[PDF] 1 Basics of Series and Complex Numbers

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Instruction: This Part II of Form 1-Z is required by Rule 257(d) of Regulation A An officer of the issuer or any other duly authorized person may sign, and must do so by typed signature The name and title of the person signing the form must be typed or printed under the signature The signatory to the filing must also manually

72 Complex arithmetic - mathcentreacuk

j2 = −1 If z 1 and z2 are the two complex numbers their product is written z1z2 Example If z1 = 5− 2j and z2 = 2+4j find z1z2 Solution z1z2 = (5− 2j)(2+4j) = 10+20j −4j − 8j2 Replacing j2 by −1 we obtain z1z2 = 10+16j −8(−1) = 18+16j In general we have the following result: www mathcentre ac uk 7 2 1 c Pearson Education Ltd 2000

Topic 8 Notes Jeremy Orlo - MIT Mathematics

f(z) = z+ 1 z3(z2 + 1) has isolated singularities at z= 0; iand a zero at z= 1 We will show that z= 0 is a pole of order 3, z= iare poles of order 1 and z= 1 is a zero of order 1 The style of argument is the same in each case At z= 0: f(z) = 1 z3 z+ 1 z2 + 1: Call the second factor g(z) Since g(z) is analytic at z= 0 and g(0) = 1, it has a

1 Basics of Series and Complex Numbers

1 1 z = 1 + z+ z2 + = X1 n=0 zn (19) is the Taylor series of f(z) = 1=(1 z) about z= 0 As mentioned earlier, the function 1=(1 z) exists and is in nitely di erentiable everywhere except at z= 1 while the series P 1 n=0 z nonly exists in the unit circle jzj

SOLUTIONS TO HOMEWORK ASSIGNMENT &# 7

z2 +z +1 z=e2πı/3 = 2πı 1 2z +1 z=e2πı/3 = 2π √ 3 (b) The only singularity of z2e1/z sin(1/z) occurs at z = 0, and it is an essential singularity Therefore the formula for computing the residue at a pole will not work, but we can still compute some of the coefficients in the Laurent series expansion about z = 0 : z2e1/z sin(1/z) = z2

Functions of a Complex Variable - MIT OpenCourseWare

1 z + 1 dz 2 −1 z is −2 − i if the path is the upper half of the circle r = 1 [Write z = ei , where varies from to 0, or from (2k + 1) to 2k , where k is any integer ] (b) Show (also by direct integration) that the value is −2 + i if the path is the lower half of the circle

Lecture 22: Inverse Functions - Furman

1+z 1−z Hence w = 1 2 log 1+z 1−z Thus we define the inverse hyperbolic tangent function by tanh−1(z) = 1 2 log 1+z 1−z We find the other inverse hyperbolic trigonometric functions in a similar manner The most important of these are sinh−1(z) = log z +(z2 +1)12 and cosh−1(z) = log z +(z2 −1)12 The derivatives are d dz sinh

7 Taylor and Laurent series - MIT Mathematics

1 w z which looks a lot like the sum of a geometric series We will make frequent use of the following manipulations of this expression 1 w z = 1 w 1 1 z=w = 1 w 1 + (z=w) + (z=w)2 + ::: (3) The geometric series in this equation has ratio z=w Therefore, the series converges, i e the formula is valid, whenever jz=wj

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c FW Math 321, 2012/12/11 Elements of Complex Calculus1 Basics of Series and Complex Numbers

1.1 Algebra of Complex numbers

A complex numberz=x+iyis composed of areal part<(z) =xand animaginary part=(z) =y, both of which arereal numbers,x,y2R. Complex numbers can be dened as pairs of real numbers (x;y) with special manipulation rules. That's how complex numbers are dened in Fortran or C. We can map complex numbers to the planeR2with the real part as thexaxis and the imaginary part as they-axis. We refer to that mapping as thecomplex plane. This is a very useful visualization. The formx+iyis convenient with the special symbolistanding as the imaginary unit dened such thati2=1. With that form and that speciali2=1 rule, complex numbers can be manipulated like regular real numbers.jzjz=x+iyz =xiyx yAddition/subtraction: z

1+z2= (x1+iy1) + (x2+iy2) = (x1+x2) +i(y1+y2):(1)

This isidentical to vector addition for the 2D vectors(x1;y1) and (x2;y2).

