiar ad Comple Numbers We first focus o the etr for B 0, where the result is ± 1. Of course, there is o real umber whose square is 1, so the result is referred to as a imagiar umber. B covetio, 1 is desigated b the letter i. All imagiar umbers ma the be represeted as bi, where b is a real umber. Our quadratic equatio calculatios for B 0 thus become ± i. But, what if we let B,, K,, B 0? Here the results are a bit more complicated, ivolvig a combiatio of real ad imagiar umbers. Such umbers are referred to as comple umbers, ad are usuall represeted as a + bi, where a ad b are real umbers that ma take o a value betwee ad +. A few poits to ote here: 1. If b 0, ou have the real umber a.. If a 0, the umber bi is said to be pure imagiar, or, more simpl, imagiar.. If both a, b 0, the the umber is said to be comple. I this case o a is referred to as the real part of the comple umber, ad is represeted as Re(). o b is termed the imagiar part of the comple umber, ad is represeted b Im().. I the cotet of comple umbers, the, the roots of a quadratic equatio ma be: o real, o pure imagiar, or o comple, i which case the occur as the comple cojugates, a + bi ad a bi.. Fiall, poits #1 ad # show that the real ad imagiar umbers are subsets of the comple umbers, suggestig that the basic mathematical operatios of additio,

3 subtractio, multiplicatio, divisio, epoetiatio, etc. ca be applied to imagiar ad comple umbers. This turs out to be the case, albeit some modificatios are ecessar. Additio ad Subtractio of Comple Numbers Additio ad subtractio of comple umbers are completel trasparet, but serve to illustrate a importat techique that will prove useful to us later o. The formula for carrig out additio of comple umbers is simpl ( a + bi) + ( c + di) ( a + c) + ( b + d )i where akd are real umbers. The correspodig formula for subtractio of comple umbers is ( a + bi) ( c + di) ( a c) + ( b d )i Note what was doe each: the real ad imagiar parts of the two comple umbers were grouped, after which summatio was carried out separatel withi each group, ieldig aother comple umber with real part equal to ( a ± c) ad imagiar part equal to ( b ± d ). Also ote that if b d 0, we re just addig or subtractig real umbers. Multiplicatio ad Divisio of Comple Numbers Multiplicatio of comple umbers is straightforward, ad should look familiar to ou from our algebra das: ( a + bi)( c + di) ac + adi + bci + bdi ( ac + bdi ) + ( ad + bc) ( ac bd ) + ( ad + bc) i with the bd comig from the fact that i 1. Note that we agai grouped like terms (real ad imagiar) to obtai the fial result. Comple divisio is similarl straightforward. Give the comple divisio problem ( a + bi) ( c + di) we first ote that we if we multipl both the umerator ad deomiator b c di : ( a + bi) ( c + di) ( ac bdi + bci adi) ( c d i ) we ca get rid of that i i the deomiator. (what term do we appl to the combiatio of c + di ad c di?) Note the value of the origial quotiet is uchaged sice 1.We the group terms to obtai ac + bd + ( bc ad ) i c + d ( ) if c + d 0. We ca the immediatel derive the fial result b separatig the real ad imagiar terms to ield the fial result: ( a + bi) ac + bd ( c + di) ( c + d ) + i ( bc ad ) ( c d ) i +

4 Like comple additio ad subtractio, comple multiplicatio ad divisio ield comple umbers of the stadard form, a real part ad a imagiar part cosistig of a real umber multiplied b i. Also ote that if b d 0 the above problems reduce to multiplicatio or divisio of real umbers. I do t epect ou to commit to memor the precedig material o comple arithmetic. I do, however, wat ou to keep i mid the techiques of groupig terms ad multiplig ( a + bi ) b ( c + di) to clea up the deomiator (this is frequetl referred to i tetbooks as ratioalizig the deomiator, although that s ot strictl speakig correct usage of the term). Both are useful algebraic tricks that will come ito pla a umber of times durig lecture ad lab sessios et semester. Epoets Ivolvig Comple Numbers We ow arrive at the reaso for our fora ito the world of comple umbers. Net semester, ou will ecouter a umber of simple models that ca ield comple solutios of the form: t ( ) ( a± bi) () t f e Equatio where e is the base of atural logarithms (.188 ) ad a ± bi is a comple cojugate pair. How ( a bi)t o earth ca we geerate somethig meaigful out of e ±, a umber raised to a power that ivolves comple umbers? That seems mid-breakigl absurd i the etreme. Well, i 18 Leohard Euler, oe of histor s pre-emiet mathematicias, showed us the wa. First, we take advatage of the fact that, as with real umbers, a epoetial term ivolvig a comple sum ca be decomposed ito the product of two epoetial terms. For eample, if we start with the a + bi member of the cojugate pair, we obtai: ( a+ bi) t at ibt e e e Equatio at That looks promisig, sice a is a real umber ad we kow how to deal with e. However, there s ibt still that pesk e term remaiig to be dealt with, so it might ot seem as though we ve gaied athig at all. But, we the recall that powers of e ca be represeted b a ifiite series: e L +, 1!!!! 0! ( ) ( ) where! 1 L 1 (i.e., factorial). If we substitute ibt for i Equatio ad recall that i 1, i 1, i 6 1, etc., we ca accomplish the followig: e ibt 6 ( ibt) ( ibt) ( ibt) ( ibt) ( ibt) ( ibt) ( ibt) 1+ ibt L +!!!! 6!! 6 ( bt) ( bt) ( bt) ( bt) ( bt) ( ) 6 6 bt 1+ ibt + i + i i + i + i i + i + i i!!!! 6!! 6 ( bt) ( bt) ( bt) ( bt) ( bt) ( bt) 1+ ibt i + + i i +L.!!!! 6!!! +L

