### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Intro Diff Equat MATH 216

UM

GPA 3.87

### View Full Document

## 6

## 0

## Popular in Course

## Popular in Mathematics (M)

This 11 page Class Notes was uploaded by Mrs. Preston Lehner on Thursday October 29, 2015. The Class Notes belongs to MATH 216 at University of Michigan taught by Staff in Fall. Since its upload, it has received 6 views. For similar materials see /class/231510/math-216-university-of-michigan in Mathematics (M) at University of Michigan.

## Reviews for Intro Diff Equat

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/29/15

I arh 216 DiFFclzcnrlm 39qunrions Review of Complex Numbers This is eshezl xewew 0f memem emoeple e mm Wm Complex mmpees ere ueei lhrmgwulmelhanellcs emi ll epplmlem le Wuculexe whee we cry mselveemeeemlel equeuee il l n m eeevemeel emi eemel le use complex mmpees m express the xiluumls lleee we mew dime pee end mulls pem me llleay 0f eemplex mmpees that wl pe ueei m Mslh m A eemplex mmpee 1 may he expmsed es en edema pen 0 mlmmpees wheel Fl e 12 e 1 ee 1 me Wee melee melelewmgeeelee ee elm eel Reg at 2 we emeee me mlmn 0 me eemplex number 1 e 1mz or y e 32 dmmes me mwmmex me complex mmpee 1 Reel me lwe complex numhels exe emel ll end only ii pee llle reel end llle lmeemsxy pens exeequel le ellexweeie 2 manequsls 4 Zpeyphfendmlyil39z zpend all 92 eeevemeel way 0 lleelene epeul eemplex numhels is le lmeene lleem es pem s m the M pleee m lles eeee ll is eellee me mm pmel es musuelsi in me lellewmg glre in me amplex pleee me lme y o e lmuemly eellei me met me ml the lme yam mum xkl37 Heme l The eemplex mmpee z a placed in the complex pleee o e pemeelly ellei llle mmmee Doing arithmetic with complex numbers Addle end muluplleeuee 0f lwe eemplex mmpees e zmji end e ewe exe rimmed by me Mowingme 0 Addition 21 2231L zgy1 yg 1 x2 z39y1 yg o Multiplication 2122 mm 7 y1y2x1y2 xgyl mm 7 ylyg z z1y2 ygyl Note that if we interpret 21 and 22 as points in the complex plane as in Figure 1 then addition of complex numbers is the same as vector addition in the plane we are just adding the real and imaginary parts componentwise On the other hand the multiplication of two complex numbers may perhaps seem different than what you might have expected it to be this is only an illusion however and when we introduce exponential forms for complex numbers later the multiplication will make perfect sense Although complex numbers obey different rules of arithmetic than do ordinary real num bers it is very important to keep in mind that the complex numbers simply generalize the notion of the real numbers Indeed we can think of the real number x as the complex number L0 z 20 Such a complex number whose imaginary part is zero is said to be purely real If we add or multiply two purely real complex numbers then according to the rules for complex arithmetic we have 1 0 2 0 1 2 0 and x102 0 x120 so in each case the result is also a purely real complex number and the real part in each case is exactly what we would have found by applying the usual rules of addition and multiplication for real numbers to the real parts This shows that all the new operations de ned for complex numbers when applied to purely real numbers give the usual familiar corresponding operations One way to think of 0 1 is as the new number 239 which is purely imaginary in the sense that its real part is zero and so my z z y is the sum of the purely real number x and the purely imaginary number z y Example According to the above arithmetic rules for complex arithmetic we have 9670 07y 9611 and 071y70 07y Combining these we deduce that 9671 9670 071y0 which is another way of writing the relation 2 z z y D Example We can calculated repeated products of a complex number 2 with itself which is what we mean by raising z to an integer power Thus by de nition really 22 22 and 23 222 and so on In particular 2392 2d 0 10 1 Using the rule for multiplication we then see that 2392 m 010 1 710 71 which veri es the fact that 239 is a square root of 71 D Example The fact that 2392 71 makes the rule for multiplication of complex numbers very easy to remember if one uses the z z z y notation lndeed just by multiplying out the individual terms 2122 1 y12 