Complex Numbers. Now we also saw that if a and b were both positive then ab = a b. For a second let s forget that restriction and do the following.

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1 Complex Numbers The last topc n ths secton s not really related to most of what we ve done n ths chapter, although t s somewhat related to the radcals secton as we wll see. We also won t need the materal here all that often n the remander of ths course, but there are a couple of sectons n whch we wll need ths and so t s best to get t out of the way at ths pont. In the radcals secton we noted that we won t get a real number out of a square root of a negatve number. For nstance - 9 sn t a real number snce there s no real number that we can square and get a NEGATIVE 9. Now we also saw that f a and b were both postve then ab = a b. For a second let s forget that restrcton and do the followng. - 9 = ( 9)( - 1) = 9-1= 3-1 Now, - 1 s not a real number, but f you thnk about t we can do ths for any square root of a negatve number. For nstance, = 100-1= = = 90-1 etc. So, even f the number sn t a perfect square we can stll always reduce the square root of a negatve number down to the square root of a postve number (whch we or a calculator can deal wth) tmes - 1. So, f we just had a way to deal wth - 1 we could actually deal wth square roots of negatve numbers. Well the realty s that, at ths level, there just sn t any way to deal wth - 1 so nstead of dealng wth t we wll make t go away so to speak by usng the followng defnton. Note that f we square both sdes of ths we get, = - 1 =- 1 It wll be mportant to remember ths later on. Ths shows that, n some way, s the only number that we can square and get a negatve value. Usng ths defnton all the square roots above become, - 9 = = 10-5 = 5-90 = 90 These are all examples of complex numbers. The natural queston at ths pont s probably just why do we care about ths? The answer s that, as we wll see n the next chapter, sometmes we wll run across the square roots of negatve 007 Paul Dawkns 5

2 numbers and we re gong to need a way to deal wth them. So, to deal wth them we wll need to dscuss complex numbers. So, let s start out wth some of the basc defntons and termnology for complex numbers. The standard form of a complex number s a+ b where a and b are real numbers and they can be anythng, postve, negatve, zero, ntegers, fractons, decmals, t doesn t matter. When n the standard form a s called the real part of the complex number and b s called the magnary part of the complex number. Here are some examples of complex numbers The last two probably need a lttle more explanaton. It s completely possble that a or b could be zero and so n 16 the real part s zero. When the real part s zero we often wll call the complex number a purely magnary number. In the last example (113) the magnary part s zero and we actually have a real number. So, thnkng of numbers n ths lght we can see that the real numbers are smply a subset of the complex numbers. The conjugate of the complex number a+ b s the complex number a- b. In other words, t s the orgnal complex number wth the sgn on the magnary part changed. Here are some examples of complex numbers and ther conjugates. complex number conjugate Notce that the conjugate of a real number s just tself wth no changes. Now we need to dscuss the basc operatons for complex numbers. We ll start wth addton and subtracton. The easest way to thnk of addng and/or subtractng complex numbers s to thnk of each complex number as a polynomal and do the addton and subtracton n the same way that we add or subtract polynomals. Example 1 Perform the ndcated operaton and wrte the answers n standard form (a) (b) ( 4+ 1) -( 3-15) (c) 5-( - 9+ ) Soluton There really sn t much to do here other than add or subtract. Note that the parentheses on the frst terms are only there to ndcate that we re thnkng of that term as a complex number and n 007 Paul Dawkns 53

3 general aren t used = 1-3 (a) = = 1+ 7 (b) = = 9+ 4 (c) ( ) Next let s take a look at multplcaton. Agan, wth one small dfference, t s probably easest to just thnk of the complex numbers as polynomals so multply them out as you would polynomals. The one dfference wll come n the fnal step as we ll see. Example Multply each of the followng and wrte the answers n standard form [Soluton] (a) ( ) (b) ( 1 5)( 9 ) (c) ( 4 )( 3) (d) ( 1 8)( 1 8) [Soluton] + + [Soluton] - + [Soluton] Soluton (a) So all that we need to do s dstrbute the 7 through the parenthess = ( ) Now, ths s where the small dfference mentoned earler comes nto play. Ths number s NOT n standard form. The standard form for complex numbers does not have an n t. Ths however s not a problem provded we recall that =- 1 Usng ths we get, 7-5+ = = We also rearranged the order so that the real part s lsted frst. (b) In ths case we wll FOIL the two numbers and we ll need to also remember to get rd of the = = = ( ) (c) Same thng wth ths one = = = ( ) (d) Here s one fnal multplcaton that wll lead us nto the next topc = = 1+ 64= 65 ( )( ) Don t get excted about t when the product of two complex numbers s a real number. That can and wll happen on occason. 007 Paul Dawkns 54