Multiplication:

z

1z2= (x1+iy1)(x2+iy2) = (x1x2y1y2) +i(x1y2+x2y1):(2)

Complex conjugate:

z =xiy(3) An overbar zor a starzdenotes thecomplex conjugateofz, which is same aszbut with the sign of the imaginary part ipped. It is readily veried that the complex conjugate of a sum is the sum of the conjugates: (z1+z2)=z1+z2, and the complex conjugate of a product is the product of the conjugates (z1z2)=z1z2(show that as an exercise).

Modulus (or Norm)

jzj=pzz =px

2+y2;(4)

This modulus is equivalent to the euclidean norm of the 2D vector (x;y), hence it obviously satisfy the triangle inequalityjz1+z2j jz1j+jz2j. However we can verify thatjz1z2j=jz1jjz2j.

Division:

z 1z

2=z1z2z

2z2=(x1+iy1)(x2iy2)x

22+y22=x1x2+y1y2x

22+y22

+ix2y1x1y2x

22+y22

:(5) All the usual algebraic formula apply, for instance (z+a)2=z2+2za+a2and more generally the binomial formula(dening 0! = 1) (z+a)n=nX k=0 n k z kank=nX k=0n!k!(nk)!zkank:(6) c

F. Walee, Math 321 Notes, UW 2012/12/112

Exercises:

1. Prove that (z1+z2)=z1+z2, (z1z2)=z1z2andjz1z2j=jz1jjz2j.

2. Calculate (1 +i)=(2 +i3).

3. Show that the nal formula for division follows from the denition of multiplication (as it

should): ifz=z1=z2thenz1=zz2, solve for<(z) and=(z).

1.2 Limits and Derivatives

The modulus allows the denition of distance and limit. Thedistancebetween two complex numberszandais the modulus of their dierencejzaj. A complex numberztends to a complex numberaifjzaj !0, wherejzajis the euclidean distance between the complex numberszand ain the complex plane. A functionf(z) is continuous ataif limz!af(z) =f(a). These concepts allow the denition of derivatives and series.

Thederivativeof a functionf(z) atzis

df(z)dz = lima!0f(z+a)f(z)a (7) whereais a complex number anda!0 meansjaj !0. This limit must be the same no matter howa!0. We can use the binomial formula (6) as done in Calc I to deduce that dz ndz =nzn1(8) for any integern= 0;1;2;:::, and we can dene theanti-derivativeofznaszn+1=(n+1)+Cfor all integern6=1. All the usual rules of dierentiation:product rule, quotient rule, chain rule,..., still apply for complex dierentiation and we will not bother to prove those here, the proofs are just like in Calc I. So there is nothing special about complex derivatives, or is there? Consider the functionf(z) = <(z) =x, the real part ofz. What is its derivative? Hmm..., none of the rules of dierentiation help us here, so let's go back to rst principles: d<(z)dz = lima!0<(z+a) <(z)a = lima!0<(a)a =?! (9) What is that limit? Ifais real, thena=<(a) so the limit is 1, but ifais imaginary then<(a) = 0 and the limit is 0. So there is no limit that holds for alla!0. The limit depends onhow a!0, and we cannot dene thez-derivative of<(z).<(z) is continuous everywhere, but nowhere z-dierentiable!

Exercises:

1. Prove formula (8) from the limit denition of the derivative [Hint: use the binomial formula].

2. Prove that (8) also applies to negative integer powerszn= 1=znfrom the limit denition of

the derivative. c

F. Walee, Math 321 Notes, UW 2012/12/113

1.3 Geometric sums and series

For any complex numberq6= 1, thegeometric sum

1 +q+q2++qn=1qn+11q:(10)

To prove this, letSn= 1 +q++qnand note thatqSn=Sn+qn+11, then solve that forSn. Thegeometric seriesis the limit of the sum asn! 1. It follows from (10), that the geometric series converges to 1=(1q) ifjqj<1, and diverges ifjqj>1,1 X n=0q n= 1 +q+q2+=11q;ijqj<1:(11) Note that we have two dierent functions ofq: (1) the seriesP1 n=0qnwhich only exists when jqj<1, (2) the function 1=(1q) which is dened and smooth everywhere except atq= 1. These two expressions, the geometric series and the function 1=(1q) are identical in the diskjqj<1,

but they are not at all identical outside of that disk since the series does not make any sense (i.e.

it diverges) outside of it. What happens on the unit circlejqj= 1? (consider for exampleq= 1, q=1,q=i, ...)jqj= 1 <(q)=(q) diverges convergesExercises:

1. Derive formula (10) and absorb the idea of the proof. What isSnwhenq= 1?

2. CalculateqN+qN+2+qN+4+qN+6+::::withjqj<1.

1.4 Ratio test

The geometric series leads to a useful test for convergence of the general series 1 X n=0a n=a0+a1+a2+(12) We can make sense of this series again as the limit of the partial sumsSn=a0+a1++an asn! 1. Any one of these nite partial sums exists but the innite sum does not necessarily converge. Example: takean= 18n, thenSn=n+ 1 andSn! 1asn! 1. Anecessarycondition for convergence is thatan!0 asn! 1as you learned in Math 222 and can explain why, but that is not sucient. Asucientcondition for convergence is obtained by c

F. Walee, Math 321 Notes, UW 2012/12/114

comparison to a geometric series. This leads to theRatio Test: the series (12) converges if lim n!1jan+1jjanj=L <1 (13) Why does the ratio test work? IfL <1, then pickanyqsuch thatL < q <1 and one can nd a (suciently large)Nsuch thatjan+1j=janj< qfor allnNso we can write jaNj+jaN+1j+jaN+2j+jaN+3j+=jaNj

1 +jaN+1jjaNj+jaN+2jjaN+1jjaN+1jjaNj+

1, then we can reverse the proof (i.e.pickqwith 1< q < LandNsuch thatjan+1j=janj> q

8nN) to show that the seriesdiverges. IfL= 1, you're out of luck. Go home and take a nap.

1.5 Power series

A power series has the form

1 X n=0c n(za)n=c0+c1(za) +c2(za)2+(15) where thecn's are complex coecients andzandaare complex numbers. It is a series in powers of (za). By theratio test, the power series converges if lim n!1 c n+1(za)n+1c n(za)n =jzajlimn!1 c n+1c n jzajR <1;(16) where we have dened lim n!1 c n+1c n =1R :(17)R ajzaj< R <(z)=(z)The power series converges ifjzaj< R. It divergesjzaj>

R.jzaj=Ris a circle of radiusRcentered ata, hence

Ris called theradius of convergenceof the power series.R can be 0,1or anything in between. But the key point is thatpower series always converge in a diskjzaj< Rand diverge outside of that disk. This geometric convergence inside a disk implies that power series can be dierentiated (and in- tegrated) term-by-term inside their disk of convergence (why?). The disk of convergence of the

derivative or integral series is the same as that of the original series. For instance, the geometric

seriesP1 n=0znconverges injzj<1 and its term-by-term derivativeP1 n=0nzn1does also, as you can verify by the ratio test.

Taylor Series

The Taylor Series of a functionf(z) aboutz=aisf(z) =f(a) +f0(a)(za) +12 f00(a)(za)2+=1X n=0f (n)(a)n!(za)n;(18) c

F. Walee, Math 321 Notes, UW 2012/12/115

wheref(n)(a) =dnf=dzn(a) is thenth derivative off(z) ataandn! =n(n1)1 is the factorial ofn, with 0! = 1 by convenient denition. The equality betweenf(z) and its Taylor series is only valid if the series converges. The geometric series

11z= 1 +z+z2+=1X

n=0z n(19) is the Taylor series off(z) = 1=(1z) aboutz= 0. As mentioned earlier, the function 1=(1z) exists and is innitely dierentiable everywhere except atz= 1 while the seriesP1 n=0znonly exists in the unit circlejzj<1. Several useful Taylor series are more easily derived from the geometric series (11), (19) than from the general formula (18) (even if you really like calculating lots of derivatives!). For instance

11z2= 1 +z2+z4+=1X

n=0z

2n(20)

11 +z= 1z+z2 =1X

n=0(z)n(21) ln(1 +z) =zz22 +=1X n=0(1)nzn+1n+ 1(22) The last series is obtained by integrating both sides of the previous equation and matching atz= 0 to determine the constant of integration. These series converge only injzj<1 while the functions on the left hand side exist for (much) larger domains ofz.

Exercises:

1. Explain why the domain of convergence of a power series is always a disk (possibly innitely

large), not an ellipse or a square or any other shape [Hint: read the notes carefully]. (Any- thing can happen on the boundary of the disk: weak (algebraic) divergence or convergence, perpetual oscillations, etc., recall the geometric series).

2. Show that if a functionf(z) =P1

n=0cn(za)nfor allz's within the (non-zero) disk of convergence of the power series, then thecn's must have the form provided by formula (18).

