Calculus with Complex Numbers, 1st Edition - download pdf or read online

By John B. Reade

ISBN-10: 0415308461

ISBN-13: 9780415308465

ISBN-10: 041530847X

ISBN-13: 9780415308472

This useful remedy explains the purposes complicated calculus with out requiring the rigor of a true research historical past. the writer explores algebraic and geometric elements of advanced numbers, differentiation, contour integration, finite and limitless genuine integrals, summation of sequence, and the basic theorem of algebra. The Residue Theorem for comparing complicated integrals is gifted in an easy means, laying the basis for additional research. A operating wisdom of genuine calculus and familiarity with advanced numbers is believed. This e-book turns out to be useful for graduate scholars in calculus and undergraduate scholars of utilized arithmetic, actual technology, and engineering.

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Extra info for Calculus with Complex Numbers, 1st Edition

Example text

This conformal property is crucial in applications to fluid dynamics. Examples 1. Prove that for all |z| = 2 2 ≤ |z − 4| ≤ 6. 2. Prove that for all |z| = 3 8 10 z2 + 1 ≤ 2 . ≤ 11 7 z +2 3. Prove that for all |z| = 4 3 5 z+i ≤ ≤ . 5 z−i 3 Complex functions 4. Prove that for all |z| = R > 2 1 1 . ≤ 2 z2 + z + 1 R −R−1 5. 6. 7. Prove that |ez | = eRe z . Find where |ez | is maximum for |z| ≤ 2 (draw a diagram). Prove that for z = x + iy | sin(x + iy)|2 = sin2 x + sinh2 y, | cos(x + iy)|2 = cos2 x + sinh2 y.

If we could prove that γ2 z dz →0 z2 + 1 as R → ∞, then we could deduce that R −R x dx = x2 + 1 γ1 z dz = z2 + 1 γ z dz − z2 + 1 γ2 z dz → πi z2 + 1 56 Evaluation of infinite real integrals as R → ∞. Which is nonsense because R −R x dx =0 x2 + 1 for all R since the integrand is odd. In fact γ2 z dz →0 z2 + 1 in this case. Its value is γ2 z dz = +1 z2 γ z dz − +1 z2 γ1 z dz = πi +1 z2 for all R > 1. 4 Integrals involving cos x, sin x Consider the integral ∞ −∞ cos x dx . 2 will be no use here because | cos z|2 = cos2 x + sinh2 y (z = x + iy) is unbounded in the upper half plane.

F (z) = 4S 2 T 2 + 2S 4 = 2 at z = 0, therefore a3 = 2/3! = 1/3. Hence we obtain tan z = z + z3 + ··· 3 Alternatively, we can write tan z = sin z z − z3 /3! + z5 /5! − · · · = cos z 1 − z2 /2! + z4 /4! − · · · = z− z5 z3 + − ··· 6 120 × 1+ = z− =z+ z2 z2 − + ··· 2 24 z5 z3 + − ··· 6 120 2 z3 + z5 + · · · . 7 Taylor expansions The Maclaurin expansion is a particular case of a more general expansion due to Taylor (1715) which represents f (z) as a series in powers of z−c for any fixed c as ∞ f (z) = an (z − c)n , n=0 where the nth coefficient an is given by the formula f (n) (c) .

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Calculus with Complex Numbers, 1st Edition by John B. Reade


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