By Artstein-Avidan S.

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Extra resources for A Bernstein-Chernoff deviation inequality, and geometric properties of random families of operators

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Z 3 2 1 –3 –2 –1 0 1 2 3 –1 –2 –3 Figure 35: The curve C and its orientation. (a) By using the Theory of Complex Functions we get 4z 3 dz = z 4 C 3 3 = 0. Alternatively it follows by using the parametric description, 1 4z 3 dz = 0 C 4 · 33 · e6iπt · 3 · 2iπ · e2iπt dt = 34 1 0 8iπ · e8iπt dt = 81 · e8iπt 1 0 = 0. (b) We get by insertion of the parametric description that 1 z dz = 0 C 3 e−2iπt · 3 · 2iπ · e2iπt dt = 18πi. (c) We get by insertion of the parametric description that C 1 dz = z 1 0 1 · 3 · 2iπ e2iπt dt = 2iπ.

2 Figure 34: The curve C with its orientation. (a) By using the Theory of Complex Functions we get 4z 3 dz = z 4 C −1 1 = (−1)4 − 14 = 0. Alternatively we apply the parametric description 1 4z 3 dz = 0 C 4 e−3iπt · (−iπ)e−iπt dt = 1 (−4iπ)e−4iπt dt = e−4iπt 0 1 0 = 1 − 1 = 0. (b) By insertion of the parametric description we get 1 z dz = C 0 e+iπt · (−iπ)e−πt dt = −iπ. (c) By insertion of the parametric description we get C 1 dz = z 1 0 1 (−iπ)e−iπt dt = −iπ. 9 Sketch the curve C of the parametric description z = 3 e2πit , t ∈ [0, 1], and indicate its orientation.

D) Since Re(z) = |z| |x| x2 + y 2 ≤1 for every z = 0, we get z Re(z) z Re(z) ≤ |z| · 1 → 0 −0 = |z| |z| for z → 0, and we conclude that this function can be extended continuously to z = 0 with the value f (0) = 0. 3 Check if the following limit values exist. in , n→+∞ nn (a) lim (b) lim in , n→+∞ (c) lim n n→+∞ 1+i n 2 . (a) We shall prove that the limit value is 0, thus |an − 0| < ε for every n ≥ N (ε). This follows easily from the following trivial estimate |an − 0| = |an | = n! n! in 1 2 3 n−1 n 1 = n = · · ··· · ≤ , n nn n n n n n n for n → +∞.

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