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**Example text**

The polynomial P (z) = z 2 + z + 1 has integer coefficients, so p = ±1 and q = 1, and z = ±1 are the only possible rational roots. A simple check shows that P (1) = 3 = 0 and P (−1) = 1 = 0, so P (z) does not have rational roots. Using the solution formula for equations of second degree we find that the roots are the complex conjugated √ 3 1 . com 32 The Complex Numbers Elementary Analytic Functions Procedure of finding rational roots. 1) Check that the polynomial (28) has (real) integer coefficients a0 , a1 , .

F (z) − f (z0 )| < ε. This is true for every z0 ∈ Ω and every ε > 0, so the claim is proved. 2, so the two definitions are equivalent. 2 is far the best one to use in theoretical consideration, so the two definitions are supplementary to each other. 1 Let Ω and ω be two open and nonempty subsets of C, and let f : Ω → ω and g : ω → C be two continuous maps. Then the composition g ◦ f : Ω → C is also continuous. Proof. 2, where an application of the equivalent (43) would become very messy in comparison.

K. By taking the limit ∆xj → 0 it follows that g1 , . . , gp at x have partial derivatives with respect to ∂f −1 xj , and that the coefficient matrix of by this limit process is − fy (x, g(x)) of continuous ∂xj elements. Hence (58) ∂g = − fy (x, g(x)) ∂xj where we consider (59) −1 ∂f , ∂xj ∂f ∂g and as columns.