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April 5, 2017 | Mathematics | By admin | 0 Comments

By Su Wang Kuo

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So suppose N x + y − 2i x y ≡ 2 αk (x + i y)k . 2 k=0 Setting y = 0, we obtain N x2 ≡ αk x k k=0 or α0 + α1 x + (α2 − 1)x 2 + · · · + α N x N ≡ 0. Setting x = 0 gives α0 = 0; dividing out by x and again setting x = 0 shows α1 = 0, etc. We conclude that α1 = α3 = α4 = · · · = α N = 0 α2 = 1, and so our assumption that N x 2 + y 2 − 2i x y ≡ αk (x + i y)k k=0 has led us to x 2 + y 2 − 2i x y ≡ (x + i y)2 = x 2 − y 2 + 2i x y, which is simply false! A bit of experimentation, using the method described above (setting y = 0 and “comparing coefficients”) will show how rare the analytic polynomials are.

Sin2 z + cos2 z = 1, c. (sin z) = cos z. * Show that a. sin( π2 + iy) = 12 (e y + e−y ) = cosh y b. | sin z| ≥ 1 at all points on the square with vertices ±(N + 12 )π ± (N + 12 )π i, for any positive integer N. c. | sin z| → ∞, as Imz = y → ±∞. Exercises 43 17. Find (cos z) . 18. Find sin−1 (2)– that is, find the solutions of sin z = 2. * Find all solutions of the equation: z ee = 1. 20. Show that sin(x + iy) = sin x cosh y + i cos x sinh y. 21. Show that the power series f (z) = 1 + z + z2 + ...

Power Series Expansion about z = α All of the previous results on power series are easily adapted to power series of the form Cn (z − α)n . By the simple substitution w = z − α, we see, for example, that series of the above form converge in a disc of radius R about z = α and are differentiable throughout |z − α| < R where R = 1/lim|Cn |1/n . ) Exercises 1. 3 by showing that for an analytic polynomial P, Py = i Px . * a. Suppose f (z) is real-valued and differentiable for all real z. Show that f (z) is also real-valued for real z.

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