4 In the fnal part of the prevous example we multpled a number by ts conjugate. There s a nce general formula for ths that wll be convenent when t comes to dscusson dvson of complex numbers. ( a+ b)( a- b) = a - ab+ ab- b = a + b So, when we multply a complex number by ts conjugate we get a real number gven by, ( a+ b)( a- b) = a + b Now, we gave ths formula wth the comment that t wll be convenent when t came to dvdng complex numbers so let s look at a couple of examples. Example 3 Wrte each of the followng n standard form. (a) 3 - [Soluton] (b) [Soluton] 9-8 (c) [Soluton] 1+ (d) 6-9 [Soluton] Soluton So, n each case we are really lookng at the dvson of two complex numbers. The man dea here however s that we want to wrte them n standard form. Standard form does not allow for any 's to be n the denomnator. So, we need to get the 's out of the denomnator. Ths s actually farly smple f we recall that a complex number tmes ts conjugate s a real number. So, f we multply the numerator and denomnator by the conjugate of the denomnator we wll be able to elmnate the from the denomnator. Now that we ve fgured out how to do these let s go ahead and work the problems. (a) ( 3-) ( -7) = = = = Notce that to offcally put the answer n standard form we broke up the fracton nto the real and magnary parts. 3 3 ( 9+ ) (b) = = = (c) ( 1 ) = = = = Paul Dawkns 55

5 (d) Ths one s a lttle dfferent from the prevous ones snce the denomnator s a pure magnary number. It can be done n the same manner as the prevous ones, but there s a slghtly easer way to do the problem. Frst, break up the fracton as follows = - = - Now, we want the out of the denomnator and snce there s only an n the denomnator of the frst term we wll smply multply the numerator and denomnator of the frst term by an ( ) = - = - = - = () The next topc that we want to dscuss here s powers of. Let s just take a look at what happens when we start lookng at varous powers of. 1 1 = = =- 1 = = =- =- 4 4 = = - 1 = 1 = = = = ( )() = = =- 1 = = =- =- ( ) () = = 1 = 1 = 1 Can you see the pattern? All powers f can be reduced down to one of four possble answers and they repeat every four powers. Ths can be a convenent fact to remember. We next need to address an ssue on dealng wth square roots of negatve numbers. From the secton on radcals we know that we can do the followng. 6= 36 = ( 4)( 9) = 4 9 = ( )( 3) = 6 In other words, we can break up products under a square root nto a product of square roots provded both numbers are postve. It turns out that we can actually do the same thng f one of the numbers s negatve. For nstance, ( )( ) 6 = - 36 = = = 3 = 6 However, f BOTH numbers are negatve ths won t work anymore as the followng shows. ( )( ) 6= 36 = = 3 = 6 =- 6 We can summarze ths up as a set of rules. If a and b are both postve numbers then, 007 Paul Dawkns 56

6 a b = ab - a b = -ab a - b = -ab ( )( ) -a -b -a -b Why s ths mportant enough to worry about? Consder the followng example. Example 4 Multply the followng and wrte the answer n standard form. ( )( ) Soluton If we where to multply ths out n ts present form we would get, ( )( ) = Now, f we were not beng careful we would probably combne the two roots n the fnal term nto one whch can t be done! So, there s a general rule of thumb n dealng wth square roots of negatve numbers. When faced wth them the frst thng that you should always do s convert them to complex number. If we follow ths rule we wll always get the correct answer. So, let s work ths problem the way t should be worked. ( )( ) = = = 6+ The rule of thumb gven n the prevous example s mportant enough to make agan. When faced wth square roots of negatve numbers the frst thng that you should do s convert them to complex numbers. There s one fnal topc that we need to touch on before leavng ths secton. As we noted back n the secton on radcals even though 9 = 3 there are n fact two numbers that we can square to get 9. We can square both 3 and -3. The same wll hold for square roots of negatve numbers. As we saw earler - 9 = 3. As wth square roots of postve numbers n ths case we are really askng what dd we square to get -9? Well t s easy enough to check that 3 s correct. ( ) 3 = 9 =- 9 However, that s not the only possblty. Consder the followng, - 3 = - 3 = 9 =- 9 and so f we square -3 we wll also get -9. So, when takng the square root of a negatve number there are really two numbers that we can square to get the number under the radcal. However, we wll ALWAYS take the postve number for the value of the square root just as we do wth the square root of postve numbers. 007 Paul Dawkns 57