3. What is the Taylor series of 1=(1z) aboutz= 0? what is its radius of convergence? does

the series converge atz=2? why not?

4. What is the Taylor series of the function 1=(1 +z2) aboutz= 0? what is its radius of

convergence? Use a computer or calculator to test the convergence of the series inside and outside its disk of convergence.

5. What is the Taylor series of 1=zaboutz= 2? what is its radius of convergence? [Hint:

z=a+ (za)]

6. What is the Taylor series of 1=(1 +z)2aboutz= 0?

7. Look back at all the places in these notes and exercises (including earlier subsections) where

we have used the geometric series for theoretical or computational reasons. c

F. Walee, Math 321 Notes, UW 2012/12/116

1.6 Complex transcendentals

The complex versions of the Taylor series for the exponential, cosine and sine functions exp(z) = 1 +z+z22 +=1X n=0z nn!(23) cosz= 1z22 +z44! =1X n=0(1)nz2n(2n)!(24) sinz=zz33! +z55! =1X n=0(1)nz2n+1(2n+ 1)!(25) converge in theentirecomplex plane for anyzwithjzj<1as is readily checked from the ratio test. These series can now serve as thedenitionof these functions for complex arguments. We can verify all the usual properties of these functions from the series expansion. In general we can integrate and dierentiate series term by term inside the disk of convergence of the power series. Doing so for exp(z) shows that the function is still equal to its derivative ddz exp(z) =ddz 1X n=0z nn!! =1X n=1z n1(n1)!= exp(z);(26) meaning that exp(z) is the solution of the complex dierential equationdf=dz=fwithf(0) = 1. Likewise the series (24) for coszand (25) for sinzimply ddz cosz=sinz;ddz sinz= cosz:(27) Another slighttour de forcewith the series for exp(z) is to use the binomial formula (6) to obtain exp(z+a) =1X n=0(z+a)nn!=1X n=0n X k=0 n k zkankn!=1X n=0n X k=0z kankk!(nk)!:(28) The double sum is over the triangular region 0n 1, 0kninn,kspace. If we interchange the order of summation, we'd have to sum overk= 0! 1andn=k! 1(sketch it!). Changing variables tok,m=nkthe range ofmis 0 to1as that ofkand the double sum reads exp(z+a) =1X k=01 X m=0z kamk!m!= 1X k=0z kk!! 1X m=0a mm!! = exp(z)exp(a):(29) This is a major property of the exponential function and we veried it from its series expansion (23) for general complex argumentszanda. It implies that if we dene as before e= exp(1) = 1 + 1 +12 +16 +124
+1120
+= 2:71828:::(30) then exp(n) = [exp(1)]n=enand exp(1) = [exp(1=2)]2thus exp(1=2) =e1=2etc. so we can still identify exp(z) as the numbereto thecomplex powerzand (29) is the regular algebraic rule for exponents:ez+a=ezea. In particular exp(z) =ez=ex+iy=exeiy;(31) c

F. Walee, Math 321 Notes, UW 2012/12/117

e xis our regular real exponential buteiyis the exponential of a pure imaginary number. We can make sense of this from the series (23), (24) and (25) to obtain e iz= cosz+isinz; eiz= coszisinz;(32) or cosz=eiz+eiz2 ;sinz=eizeiz2i:(33) These hold for any complex numberz. [Exercise: Show thateizisnotthe conjugate ofeizunless zis real]. Forzreal, this isEuler's formulausually written in terms of a real anglee i= cos+isin:(34)

This is arguably one of the most important formula in all of mathematics! It reduces all of trigonom-

etry to algebra among other things. For instanceei(+)=eieiimplies cos(+) +isin(+) =(cos+isin)(cos+isin) =(coscossinsin) +i(sincos+ sincos) (35) which yields two trigonometric identities in one swoop.

Exercises:

1. Use series to compute the numbereto 4 digits. How many terms do you need?

2. Use series to compute exp(i), cos(i) and sin(i) to 4 digits.

3. Express cos(1 + 3i) in terms of real expressions and factors ofithat a 221 student might

understand and be able to calculate.

4. What is the conjugate of exp(iz)?

5. Use Euler's formula and geometric sums to derive compact formulas for the trigonometric

sums

1 + cosx+ cos2x+ cos3x++ cosNx=? (36)

sinx+ sin2x+ sin3x++ sinNx=? (37)

6. Generalize the previous results by deriving compact formulas for the geometric trigonometric

series

1 +pcosx+p2cos2x+p3cos3x++pNcosNx=? (38)

psinx+p2sin2x+p3sin3x++pNsinNx=? (39) wherepis an arbitrary real constant.