5 Net, takig a cue from comple arithmetic, we group real ad imagiar terms to obtai: e ibt 1 6 ( bt) ( bt) ( bt) ( bt) ( bt) ( bt) L!! 6! L ibt i i i!!! 6 ( bt) ( bt) ( bt) ( bt) ( bt) ( bt) 1 + L + i bt + i L!! 6!!!! Equatio which ma ot seem like much of a gai, si ce it s still a com ple umber. Fortuatel ad here s the cool part Euler recogized that, like e, cos ad si also have ifiite series equivalets: 6 cos 1 + L!! 6! ad 0 ( 1) ( ) si +!!! L! 1 1 ( 1) 1 ( 1)! Equatio a Equatio b Take together, Equatio a ad b allow us to rewrite Equatio as the sum of sie ad cosie terms: e ibt 6 ( bt) ( bt) ( bt) ( bt) ( bt) ( bt) L L i bt + +!! 6!! 1!! 1 cos bt cos bt + i si bt si bt i This etremel importat result is simpl a modificatio of Euler s formula ( e cos + i si ), the most remarkable formula i mathematics (R. Fema, 19), with ibt substituted for. If we had istead started our derivatio with the a bi cojugate, we would have arrived at e ibt cosbt i si bt meaig the two solutios for Equatio are ad ( a+ bi) t at () t e e ( cos bt + isi bt) ( a+ bi) t at () t e e ( cosbt i si bt) The et step, which we wo t detail, combies those two solutios ad leads to the ultimate solutio: at () t e ( c bt c si bt) 1 cos + Equatio 6 where a ad b are the real compoets of the comple epoet i Equatio, ad c1 ad c are costats whose values are determied b the iitial coditios specified i the model. The ed result is that a comple epoetial fuctio (Equatio ) has bee coverted ito a purel real-valued fuctio that s easil iterpreted. Thak ou, Professor Euler!

6 What I Wat You to Take From the Precedig The precedig ma seem a bit much, but I believe it s importat to epose ou to the uderlig mathematic s of topics we ll be studig. What I wat ou to remember is a little less itese: Equatios ivolvig a comple epoetial ca be represeted as real sie ad c osie fuctios. Because sie ad cosie fuctios are periodic, such equatios w ill ehibit periodic (cclic) behavior. Sice a umber of the models we ll be workig with i this course ca, uder realistic coditios, ield solutios of comple epoetial form, those models will ehibit periodic behavior. We will delve etesivel ito periodic fuctios ad the ramificatios of Equatio 6 durig the upcomig semester, because models that lead to solutios of the form represeted b Equatio have importat applicatios i a wide variet of biological models. For ow, we ll cotet ourselves with a look at graphs of Equatio 6 for c c 1, b 6, ad a -0.1, 0, or 0.1: 1 a 0 a 0.1 a Time The importat features to ote i this figure are that (i) the solutios oscillate, ad (ii) the sig of a determies whether the oscillatios amplitude decreases, remais costat, or icreases with time. Periodic Fuctios As a aside, it s worth otig that the term periodic has a precise defiitio i mathematics. A fuctio f is said to be periodic if ad ol if there eists some iterval, L, such that ad ( ) ( i + L ) f ( i ) f f ( ) f ( ), 1,,, K i+ L i where f ( ) represets the th-order derivative of f ( ). I others words, i order for a fuctio to be trul periodic, both the value of the fuctio ad of all derivatives of the fuctio must have the same value at begiig of the iterval ad at the ed. We ca illustrate this with a couple eamples. First, cosider the sie fuctio: Agle ( radias )

7 Note that a give value of the sie fuctio repeats a umber of times over the iterval from 8 to +8, but that the value ad the derivatives of the fuctio repeat themselves ol at itervals of π ( 6. 8). The sie fuctio is therefore termed π periodic. Now, tr to determie the period of the followig fuctio: Agle ( radias ) Ca the period be determied from the data give to ou?

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