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
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
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
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 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
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
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
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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Forme algébrique des nombres complexes
1 −2i −z (1 −iz)2 +e−i θ(1 −ix)2(3 +2i) Module Soit z ∈ C On pose z =a+ib où a et b sont deux réels Le module de z est z= √ a2 +b2 Pour tout nombre complexe z =a+ib, a et b réels, zz =z2 =a2 +b2 Pour tout nombre complexe non nul z, 1 z = z z2 Propriétés de calculs « Le module marche bien avec la multiplication » : Pour tous nombres complexes z et z′, z×z
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Chapitre2 NOMBRESCOMPLEXES Enoncédesexercices
1 LESBASIQUES CHAPITRE2 NOMBRESCOMPLEXES Exercice2 30Résoudre(z−i)n=zn Exercice2 31Soienta,betctroiscomplexesdemodule1 telsqueac=−1,montrerqueTaille du fichier : 841KB
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Les nombres complexes - unicefr
1) z = 4 −3i 2) z = 1 +2i 3) z = −2 +i Exercice26 Trouver une forme trigonométrique de chacun des nombres complexes suivants : 1) z = (1 −i)2 2) z = 1 −i √ 3 1 +i 3) z = (√ 3 +i)9 (1 +i)12 Exercice27 On donne les nombres complexes suivants : z1 = √ 6 −i √ 2 2 et z2 = 1 −i 1) Donner le module et un argument de z1, z2 et z1 z2 2) Donner la forme algébrique de z1 z2 3) En
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Chapitre1:Lescomplexes
1 z 2 =⇒cos3(x)= 1 3 z+ 1 3 Avec le binôme,onobtient cos3(x) = 1 23 z3+3z2× 1 z +3z× 1 z2 + 1 z3 = 1 23 z3+ 1 z3 +3 z+ 1 z = 1 22 1 2 z3+ 1 z3 +3× 1 2 z+ 1 z = cos(3x)+3cos(x) 4 Demêmesinx= z− 1 z 2i d’où sin3(x) = 1 23i3 z3−3z2× 1 z +3z× 1 z2 − 1 z3 =− 1 22 1 2i z3− 1 z3 −3× 1 2i z+ 1 z = − sin(3x)−3sin(x) 4
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Nombres complexes 1 Forme cartésienne, forme polaire
z2 +z+1 =0 ; z2 (1+2i)z+i 1 =0 ; z2 p 3z i=0 ; z2 (5 14i)z 2(5i+12)=0 ; z2 (3+4i)z 1+5i=0 ; 4z2 2z+1 =0 ; z4 +10z2 +169 =0 ; z4 +2z2 +4 =0: Indication H Correction H Vidéo [000031] 3 Racine n-ième Exercice 8 Calculer la somme S n =1+z+z2 + +zn Indication H Correction H Vidéo [000047] Exercice 9 1 Résoudre z3 = 1 et montrer que les racines s’écrivent 1, j, j2 Calculer 1+ j+ j2 et en Taille du fichier : 203KB
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NOMBRES COMPLEXES - Free
Ch4 : Nombres complexes (TS) - 1/18 - NOMBRES COMPLEXES I INTRODUCTION ET DEFINITION Tous les nombres positifs ont une racine carrée, par exemple, 9 a pour racine 3 Taille du fichier : 195KB
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Partie A Partie B
2) Soit z ∈ C z3 −(4 +i)z2 +(7 +i)z−4 = 0 ⇔ (z −1)(z −2−2i)(z −1 +i) = 0 ⇔ z = 1 ou z = 2 +2i ou z = 1 −i En notant S l’ensemble des solutions de l’équation (E), on a donc S = {1,2 +2i,1 −i} Partie B 1) Voir graphique à la fin de l’exercice 2) 2 +2i 1 −i = 2(1 +i)2 (1 −i)(1 +i) = 2(1 +2i −1) 12 +12 = 2i
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Exo7 - Exercices de mathématiques
1, z, z2, z3 et z4 sont les cinq racines cinquièmes de 1 dans C Par suite, 1+z+z2 +z3 +z4 =0 Mais alors a+b=z+z2 +z3 +z4 = 1 et ab=(z+z4)(z2 +z3)=z3 +z4 +z6 +z7 =z+z2 +z3 +z4 = 1 (car z5 =1): 4 a et b sont donc les solutions de l’équation X2 +X 1 = 0 dont les racines sont 1+ p 5 2 et p 2 Enfin, puisque 2p 5 20; p 2, on a a>0 Par suite, cos 2p 5 = 1+ p 5 4 et cos 4p 5 = 1 p 5 4 D Taille du fichier : 262KB
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Exercices de préparation au contrôle du jeudi 25 octobre
(1+z)(1+z2) 1+z f (z)=1+z2 f (i)=1+i2=1 1=0 f (i 1)=1+(i 1)2=1+i2 2i+1=1 2i f (2+i)=1+(2+i)2=1+4+4i+i2=1+4+4i 1=4+4i b) On résout dans { 1} 2f (z)=0 ⇔ 1+z =0 (au moment de résoudre, on est obligé de penser à la factorisation) f (z)=0 2⇔ z = 1 ⇔ z=i ou z= i S={ i;i} Exercice 3 : 1) z2+2z+3=0 (E 1) 2) z 2= z 1 (E 2) 3) (1+2i)z( i 1)=iz
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Zadania o liczbach zespolonych - mathuwbedupl
Z pierwszego r´ownania a = 3b, wiec po podstawieniu do drugiego r´ownania 3, b + 5b = 8, skad, b = 1 i a = 3b = 3 Odp a = 3 i b = 1 c) Obliczamy (4−3i) 2= 16−24i+9i = 16−24i−9 = 7−24i, (1+i)2 = 1+2i+i2 = 1+2i−1 = 2i Teraz zapisujemy lewa stron, e r´owno´sci w
Im z = 1. 2i(z ? ¯z). Thus z is real if and only if ¯z = z and pure imaginary if and only if ¯z = ?z. More importantly we have the following formulas
Figure 1: A complex number z and its conjugate ¯z in complex space. Eg 5.2.1 Given that z1 =3+4i z2 = 1 ? 2i
e?zt dt (Re z > 0). Ans: (a) ?. 1. 2 ? i ln 4;. (b). ?3. 4. + i. 4. ;. (c). 1 z . Solution: (a) We have. ? 2. 1 (1t ? i). 2 dt = ?. 2. 1 ( 1t2 ?. 2i.
