By Hugo. Rossi

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**Example text**

6]. 7. Given complex numbers c1 , . . , cn not all equal to 0, show that z n + c1 z n−1 + · · · + cn = 0 =⇒ 1 |z| < 2 max |cj | j . 8. 2. Let (u, v) = (0, 0) ∈ R2 , let r = ψ ∈ [0, 2π) be determined by u v cos ψ = , and sin ψ = . r r Then, cos θ sin θ − sin θ cos θ) u v = cos θ sin θ − sin θ cos θ) √ r cos ψ r sin ψ =r cos θ cos ψ − sin θ sin ψ sin θ cos ψ + cos θ sin ψ =r cos(θ + ψ) . 6. 13), we obtain n (1 + i)n = 2 2 (cos nπ nπ + i sin ). 6. Solutions 33 Remark. 6) n (α + β)n = k=0 n k αk β n−k .

19. We have |1 − Bw (z)| = 1 − z−w |w + w| = . 33) with z + w instead of z) |z + w| ≥ Re (z + w) > Re z. 20. We have (z − w)(v − w) (1 − zw)(1 − vw) (1 − zw)(1 − vw) − (z − w)(v − w) = (1 − zw)(1 − vw) (1 − zv)(1 − |w|2 ) = , (1 − zw)(1 − vw) 1 − bw (z)bw (v) = 1 − and hence we obtain the required identity. 48) will hold in particular when z, v and w belong to D. 1. Let w ∈ D. The function 1 − bw (z)bw (v) 1 − zv is positive deﬁnite in Ω = D. 21. We have (z − w)(v − w) (z + w)(v + w) (z + w)(v + w) − (z − w)(v − w) = (z + w)(v + w) 2(z + v)(Re w) = , (z + w)(v + w) 1 − Bw (z)Bw (v) = 1 − and hence 1 − Bw (z)Bw (v) 2Re w = .

14. 27) 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 ).