This article gives the statement, and possibly proof, of a basic fact in complex analysis.
View a complete list of basic facts in complex analysis
Statement
Suppose
is an open subset, and
is complex-differentiable at every point. Suppose
is a triangle lying completely inside
. Then, we have:
Proof
Define:
Note that
can be split into four congruent triangles by joining the midpoints of the sides. Since
is the sum of
evaluated for each of the pieces, at least one of the pieces
must satisfy:
Pick this triangle, and again divide it into four triangles. Repeating this process we get a descending sequence of triangles, each having side-lengths half of its predecessor:
and with the further property that:
On the other hand, we have that:
Now let
be the unique point in the intersection of all the
s. Since
is differentiable at the point
, we have:
Define now:
Then
. Observe that:
Thus, we get:
Applying the length estimates and size estimates now gives the result. Fill this in later