By John B. Reade

ISBN-10: 0415308461

ISBN-13: 9780415308465

ISBN-10: 041530847X

ISBN-13: 9780415308472

This sensible therapy explains the functions advanced calculus with out requiring the rigor of a true research historical past. the writer explores algebraic and geometric features of advanced numbers, differentiation, contour integration, finite and countless genuine integrals, summation of sequence, and the elemental theorem of algebra. The Residue Theorem for comparing complicated integrals is gifted in an easy approach, laying the basis for extra research. A operating wisdom of actual calculus and familiarity with advanced numbers is thought. This publication comes in handy for graduate scholars in calculus and undergraduate scholars of utilized arithmetic, actual technology, and engineering.

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**Example text**

8. 9. Find where | sin z| is maximum for |z| ≤ 1 (draw a diagram). Prove that all points z satisfying z+1 =2 z+4 lie on a circle. Find its centre and radius. 1 Differentiability and continuity For a real function f (x) of a real variable x the derivative f (x) is deﬁned as the limit f (x + h) − f (x) . 1) f (x + h) − f (x) h is the gradient of the line P Q which converges to the tangent at P as Q → P . So f (x) is the gradient of the tangent at P . For example, if f (x) = x 2 then we have f (x + h) − f (x) (x + h)2 − x 2 x 2 + 2xh + h2 − x 2 = = h h h 2 2xh + h = 2x + h, = h which → 2x as h → 0.

We say f (z) has a pole at z = c if the principal part has only ﬁnitely many non-zero terms. If the principal part has inﬁnitely many non-zero terms we say f (z) has an essential singularity. The order of a pole is the largest n for which a−n = 0. A pole of order 1 is called a simple pole. A pole of order 2 is called a double pole. The residue of f (z) at z = c is the coefﬁcient a−1 in the Laurent expansion at z = c. For example, f (z) = e1/z has an essential singularity at z = 0, since the Laurent expansion at z = 0 is e1/z = 1 + 1 1 1 + + ··· .

34 Derivatives We get the Laurent expansion at z = −i by putting t = z + i and expanding in terms of t. This time we have 1 t 1 1 + + + ··· , =− 2it 4 8i 1 + z2 which shows that f (z) also has a simple pole at z = −i, but now with residue −1/2i. Notes For a proof of Theorem 2, see, for example, Knopp (1945) page 30. Neither Taylor nor Maclaurin gave a rigorous proof of the validity of their expansions. They are not valid in general, even for functions with derivatives of all orders. An interesting example is the function f (x) = e−1/x , 2 which (if we assume f (x) = 0 at x = 0) has f (n) (0) = 0 for all n, so has a Maclaurin expansion which vanishes identically, therefore cannot = f (x) at any x = 0.

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

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