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Additional info for The theory of the imaginary in geometry
N} p(zi) = 0 where a0, a1, a2, ... an are complex coefficients. For real coefficients, the zeros are whether real or pairs of conjugate complex numbers. The proof is by supposing that p(z) has not any zero. In this case f(z) = 1/p(z) is analytic and bounded (because p(z)→ 0 for z→ ∞) in the whole plane. From the Liouville’s theorem f(z) and p(z) should be constant becoming in contradiction with the fact that p(z) is a polynomial. In conclusion p(z) has at least one zero. According to the division algorithm, the division of the polynomial p(z) by z − b decreases the degree of the quotient q(z) by a unity, and yields a complex number r as remainder: p(z) = (z − b) q(z) + r The substitution of z by b gives: p(b) = r That is, the remainder of the division of a polynomial by z − b is equal to its numerical value for z = b .
If f(x) is bounded we have: f(x)< M The derivative of f(x) is always given by: f' (z ) = f (t ) 1 2 π e 12 ∫ (t − z ) 2 dt C Following the circular path t − z = r exp(e12ϕ ) we have: f' (z ) = 1 2π r 2π ∫ f (r exp(e ϕ )) exp(− e ϕ ) dϕ 12 Using the inequality f' (z ) ≤ 12 0 1 2π r ∫ f ( z ) dz ≤ ∫ 2π ∫ 0 f ( z ) dz , we find: f (r exp(e12ϕ )) dϕ ≤ 1 2π r 2π M ∫ M dϕ = 2π r 0 TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 25 Since the function is analytic in the entire plane, we may take the radius r as large as we wish.
Z + 2z − 8 ∞ 1 and its analytic function. 11Calculate the Lauren series of 2 and the annulus of convergence. 12 Prove that if f(z) is analytic and does not vanish then it is a conformal mapping. TREATISE OF PLANE GEOMETRY THROUGH GEOMETRIC ALGEBRA 27 4. TRANSFORMATIONS OF VECTORS The transformations of vectors are mappings from the vector plane to itself. Those transformations preserving the modulus of vectors, such as rotations and reflections, are called isometries and those which preserve angles between vectors are said to be conformal.