Revision Lectures Trinity Term 2012 Lecturer: F Hautmann CP3

Transcription

1 Revision Lectures Trinity Term 2012 Lecturer: F Hautmann CP3 Complex Numbers and Ordinary Differential Equations Lecture times: Monday week 2 at 11 am Tuesday week 2 at 11 am Venue: Martin Wood Lecture Theatre Slides will be posted on lecture webpage: www-thphys.physics.ox.ac.uk/people/francescohautmann/cp4rev/

2 I. Complex numbers 1 Basic algebra and geometry of the complex plane 2 Elementary functions of complex variable 3 De Moivre s theorem and applications 4 Curves in the complex plane 5 Roots of complex numbers and polynomials

3 1. THE COMPLEX PLANE C = set of complex nos with +, operations Im z b z = a+ib, a = Rez, z i 2 = 1 b = Imz ρ = mod( z)= z ρ θ a Re z θ=arg(z) b z z = a+ib = ρe iθ = ρ(cosθ+isinθ) z = a ib = ρe iθ = ρ(cosθ isinθ)

4 2. Elementary functions on C Complex polynomials and rational functions defined by algebraic operations in C Complex exponential: e z = e x+iy = e x (cosy +isiny) complex trigon. and hyperb. fctns in terms of exp. e.g. sinz = (e iz e iz )/(2i) sinhz = (e z e z )/2 Complex logarithm lnz: e lnz = z lnz = ln z +i(θ +2nπ), n = 0,±1,... ( multi-valued) complex powers: z α = e αlnz (α complex)

5 3. DE MOIVRE S THEOREM de Moivre s theorem and trigonometric identities z n =(re ı! ) n =r n e in! =r n (cosn! +isinn!) For r=1 (e ı! ) n =(cos! +ısin!) n =cosn! +ısinn!. De Moivre e.g. n=2 : 2 2! =! "! cos 2 cos sin sin 2! = 2 cos! sin!

6 June Express tan3θ in terms of tanθ. DeMoivre cos3θ +isin3θ = (cosθ +isinθ) 3 = cos 3 θ +3icos 2 θsinθ 3cosθsin 2 θ isin 3 θ Then : tan3θ = sin3θ cos3θ = 3cos2 θsinθ sin 3 θ cos 3 θ 3cosθsin 2 θ = 3tanθ tan3 θ 1 3tan 2 θ June Prove that sin 5 θ = 1 24(sin5θ 5sin3θ +10sinθ) sin 5 θ = ( e iθ e iθ 2i ) 5 = [e 5iθ 5e 3iθ +10e iθ 10e iθ +5e 3iθ e 5iθ ]/(i 2 5 ) = [sin5θ 5sin3θ +10sinθ]/2 4

7 9.a) Prove the identity N 1 n=0 June 2008 cos(nx) = sin(nx/2) sin(x/2) cos((n 1)x/2). N 1 n=0 cos(nx) = Re N 1 n=0 e inx = Re ( e inx n=0 [ (1 e inx = Re ) ] e inx n=0 = Re einx/2 [ 2isin(Nx/2)] e ix/2 [ 2isin(x/2)] = sin(nx/2) sin(x/2) cos((n 1)x/2) n=n e inx }{{} e inx n=0 e inx = Re 1 einx 1 e ix = sin(nx/2) sin(x/2) ) Re e i(n 1)x/2

8 The complex logarithm lnz ln e z = z = z e ı! =e ln z e ı! =e ln z +ı!! lnz =ln z +ıarg(z) Need to know! including 2" n phase ambiguity in z lnz = ln z +i(θ+2πn), n integer different n different branches of the logarithm n = 0: principal branch ln z example of multi-valued function

9 Note: It follows that the formula ln(z 1 z 2 ) = lnz 1 +lnz 2 is valid, once the branch of lnz 1 and lnz 2 is chosen, only for a given choice of the branch for ln(z 1 z 2 ). In general, ln(z 1 z 2 ) = lnz 1 +lnz 2 +2πik forsomek Z June Prove the above formula. Proof. Let lnz 1 = w 1, lnz 2 = w 2, and ln(z 1 z 2 ) = w 3. Then e w 3 = e ln(z 1z 2 ) = z 1 z 2 = e w 1 e w 2 = e w 1+w 2 whichmeans w 3 = w 1 +w 2 +2πik forsomek Z, thatis, ln(z 1 z 2 ) = lnz 1 +lnz 2 +2πik.

10 Example: evaluation of ln( 1) ln( 1) = lne iπ = ln1+i(π +2πn) = iπ +2πin, n integer ln( 1) = iπ for n = 0 (principal branch) Note: with z 1 = 1, z 2 = 1, the formula ln(z 1 z 2 ) = lnz 1 +lnz 2 is satisfied, once the principal branch value ln( 1) = iπ is taken, only for the branch in which ln( 1) 2 = ln1 = 2πi.