7. The formula (35) leads to the well-known double angle formula cos2= 2cos21 and

sin2= 2sincos. They also lead to the triple angle formula cos3= 4cos33cosand sin3= sin(4cos21). These formula suggests that cosnis a polynomial of degreen in cosand that sinnis sintimes a polynomial of degreen1 in cos. Derive explicit formulas for those polynomials. [Hint: use Euler's formula foreinand the binomial formula]. The polynomial for cosnin powers of cosis theChebyshevpolynomialTn(x) with cosn= T n(cos). c

F. Walee, Math 321 Notes, UW 2012/12/118

1.7 Polar representation

Introducing polar coordinates in the complex plane such thatx=rcosandy=rsin, then using Euler's formula (34), any complex number can be writtenz=x+iy=rei=jzjeiarg(z):(40) This is thepolar formof the complex numberz. Its modulus isjzj=rand the angle= arg(z)+2k is called thephaseofz, wherek= 0;1;2;:::is an integer. A key issue is that for a givenz, its phaseis only dened up to an arbitrary multiple of 2since replacingby2does not changez. However the argument arg(z) is a function ofzand therefore we want it to be uniquely dened for everyz. For instance we can dene 0arg(z)<2, or 1z2=r1ei1r2ei2=r1r2ei(1+2);(41) z 1z

2=r1ei1r

2ei2=r1r

2ei(12):(42)

1.8 Logs

The power series expansion of functions is remarkably powerful and closely tied to the theory of functions of a complex variable.A priori, it doesn't seem very general, how, for instance, could we expandf(z) = 1=zinto a series inpositivepowers ofz 1z =a0+a1z+a2z2+?? We can in fact do this easily using the geometric series. For anya6= 0 1z =1a+ (za)=1a 11 + zaa =1X n=0(1)n(za)na n+1:(43) Thus we can expand 1=zin powers ofzafor anya6= 0. That (geometric) series converges in the diskjzaj0, that disk where the series converges can be made as big as one wants but it can never include the origin which of course is the solesingular pointof the function 1=z. Integrating (43) fora= 1 term by term yields lnz=1X n=0(1)n(z1)n+1n+ 1(44) as the antiderivative of 1=zthat vanishes atz= 1. This looks nice, however that series only converges forjz1j<1. We need a better denition that works for a larger domain in thez-plane. c

F. Walee, Math 321 Notes, UW 2012/12/119

The Taylor series denition of the exponential exp(z) =P1 n=0zn=n! is very good. It converges for allz's, it led us to Euler's formulaei= cos+isinand it allowed us to verify the key property of the exponential, namely exp(a+b) = exp(a)exp(b) (whereaandbare anycomplex numbers), from which we deduced other goodies: exp(z)ezwithe= exp(1) = 2:71828:::, and e z=ex+iy=exeiy. What about lnz? As for functions of a single real variable we can introduce lnzas the inverse of e zor as the integral of 1=zthat vanishes atz= 1.

1.8.1lnzas the inverse ofez

Givenzwe want to dene the function lnzas the inverse of the exponential. This means we want to nd a complex numberwsuch thatew=z. We can solve this equation forwas a function of zby using the polar representation forz,z=jzjeiarg(z), together with the cartesian form forw, w=u+iv, whereu=<(w) andv==(w) are real. We obtain e w=z,eu+iv=jzjeiarg(z); ,eu=jzj; eiv=eiarg(z);(why?) ,u= lnjzj; v= arg(z) + 2k;(45) wherek= 0;1;2;Note thatjzj 0 is apositive real numberso lnjzjis our good old natural log of a positive real number. We have managed to nd the inverse of the exponential e w=z,w= lnjzj+iarg(z) + 2ik:(46) The equationew=zforw, givenz, has an innite number of solutions. This make sense since e w=eueiv=eu(cosv+isinv) is periodic of period 2inv, so ifw=u+ivis a solution, so is u+i(v+ 2k) for any integerk. We can take anyoneof those solutions as our denition of lnz, in particularlnz= ln jzjeiarg(z) = lnjzj+iarg(z):(47) This denition is unique since we assume that argzis uniquely dened in terms ofz. However dierent denitions of argzlead to dierent denitions of lnz. Example:If arg(z) is dened by 0arg(z)<2then ln(3) = ln3 +i, but if we dene instead arg(z)< then ln(3) = ln3i. Note that you can now take logs of negative numbers! Note also that the lnzdenition ts with our usual manipulative rules for logs. In particular since ln(ab) = lna+ lnbthen lnz= ln(rei) = lnr+i. This is the easy way to remember what lnzis.

