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# Calc III for Comput Sci MATH 2605

GPA 3.62

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This 0 page Class Notes was uploaded by Chelsea Nolan MD on Monday November 2, 2015. The Class Notes belongs to MATH 2605 at Georgia Institute of Technology - Main Campus taught by John Elton in Fall. Since its upload, it has received 17 views. For similar materials see /class/233951/math-2605-georgia-institute-of-technology-main-campus in Mathematics (M) at Georgia Institute of Technology - Main Campus.

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Date Created: 11/02/15

Lesson 3 Roots of complex numbers some more algebraic properties for complex numbers To nd the n3911 roots of a complex number a we have to solve the equation 2quot a Suppose that a is not 0 Write a in polar form a rcos9 i sine Then from what we said last time one solution is clearly zo rlquotcos9n i sin9n that is divide the angle by 11 because raising this to the nth power gives a magnitude of rlr n r and an argument of n9n 9 as required But this is not the only solution consider the numbers zk r quot cos6 2 kisin6 27 k 01 2 nl Compute n n 26quot rcos6 2nki sin6 271k which is the same as a because adding a multiple of 211 to 9 does not affect the cos or sin So these 11 different complex numbers are all n3911 roots of a And this is all there is if we let k n in the formula for zk above 6 t9 we get 2 r quotcosi 27f is1ni 2a wh1ch1s the same as 20 because s1n and n n cos are 2nperiodic so this is nothing new Similarly Zn1 21 etc So those values 20 21 zn1 are the only 113911 roots of a In summary There are n n h roots of any complex number a 0 They all have the same magnitude anal their arguments are equally spaced with an angle of271n between them one of them makes an angle argan with the real axis Geometrically the nth roots are the vertices of a regular polygon with n sides centered at the origin You should draw a picture for yourself as an example When a l the n roots are called the nth roots of unity Let n cos211n i sin211n which is a complex number of magnitude 1 making an angle 2nn with the real axis and is one of the n3911 roots of 1 Then all the nth roots of unity are l n 032 of Every time you multiply by on you simply rotate counterclockwise by angle 21tn and after 11 steps you are back to 1 Picture of the 6th roots of unity 2 D Now let s develop some more facts about conjugation and absolute value magnitude Rez 22 Imz 2 2 2 21quot sinceXiyX7iy2xandxiyXiy2iy z 2 iff Imz 0 that is z is real NH 2 zlz2 zlzz Proof x1 iylx2 iy2 x1x2 iy1y2x1 iy1x2 iy2 2122 a Z Proof similar to above just write out in terms of real and imaginary parts Follows from above by multiplying both sides by II N H3 i zquot 2quot Follows from multiplication rule above repeated A useful consequence of the above facts is the following Let a0 a1 an be real numbers and consider the polynomial equation anzquot an1zquot391 alz a0 0 If the complex number w is a solution to this equation so is its conjugate w That is complex roots of a polynomial with real coe icients occur in conjugate pairs Proof take the compleX conjugate of both sides of anwn an1wquot391 a1w a0 0 Using the above facts about conjugation and that the ak s are real we get an Wquot aHWquot391 aIW a0 0 so W is a solution also For absolute value magnitude recall l 2 l2ZE So lzizz 2 ZIZZZIZZ 212122 22 l 21 ill 22 l2 a so lzlz2 l 21 H 22 l The absolute value of the product is the product of absolute values This is true for any Z number of factors For quotients note 22 21 so taking absolute values of both 22 Z z 2 Sides l 22 l 1 l 21 l so 71 1 l the absolute value ofthe quotient is the quotient ZZ ZZ of the absolute values For sums it is messier Compute i21222zl2EZ21Z212222225l21VH2V2Re212 And l 21 z2 l2l 21 2 l 22 2 2Rezlzz so by adding we get the identity l 21 z2 2 l 21 z2 l22 l 2102 l Z2 2 which is called the parallelogram identity Ifyou draw the complex numbers 21 and 22 and the two diagonals of the parallelogram with those numbers as adjacent sides you see that it says that the sum of the squares of the lengths of the diagonals of a parallelogram is equal to the sum of the squares of the lengths of the four sides of the parallelogram a fact from elementary geometry which could be proved easily with the law of cosines Recall as we mentioned above that lzl 7 22 is the distance between the points 21 and 22 ZZ39ZI z 2 ZlZz r 21 Now for a few inequalities Ifz x iy then le x2 yz so x2 S le and y2 S lzlz so lxl S lzl and lyl S lzl the hypotenuse of a right triangle is longer than either adjacent side So we get