By V. B. Alekseev

Translated by way of Sujit Nair

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**Example text**

A + b) = ϕ(a) + ϕ(b) and ϕ(ab) = ϕ(a)ϕ(b). Two fields are called isomorphic if its possible to establish an isomorphism between them. If in a field only the operations of addition and multiplication considered, then all isomorphic fields have identical properties. Therefore, just as in the case of groups, isomorphic fields cannot be distinguished. As we saw in the previous paragraph, in the field of the complex numbers there is only element i such that i2 = −1. The following problem shows that the addition of this element to the field of real numbers leads to the field of complex numbers.

Let the point z0 and an arbitrary real number ε > 0 be given. We have to choose this real number δ > 0 such that for all numbers z which satisfy the condition |z − z0 | < δ the inequality ε |f (z) − f (z0 )| = |2z − 2z0 | < ε is satisfied. It is not difficult to see that it is possible to choose δ = 2 (independent of point z0 ). , |2z − 2z0 | < ε. Therefore, the function f (z) = 2z is continuous at any point z0 . In particular, it is continuous for all real values of the argument z. Therefore, if we limit ourselves to only real valued arguments, we see that the function with real argument f (x) = 2x is continuous for all real values x.

Using this angle, it is easy to obtain the assertion of theorem 6: it suffices to consider ϕ(t) = ϕ0 + ϕ1 (t) where ϕ1 (t) is the angle swept by the part of this curve from z(0) to z(t). 7 Continuous curves 53 1 O 1 O x x -2i -2i Fig. 6. O 1 O 1 Fig. 7. Fig. 8.

### Abel’s Theorem in Problems & Solutions by V. B. Alekseev

by Brian

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