Z22 12 W192 mzyi 229192 2 and then using 2392 71 we get 2122 1M 192 Z192 291 which is the rule for multiplication of complex numbers D It is easy to check directly from the de nitions given of addition and multiplication of complex numbers that all of the familiar algebraic properties that we are familiar with hold for complex numbers too In other words complex arithmetic obeys the following rules 0 Commutative Law of Addition 21 22 22 21 o Associative Law of Addition 21 22 23 21 22 23 o Commutative Law of Multiplication 2122 2221 o Distributive Law 2122 23 2122 2123 0 Unique Additive Identity 0 00 z 0 0 z z 0 Unique Multiplicative Identity 1 10 z 1 1 z z o Additive inverse 72 is 7y is 7 W z 72 0 o Multiplicative inverse For every complex number 2 x y 31 0 there exists a complex number w u1 such that xyuv uvxy 1 0 7 It turns out that the multiplicative inverse of a nonzero complex number 2 Ly is the complex number 9 which we denote by 12 Now we can de ne the quotient of two complex numbers H H 21 21 2139 22 22 22 Example The multiplicative inverse ofz39 0 1 is according to the above formula 1 0 1 739 Therefore we have 2 7 3i 1i because i2 71 D Why bother with complex numbers at all Complex numbers were originally invented as an extension of real numbers in order to have a number system in which all polynomials have roots For example the equation x2 7 3x 2 0 has two real solutions z 1 or z 2 But the similar looking quadratic equation p2 7 3x 3 0 does not have any real roots at all However if we are willing to accept complex numbers as roots then this quadratic equation also has two roots namely the complex roots 32i 2 and 327i 2 More generally for an 31 0 and other given numbers a0 an1 the polynomial equation anp alpl a0 0 of degree n does not necessarily have any real solutions However complex numbers enjoy the property that if a0a1 an are complex numbers and an 31 0 then anz alzl a0 0 always has n solutions although not all roots are necessarily distinct This fact is known as the Fundamental Theorem of Algebra In other words to nd the solutions of a polynomial equation you never need to look further than the complex numbers remarkably this is true even if the coef cients ak are themselves generalized from real numbers to complex numbers This is emactly why we need complecc numbers in a course on di erential equations like Math 216 they are necessary to give us all of the roots of the characteristic polynomial that arises from seeking eccponential solutions proportional to e of a constant coe cient di erential equation or system of di erential equations 1 1 3 3 1 5 273i 7 i17i7 i i27 7 i 2 2 Some additional terminology for complex numbers Associated with each complex number 2 is a positive number called the absolute value or modulus of z and written as the de nition in terms of the real and imaginary parts of z z iy is lzl xxz yZ If we visualize 2 as a point in the complex plane as in Figure 1 the modulus of z is just the distance from the point my to the origin 00 See Figure 2 Unless 21 and 22 y1mz i xRe z Figure 2 The modulus of a complex number 2 z iy are purely real an inequality like 21 lt 22 has no meaning because somehow both real and imaginary parts would have to be compared But the inequality lzll lt l22l does have meaning according to Figure 2 it means that 21 is closer to 00 than 22 is Example According to the de nition l1 2l V12 22 D Next for each complex number 2 there is another complex number called the complm conjugate of z and denoted by 7 or 2 The complex conjugate of z Ly z W is de ned by 3 mfg xiz39y so taking the complex conjugate of a complex number 2 amounts to changing the sign of its imaginary part Geometrically this amounts to re ection of the point representing 2 in the complex plane through the real axis as shown in Figure 3 FIWZ XR2z Figure 3 The complex conjugate of a complex number 2 Example According to the de nition T 10 10 1 Similarly 2 01 0 71 7239 D Example The following identities are easy to establish using the de nition 21 i 22 2 1 i 2 27 212271 2 27 21 T 7 2f 22 T 2 39 Also 5 z in other words the complex conjugate of the complex conjugate of any number 2 is z itself Why are there two different notations for the complex conjugate of a complex