(z ? 2i)2(z + 2i)2 . Let f(z) = 1. (z + 2i)2 . Clearly f(z) is analytic inside C. So by Cauchy's formula for derivatives: ?. C. 1. (z2 + 4)2.
1. 0 t2(1 + i)2(1 + i)dt = 2i(1 + i). 3 . Example 3.2. Compute. ? ? z dz analytic function on an open region A and ? is a curve in A from z0 to z1 then.
In this notation the sum and product of two complex numbers z1 = x1 + iy1 and z2 = x2 + iy2 is given For example
http://www.math.pitt.edu/~sparling/053/156053/156053quizzes/156053e1s.pdf
1. Compute real and imaginary part of z = i ? 4. 2i ? 3 . 2. Compute the absolute value and the conjugate of z = (1 + i)6.
The geometric series in this equation has ratio z/w. Therefore the series converges
z?2 + 1 z?1?2i is also analytic at zero the Laurent series of f(z) has the form f(z) = 3 z + 1 z2 + ? ? n=0 anz n We conclude that f(z) has a pole of order 2 at z0 = 0 and residue Res z=0 f(z) = 3 Similarly f(z) has simple poles (order 1) at z1 = 2 and z2 = 1 +2i with Res z=2 f(z) = 5i Res z=1+2i f(z) = 1 3 e1/z = ? ? n=0 z
we have that (z+i)g(z) = 1 z i which is analytic and nonzero at z= i Hence g(z) has a simple pole at z= i The residue is thus Res i(g) = lim z! i(z+ i)g(z) = 1 2i = 1 2 i 8 g(z) = ez z3 at z= 0 Ans Res 0(g) = 1 2 Solution Using the power series for ez we see that the Laurent series for g(z) about z= 0 is ez z 3 = 1 + z+ 1 2! z2 + 1
(3) By completing the square z4 4z2 + 4 2i= (z2 2)2 2iand so the equation is equivalent to (z2 2)2 = 2i= (1 + i)2;which is equivalent to z2 2 = 1 + i or z2 2 = 1 i; that is z2 = 3 + i or z2 = 1 i: Hence the solutions are the square roots of 3 + iand 1 i:They are in the form of z= z 1; z= z 2; where z2 1 = 3 + iand z2 2 = 1 ican be found in
(z2+1)(z2+2z+3) f(z) has two poles iand 1 + i p 2 in the upper half plane which are both simple Then Res(f(z);i) = 1=8 i=8 and Res(f(z); 1 + i p 2) = 1=8 The Residue Theorem shows that Z 1 1 dx (x2 + 1)(x2 + 2x+ 3) = 2?i( 1=8 i=8 + 1=8) = ? 4 provided we can show that lim R!1 Z C R f(z)dz= 0 where C R is the semicircle jzj= Rwith =z 0 To
Conjugate of A Complex Number
Conjugate of a complex number z = x + iy is denoted by; The geometrical representation of the complex number is shown in the figure given below:
Modulus of Complex Number
Modulus of the complex number is the distance of the point on the argand plane representing the complex number z from the origin. Let P is the point that denotes the complex number z = x + iy. Then OP = |z| = ?(x2 + y2).
How do you calculate I2 C 1 z2dz?
I2 = ? C 1 z2dz. First, in the applet select the function f (z)=1/z^2. Then analize the values of I2 in the following cases: C is any contour from z0 = ? i to z1 = i. What happens when you select Line Segment in the applet? What happens when you select Semicircles?
How to evaluate C1 ZDZ?
Now let’s evaluate ? C1 zdz, where C is the circle x = cost, y = sint, with 0 ? t ? 2?. Figure 2: C: z(t) = cost + isint = eit, with 0 ? t ? 2?. and, z ? (t) = ieit. Thus ? C1 zdz = ?2? 0 (e ? it)ieitdt = i?2? 0 dt = 2?i. Use the following applet to explore numerically the integral Line segments. Semicircles.
What is a contour integral f(z) from a fixed point z0 to Z1?
Although the value of a contour integral of a function f(z) from a fixed point z0 to a fixed point z1 depends, in general, on the path that is taken, there are certain functions whose integrals from z0 to z1 have values that are independent of path, as you have seen in Exercises 2 and 3.
Is z t a real integral of a complex-valued function?
Since C is a contour, z? (t) is also piecewise continuous on a ? t ? b; and so the existence of integral ( 1) is ensured. The right hand side of ( 1) is an ordinary real integral of a complex-valued function; that is, if w(t) = u(t) + iv(t), then in terms of its real and imaginary parts, as well as the differential
1 Review of complex numbers