By Emil G. Milewski Ph.D. Chief Editor

For college students in arithmetic, engineering, and physics. contains accomplished insurance of advanced numbers, set idea, mapping, services, Cauchy-Riemann stipulations, energy sequence, Taylor sequence, Green's theorem, Laurent expansions, singularities, residues, modifications, and various medical purposes.

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3) is a real number, therefore the left-hand side also has to be a real number, z GR. Thus, we can drop the absolute value syrnbol and·write s, then (2) (3) 2z = z 2 + 1 (4) r (5) t· or z 2 - 2z + 1 (z - 1) 2 o ( 4) The only solution is (6) z =_ 1 (5) (7) +36 = 1136 Prove the theorem known as the triangle inequality. e. (9) (10) (1) (11) (12) (13) (14) Solution: ! (15) leads + Z2Z2 + z, eq. (5) to any finite number of complex numbers ~ zK 1· -< K=l ~ lz K 1 (6) Hence, 1K=l e PROBLEM 1-24 Prove that if Solution: z1 and z2 are complex numbers, 11z11 - 1z2 l ~ 1z1 - z2 I I 'ª then Eq.

Now, if there a FIELD OF COMPLEX NUMBERS a PROBLEM 1-16 and hence An important concept in mathernatics is the notion of a field. A set S on which two operations of addition and multiplication are defined is called a field if the operations satisfy the following axioms: I. II. The set of comJ operations of :: Problem 1-6. ~ a e e and b G e and 1-8, we shc mutative, assoc O + iO = O, be< If a G S, b G S, then a+b G S and ab G S. Commutative Laws. If a G S, b G S, then a+b=b+a 1 ¡. ab = ba l,, ,. III.

6. Im . 3. The sets Re (1) g. 4 ) • For p = O , Fig. 6 This conclusion can be reached without long calculations. I t follows that lz 2 -ll = f (z+l)(z-l)f lz • 2 8. ~e = fz+lf fz-11 = s -ll is the product of the distances of z Let lz2-zl = 1 and -1. Tben = 2. 1 z2-zl f rom (5) 1z(z-1) J = lzl lz-11 = 2 (6) is the product of the distances of z from O and l. (6) represents the lemniscate. (1), we obtain l 2 z+1¡ z-1 = (z+l}~+ll =4 z-1 - (z+l)(z+l) ... so that (2) z-1 or "' (1) 2 5 = 4(z-l)(z-1) (3) 5 - (4) zz - 3 z - 3 z + 1 = o.

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