§3. First concepts of analytic functions
169
Corollary 1 (addition formula for the exponential series).
(22.5)
exp(x). exp(y) = exp(x + y),
for all x,y E C.
Since in general exp(z) = L z[n1, the term Wn of (3) is then indeed equal
to the sum of the products x[Ply[ql for all pairs of integers p, q ?: 0 such that
p + q = nj but this is just (x + y)[nl by the binomial formula of nO Ij since
there is no problem of convergence, the result follows. In fact, the corollary
consists of passing to the limit over n in the formula
which follows from the binomial formula.
Corollary 2. The product of two functions J and 9 defined and analytic in
an open subset G oj C is analytic in G, and
(lg)' = J'g + Jg'.
The given functions J and 9 are, by hypothesis, represented on a neighbourhood of any a E G by power series in z - aj so their product is too.
To show that the derivative of J 9 is given by the "obvious" formula -
one could establish this directly, and also for the holomorphic functions, on
defining the derivatives by passage to the limit -, one remarks that on a
neighbourhood of a z E G, J and 9 are given by Taylor's formula (19.6)
J(z+h)
g(z+h)
J(z) + J'(z)h + J"(z)h 2 /2! + ... ,
g(z) + g'(z)h + g"(z)h 2 /2! + ... .
Since these power series in h converge and represent the left hand sides for
Ihl sufficiently small, one has
J(z + h)g(z + h) = eo(z) + cl(z)h + c2(z)h 2 /2! + ...
where the coefficients cn(z) are given by Theorem 16. In particular,
Cl(Z) = J(z)g'(z) + J'(z)g(z),
and since Cl (z) must be the derivative at the point z of the function J 9 by
Taylor's formula for the latter, the result follows.
Of course the reader would like to know if the quotient of two analytic
functions is again analytic or, what comes to the same by the preceding
theorem, if the recipocal 1/ J(z) of an analytic function is again analytic.
This is so, provided, as always, that one does not divide by O.
169
Corollary 1 (addition formula for the exponential series).
(22.5)
exp(x). exp(y) = exp(x + y),
for all x,y E C.
Since in general exp(z) = L z[n1, the term Wn of (3) is then indeed equal
to the sum of the products x[Ply[ql for all pairs of integers p, q ?: 0 such that
p + q = nj but this is just (x + y)[nl by the binomial formula of nO Ij since
there is no problem of convergence, the result follows. In fact, the corollary
consists of passing to the limit over n in the formula
which follows from the binomial formula.
Corollary 2. The product of two functions J and 9 defined and analytic in
an open subset G oj C is analytic in G, and
(lg)' = J'g + Jg'.
The given functions J and 9 are, by hypothesis, represented on a neighbourhood of any a E G by power series in z - aj so their product is too.
To show that the derivative of J 9 is given by the "obvious" formula -
one could establish this directly, and also for the holomorphic functions, on
defining the derivatives by passage to the limit -, one remarks that on a
neighbourhood of a z E G, J and 9 are given by Taylor's formula (19.6)
J(z+h)
g(z+h)
J(z) + J'(z)h + J"(z)h 2 /2! + ... ,
g(z) + g'(z)h + g"(z)h 2 /2! + ... .
Since these power series in h converge and represent the left hand sides for
Ihl sufficiently small, one has
J(z + h)g(z + h) = eo(z) + cl(z)h + c2(z)h 2 /2! + ...
where the coefficients cn(z) are given by Theorem 16. In particular,
Cl(Z) = J(z)g'(z) + J'(z)g(z),
and since Cl (z) must be the derivative at the point z of the function J 9 by
Taylor's formula for the latter, the result follows.
Of course the reader would like to know if the quotient of two analytic
functions is again analytic or, what comes to the same by the preceding
theorem, if the recipocal 1/ J(z) of an analytic function is again analytic.
This is so, provided, as always, that one does not divide by O.