1.8.2 Complex powers

As for functions of real variables, we can now dene general complex powers in terms of the complex log and the complex exponential a b=eblna=eblnjajeibarg(a);(48) be careful thatbis complex in general, soeblnjajis not necessarily real. Once again we need to dene arg(a) and dierent denitions can actually lead to dierent values forab. In particular, we have the complex power functions z a=ealnz=ealnjzjeiaarg(z)(49) c

F. Walee, Math 321 Notes, UW 2012/12/1110

and the complex exponential functions a z=ezlna=ezlnjajeizarg(a):(50) These functions are well-dened once we have dened a range for arg(z) in the case ofzaand for arg(a) in the case ofaz.

Once again, a peculiar feature of functions of complex variables is that the user is left free to choose

whichever denition is more convenient for the particular problem under consideration. Note that dierent denition for the arg(a) provides denitions forabthat donotsimply dier by anadditive multipleof 2ias was the case for lnz. For example (1)i=eiln(1)=earg(1)=e2k for somek, so the various possible denitions of (1)iwill dier by amultiplicative integer power ofe2.

1.8.3 Roots

Thefundamental theorem of algebrastates that anynth order polynomial equation of the formcnzn+cn1zn1++c1z+c0= 0 withcn6= 0 always hasnroots in the complex plane. This can be stated as saying that there always existncomplex numbersz1,:::,znsuch that c nzn+cn1zn1++c0=cn(zz1)(zzn):(51) The numbersz1,:::,znare therootsorzerosof the polynomial. These roots can be repeated as for the polynomial 2z24z+ 2 = 2(z1)2. This expansion is calledfactoringthe polynomial. The equation 2z22 = 0 has two real rootsz=1 and 2z22 = 2(z1)(z+ 1). The equation

3z2+3 = 0 has no real roots, however it has two imaginary rootsz=iand 3z2+3 = 3(zi)(z+i).

The equationzna= 0, withacomplex andna positive integer, therefore hasnroots. We might be tempted to write the solution asz=a1=nbut what does that mean? According to our denition ofababove, we havea1=n=e(lna)=nwhich depends on the argument ofasince lna= lnjaj+iarg(a). When we dene a function we need to make the denition unique, but here we are looking forall the roots. This means that we have to considerallpossible denitions of arg(a). Here's a correct way to think about this. Using the polar representationz=rei z n=rnein=a=jajei(arg(a)+2k))r=jaj1=n; =arg(a)n +k2n (52) wherek= 0;1;2;:::The modulijzj=randjajare positive real numbers, sojaj1=nis our good old root function giving a positive real value, but= arg(z) has many possible values that dier by a multiple of 2=n. Whennis a positive integer, this yieldsndistinct values ofmodulo 2, yieldingndistinct values forz. It is useful to visualize these roots. They are all equispaced on the circle of radiusjaj1=nin the complex plane.

Exercises:

1. Find all the roots, visualize and locate them in the complex plane and factor the corresponding

polynomial (i)z4= 1, (ii)z4+ 1 = 0, (iii)z2=i, (iv) 2z2+ 5z+ 2 = 0.

2. Investigate the solutions of the equationzb= 1 when (i)bis a rational number,i.e.b=p=q

withp,qintegers, (ii) whenbis irrationale.g.b=, (iii) whenbis complex,e.g.b= 1 +i. Visualize the solutions in the complex plane if possible.

3. Ifaandbare complex numbers, what's wrong with saying that ifw=ea+ib=eaeibthen

jwj=eaand arg(w) =b+ 2k? Isn't that what we did in (45)? c

F. Walee, Math 321 Notes, UW 2012/12/1111

2 Functions of a complex variable

2.1 Visualization of complex functions

A functionw=f(z) of a complex variablez=x+iyhas complex valuesw=u+iv, whereu;v are real. The real and imaginary parts ofw=f(z) are functions of the real variablesxandy f(z) =u(x;y) +iv(x;y):(53)

For example,w=z2= (x+iy)2is

z

2= (x2y2) +i2xy(54)

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