that lzl S Rez S lzl and lzl S Imz S lzl Note Rez lzl iffa is real and Z 0 If this is applied to the identity for the sum above we get l2122 l2l21 l2 lZz l2 2ReZIZzSl 21 l2 lZz l2 2l 2122 ll 21 l2 l 22 l2 2l21H22 ll21llzz DZ Taking square roots we get the important triangle inequality lzl 22 S lzll lzzl If you draw this it simply says that the third side of a triangle is no longer than the sum of the lengths of the other two sides If you check the proof you see that equality holds iff 21 22 Z 0 which is equivalent to 2122 Z 0 ifzz is not 0 that is 21 is a positive real number times 22 which says that 21 is parallel to z and points in the same direction note that multiplication of a complex number by a positive real number does not change the argument of the complex number only the length like scalar multiplication of a vector The triangle inequality may be extended to any number of terms to say that the absolute value of the sum is less than or equal to the sum of the absolute values In the triangle inequality replace 2 by z 7 21 on both sides and you get lzzl S lzll lzz 721 or lzzl lzll S lzz 721 Reversing 21 and 22 gives lzll lzzl S lzl 7 22 so combined l lzzl lzll l S l2 7 21 which says that the ali erence ofthe lengths oftwo sides of a triangle is less than or equal to the length of the third side Next lesson we will take a breather on new material and just do examples and give you some practice problems to clarify and solidify what we have done so far Lesson 3 Roots of complex numbers some more algebraic properties for complex numbers To nd the n3911 roots of a complex number a we have to solve the equation 2quot a Suppose that a is not 0 Write a in polar form a rcos9 i sine Then from what we said last time one solution is clearly zo rlquotcos9n i sin9n that is divide the angle by 11 because raising this to the nth power gives a magnitude of rlr n r and an argument of n9n 9 as required But this is not the only solution consider the numbers zk r quot cos6 2 kjisin 6 272k k 01 2 nl Compute n n 26quot rcost9 27zki sint9 272k which is the same as a because adding a multiple of 211 to 9 does not affect the cos or sin So these 11 different complex numbers are all nth roots of a And this is all there is if we let k n in the formula for zk above t9 t9 we get 2 r quot cos 27239is1n 2a wh1ch1s the same as 20 because s1n and n n cos are 2nperiodic so this is nothing new Similarly Zn1 21 etc So those values 20 21 zn1 are the only 113911 roots of a In summary There are n n h roots of any complex number a 0 They all have the same magnitude anal their arguments are equally spaced with an angle of 2 Mn between them one of them makes an angle argan with the real axis Geometrically the nth roots are the vertices of a regular polygon with n sides centered at the origin You should draw a picture for yourself as an example When a l the n roots are called the nth roots of unity Let n cos211n i sin211n which is a complex number of magnitude 1 making an angle 2nn with the real axis and is one of the nth roots of 1 Then all the nth roots of unity are l n 032 of Every time you multiply by on you simply rotate counterclockwise by angle 21tn and after 11 steps you are back to 1 Picture of the 6th roots of unity 2 D Now let s develop some more facts about conjugation and absolute value magnitude Rez 22 Imz 2 2 2 21quot sinceXiyX7iy2xandxiyXiy2iy z 2 iff Imz 0 that is z is real NH 2 zlzzzlzz Proof x1 iylx2 iy2 x1x2 iy1y2x1 iy1x2 iy2 2122 2 1 Proof similar to above just write out in terms of real and imaginary parts U 2 quot 2quot Follows from multiplication rule above repeated Follows from above by multiplying both sides by N IHN N IN A useful consequence of the above facts is the following Let a0 a1 an be real numbers and consider the polynomial equation anzquot an1zquot391 alz a0 0 If the complex number w is a solution to this equation so is its conjugate w That is complex roots of a polynomial with real coe icients occur in conjugate pairs Proof take the compleX conjugate of both sides of anwn an1wquot391 a1w a0 0 Using the above facts about conjugation and that the ak s are real we get an Wquot aHWquot391 a1W a0 0 so Wis a solution also For absolute value magnitude recall l 2 l2ZE So l 2zz 2222 l 2 2 V so 2122 l Zl Zz The absolute value of the product is the product of absolute values This is true for any Z number of factors For quotients note 22 21 so taking absolute values of both 22 z 2 Sides l 22 l l 21 l so 1 l ll the absolute value ofthe quotient is the quotient Z Z1 22 of the absolute values For sums it is messier Compute

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