11 June 2007 Find all possible values of ln(z) ln( z) where z is a complex number. z = z e iθ lnz = ln z +i(θ+2πn), n integer ln( z) = ln z +i(π +θ +2πm), m integer ln(z) ln( z) = iπ +2i(n m }{{} )π = i(2k 1)π k

12 4. CURVES IN THE COMPLEX PLANE locus of points satisfying constraint in complex variable z Curves can be defined either directly by constraint in complex variable z ex. : z = R or by parametric form γ(t) = x(t)+iy(t) ex. : x(t) = Rcost y(t) = Rsint, 0 t < 2π

13 September 2009 Sketch the loci in the complex plane representing the conditions Re(z 1 ) = 1 and Im(z 1 ) = 1. Im z 1/2 i/2 Re z

14 September 2011 Draw the set of points in the complex plane such that z 3i + z +3i = 10. Im z 3i 4 4 Re z 3i

15 June 2007 Sketch the curve in the complex plane z = a t e ibt, 6 t 6 where a and b are real constants given by a = 1.1, b = π/3. Im z Re z

16 Def.: 5. ROOTS OF COMPLEX NUMBERS A number u is said to be an n-th root of complex number z if u n = z, and we write u = z 1/n. Th.: Every complex number has exactly n distinct n-th roots. Let z = r(cosθ +isinθ); u = ρ(cosα+isinα). Then r(cosθ +isinθ) = ρ n (cosα+isinα) n = ρ n (cosnα+isinnα) ρ n = r, nα = θ+2πk (k integer) Thus ρ = r 1/n, α = θ/n+2πk/n. So u = z 1/n = r 1/n [cos n distinct values for k from 0 to n 1. (z 0) ( θ n + 2πk n ) +isin ( θ n + 2πk n )], k = 0,1,...,n 1

17 !! 3..&),.4,-./0*.1(#/), n P( z) " a z + a z +! + a. n n! 1 n! 1 0 P( z = z i ) = 0!z i is a root!"#$#%&'$()(*+,#,-./0*.1(#/,20,(&),$..&), a z + a z +! + a = a ( z! z)( z! z )!( z! z ) n n! 1 n n! 1 0 n 1 2 n n n n! 1 n n( " j! ( 1) # j) j= 1 j= 1 =! + +!. a z z z z n!.1-#$(*+,%.'5%('*&),.4,6 *78, #*9,6 :, an! 1 a0 =! z ; = (! 1) a a n " n # j n j z

18 June Determine the real numbers x and y which satisfy 100 k=0 e iπk/3 = x+iy. Observe that e iπk/3 = e 2iπk/6 are the sixth roots of unity for k = 0,1,...,5 5 e iπk/3 = 0 k=0 The same applies for subsequent sets of 6 values of k. Thus the first 6 16 = 96 terms of the sum give zero, and the last five must equal minus the root with k = 5, so that 100 k=0 e iπk/3 = 100 k=96 e iπk/3 = e iπ5/3 = e iπ2/3 = 1 2 +i 3 2 x = 1 2, y = 3 2 Alternatively : 100 e iπk/3 = e iπk/3 e iπk/3 = (1 e iπ101/3) e iπk/3 k=0 k=0 k=101 }{{} e iπ101/3 k=0 e iπk/3 k=0 = 1 ei5π/3 1 e iπ/3 = ei5π/6 sin(5π/6) e iπ/6 sin(π/6) = e iπ2/3 = 1 2 +i 3 2

19 June b) Find all the solutions of the equation ( ) n z +i = 1, z i and solve z 4 10z 2 +5 = 0. ( ) n z +i = 1 = e i(π+2nπ), N integer z i z +i z i = ei(π/n+2nπ/n), N = 0,1,...,n 1 Then : z = i ei(π/n+2nπ/n) +1 e i(π/n+2nπ/n) 1 = i cos[π(1+2n)/(2n)] i sin[π(1+2n)/(2n)] = cotg π(1+2n) 2n. Forn = 5 : (z+i) 5 = (z i) 5 z(z 4 10z 2 +5) = 0. Thenthe4rootsofz 4 10z 2 +5 = 0 are cotg π 10,cotg 3π 10,cotg 7π 10,cotg 9π 10.

20 September 2010 September 2010

21 a) DeMoivre : (cosθ +isinθ) n = e inθ = cosnθ +isinnθ b) cos6θ +isin6θ = e i6θ = (e i2θ ) 3 = (cos2θ +isin2θ) 3 cos6θ = cos 3 2θ 3cos2θsin 2 2θ = 4cos 3 2θ 3cos2θ 3 4x 3 3x = 2 Then : x = cos2θ cos6θ = 3 2 6θ = π/6 x = cos2θ = cos(π/18) 6θ = 11π/6 x = cos2θ = cos(11π/18) 6θ = 13π/6 x = cos2θ = cos(13π/18) c) ( ) n z +1 = i z +1 z 1 z 1 = ei(π/(2n)+2nπ/n), N = 0,1,...,n 1 Then z = ei(π/(2n)+2nπ/n) +1 e i(π/(2n)+2nπ/n) 1 = cos[π(1+4n)/(4n)] i sin[π(1+4n)/(4n)] = i cotg π(1+4n) 4n Forn = 3 : (z +1) 3 = i(z 1) 3 z 3 +3iz 2 +3z +i = 0. Thenthe3roots of z 3 +3iz 2 +3z +i = 0 are i cotg π 12, i cotg 5π 12, i cotg 3π 4.

22 !"#$"%&"'()**+( u =z 3, u 2 +u+1=0, u =!1± 3i 2 =e i(2"/3+n"), n =0,1 z =u 1/3 =e i(2!/9+n!/3+2m!/3), n =0,1, m =0,1,2 2b. Solve the equation z 6!15z 4 +15z 2!1=0 ( z+i) 6 =z 6 +6iz 5!15z 4!20iz 3 +15z 2 +6iz!1 ( x+ y) ( x+ y) 1 ( x+ y) 2 ( x+ y) 3 ( x+ y) 4 ( x+ y) ( )=0 " z+i z!i ( z+i) 6 + ( z!i) 6 =2 z 6!15z 4 +15z 2!1 u =e i (!/6+2!n/6), n =0,1,2,3,4,5 z!i =u z+i ( ) " z = i( 1+u) ( 1!u) ( )6 ( ) 6 #u6 =!1=e i$