number Generally the bar notation is easier to read unless it gets in the way of a dot on the 239 or unless it is easily confused with the line separating the numerator and denominator of a fraction for these situations we have the option of using the star superscript D Example Note that if z my z W then 22 w my 2amp0 2 Ree 2 7 7 9571 7 967 7y 0721 2 1m2 That is7 we have that 1 1 Rez z and lmz 7 z This gives us a simple way to express the real and imaginary parts of z in terms of z and its complex conjugate D Example Again7 directly from the de nition of complex conjugation7 22 z yxiz y 2 7 y zyy2 2 y2 lzlz where in the last step we used the de nition ofthe modulus of 2 It follows that by multiplying the numerator and denominator of 12 by 2 we get 2 This is one way to deduce the formula we gave earlier for 12 D Polar form for complex numbers Points in the complex plane can be identi ed by their Cartesian coordinates my7 or by their polar coordinates 736 as indicated in Figure 4 Elementary trigonometry tells us that y1mz xRez Figure 4 The polar coordinates of the complex number 2 the Cartesian and polar coordinates are related by z rcost9 and y rsin0 We may therefore write the complex number 2 in terms of its polar coordinates in the following way 2 z W rcost9 2397quot sin0 rcos0 z sin6 Geometrically7 it is easy to see that the modulus of z is the same thing as the polar coordinate 7 however it is also easy to see this from the above formula using the trigonometry identity cost92 sin62 1 lzl M2 cost92 r2 sin62 TV cos2 6 sin26 r Example For 2 1 7239 we have 7 and 6 in4 and therefore 17 cosig lsinig The angle 6 is not unique but its possible values differ from each other by multiples of 27139 For example 6 27m 7 7r4 works too for any n 0 i1i2 D The angle 6 is called the argument or phase angle of z and is denoted 6 argz Again thinking geometrically it is easy to see that for a complex number to have an argument it must be nonzero which is the same thing as saying that its modulus is nonzero or that r 31 0 Since for z 31 0 there are many values of argz it is useful to de ne a particular value called the principal value of argz and denoted by Argz as the unique value of 6 between in and 7r in lt Argz S 7139 If we are given the polar coordinates r 6 of a complex number 2 then it is straight forward to calculate the corresponding Cartesian coordinates my r cos6rsin6 Finding the polar coordinates given the Cartesian coordinates is a little more tricky Of course it is easy to nd 7 2 yz but then we need to nd an angle 6 so that cos6 and sin6 It also follows from these relations that the angle 6 we seek satis es 7 y 1m2 tan6 i E i W The way to nd 6 is to use the inverse trigonometric functions however in doing so you need to make sure that the resulting angle is in the correct quadrant given the signs of z and y Example Suppose that z z ly 71 239 To nd the polar coordinates of 2 we rst calculate r 71 12 Then the angle 6 we seek satis es cos6 l and sin6 1 or if we combine these tan6 71 If we apply the arctangent function to solve for 6 we get in4 which is indeed an angle whose tangent is 71 However it is not an angle whose sine is To get the right answer we recall that the arctangent function is only de ned up to integer multiples of 7139 and therefore 6 37r4 is also an angle whose tangent is 71 In this case we also see that indeed cos37T4 il and that sin37r4 1 so that indeed 6 argz 37r4 Moreover since in lt I m4 S 7139 we also have Argz 37r4 B One of the reasons for introducing polar coordinates for complex numbers is that it gives a simple geometrical interpretation to the process of complex multiplication Indeed if 21 r1cos61 lsin61 and 22 r2cos62 lsin62 then 2122 r1r2cos01 z sin01cos02 z39sin62 rlrg cost91 cost92 7 sin61 sin02 239 sin01 cos62 cos61 sin02 r1r2cos01 02 z39sin61 62 We used two trigonometric identities in the last step above to arrive at a formula in terms of the sum of the angles 91 02 In particular this calculation shows that lzlzgl lzll l22l and arg2122 arg21 arg22 That is when we multiply two complex numbers we multiply their moduli and add their arguments Note also that if r 31 0 then 1 rcos0 7 z sin0 1 1 39 fl cos 7 z sin 7 Z rcos0 2 sin0 rcos0 7 z sin0 