By Stefan Hildebrandt

ISBN-10: 3540439706

ISBN-13: 9783540439707

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Extra resources for Analysis 2 (Springer-Lehrbuch)

Example text

We have |2 + z n + z 5n | ≥ |2 − |z n + z 5n ||. For |z| < 1 we have |z n + z 5n | ≤ 2|z| < 2, and so 2 − |z n + z 5n | = 2 − |z n + z 5n | ≥ 2 − 2|z| = 2(1 − |z|), and hence the result. 17. It suﬃces to apply the triangle inequality to z1 z2 − 1 = (z1 − 1)(z2 − 1) + (z1 − 1) + (z2 − 1), and then add 1 to both sides of the obtained inequality. 18. 41) with z = |a| and w = −|b|. 19. Let z, w ∈ D. 37) with z1 = 1 and z2 = zw we have |1 − zw| ≥ 1 − |zw| > 0. 40 Chapter 1. 32): |1 − zw|2 − |z − w|2 = (1 − |z|2 )(1 − |w|2 ), ∀z, w ∈ C.

6. 13), we obtain n (1 + i)n = 2 2 cos nπ nπ + i sin . 1) Remark. 6) n (α + β)n = k=0 n k αk β n−k . We then obtain n (1 + i)n = ik k=0 n k (−1)p = p n 2p (−1)p +i p such that n 2p + 1 . 4. 8. Since d arctan u 1 = 2 du u +1 we have and d arctan 1/u 1 1 1 =− 2 , =− 2 du u 1 u +1 + 1 u2 d(arctan u + arctan 1/u) = 0, u = 0. du Thus the function arctan u + arctan 1/u is constant on (−∞, 0) and (0, ∞). Its value on each of these intervals is computed with the choices u = ±1. 36 Chapter 1. 9. (a) The formula for θ follows from the deﬁnition of arctan.

ZN , ⎞ ⎛ N N | z |2 = =1 N ⎜ |z |2 + 2 Re ⎝ =1 ⎟ z zk ⎠ . 14. 28) for w and −w and adding both identities. To prove the following two identities, one proceeds as follows: We have |1 + zw|2 = (1 + zw)(1 + zw) = 1 + zw + zw + |z|2 |w|2 , |1 − zw|2 = (1 − zw)(1 − zw) = 1 − zw − zw + |z|2 |w|2 , and |z − w|2 = (z − w)(z − w) = |z|2 − zw − wz + |w|2 . Thus |1 + zw|2 + |z − w|2 = 1 + zw + zw + |z|2 |w|2 + |z|2 − zw − wz + |w|2 = 1 + |z|2 |w|2 + |z|2 + |w|2 = (1 + |z|2 )(1 + |w|2 ), and |1 − zw|2 − |z − w|2 = 1 − zw − zw + |z|2 |w|2 − (|z|2 − zw − wz + |w|2 ) = 1 + |z|2 |w|2 − |z|2 − |w|2 = (1 − |z|2 )(1 − |w|2 ).