ZltCOSlt6gt72S1Hlt6gtgt i r l 6 6 7 so lz ll and argz 1 7 argz Note also that arg3 7 argz Exponential form of a complex number Euler s formula Recall that the in nite power series expansion de ning 6w is w2 w3 5w1w7 It turns out that both sides make sense when w is a complex number and in particular if w 20 where 6 is a real angle In this case the series expansion becomes 63 7 25 62 63 13 Z9 cost9 z sin0 92 5 9 711L207 where at the end we grouped the purely real and purely imaginary terms and recalled the in nite power series expansions of cost9 and sin0 for real values of 0 The remarkable formula we have found in this way 6w cost9z39sint9 is known as Euler s formula If we use Euler7s formula we can express 2 in terms of its polar coordinates in an even simpler form 2 rcos0 2 sin0 T619 8 where r is the absolute value of z and 6 argz is the argument This is called the ep0nential form of the complex number 2 Now we can see that we can also view the effect of multiplying two complex numbers that are expressed in terms of their polar coordinates multiplication of the moduli and addition of the arguments as being a simple consequence of the rules for multiplying exponentials 2122 newline 2 Tlrgelwl92gt 1 1 1 79 2W5 7 25119142 22 T2 Note that for any n 0 i1 i2 we have 2 WWW also 7 re The exponential form of a complex number 2 makes it easy to compute powers of 2 Indeed z rewYL rneme Combining this with Euler7s formula we have DeMoivre s Theorem which gives the real and imaginary parts of any power of a complex number of modulus one Also and cos6 z sin6 cosn6 z sinn6 The proof of DeMoivre7s theorem is basically one line long cos6 z sin6 89 eme cosn6 z sinn6 Euler7s formula was used in the rst and last steps Example Writing out DeMoivre7s formula for n 2 and doing the multiplication gives cos26 z sin26 cos6 z sin62 cos62 7 sin62 22 sin6 cos6 Since the real and imaginary parts on both sides must be equal we get cos26 cos62 7 sin62 and sin26 2 sin6 cos6 This example shows that DeMoivre7s Theorem provides an easy way to remember the multiple angle trigonometry identities D Finding roots of complex numbers is also made easy using the exponential form For example 212 Temp2 T1261927 but since 6 may be replaced by 6 27Tk for any integer k without changing 2 we have more generally that rei627rk12 HagueHm 39 The right hand side gives only two possible answers as k ranges over all possible integers These two complex numbers are the two square roots of z The nth roots of a complex number 2 are calculated in exactly the same way Zln Tln6i627rkn Tlnei6n27rkn7 9 and now we see that the right hand side gives 71 distinct values as k ranges over the integers Therefore every nonzero complex number 2 has n distinct nth roots Example The distinct cube roots of 1 are 113 1 or 113 627quot or 113 641173 They are illustrated in the complex plane in Figure 5 D h Figure 5 The cube roots of z 1 in the complex plane Problems 1 Compute the following express in the form x W a 2 3239 5 7239 b 2 32392 c 2 3239 75 i 7239 d 203 i 2239 2239 71 e 56i77r63 f V 6113 Give two answers 2 Solve the following for 2 expressed as x W a 2272220 b 22 z 1 0 3 Compute the following if z z W express in terms of z andor y c d gt lt a mz39z ez C7 VVVV Z WW e ilmz N r U 03 5 00 0 H Locate the numbers 21 22 and 21 7 22 vectorially sketch the plot a 21 22 and 22 2372 b 21 732 and 223 c 21 732 and 22 142 Simplify the following a 7 32 b a C W 7 Sketch the set of points in the complex plane determined by the following a l27172l 1 b Re7 7 2 2 Solve the following simultaneous equations for 21 and 22 a 21222 21222 Determine Argz for each of the following a z 7 72lt1 m b 2 272 7 22 c z 7 7256 Use the exponential form to compute the following a 217 mm 239 7 b 5212 c 71 27 Use DeMoivre7s Theorem to prove the following trigonometric identities a cos3t9 cos63 7 3 cost9 sin02 b sin3t9 3 cost92 sin0 7 sin63 Hint Expand cost 2sin 93 and use DeMoivre7s Theorem Find all the roots for the following and exhibit them geometrically a 22312 7

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "I used the money I made selling my notes & study guides to pay for spring break in Olympia, Washington...which was Sweet!"

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.