The Real And Complex Number Systems

and they hold for all z 0if is a negative integer.)
Solution: It is clear from definition of differentiability.
(iii) Compute the derivative f z in (a), (b), (f), (g), (h), assuming it exists.
Solution: Sincef z u x iv x , if it exists. So, we know all results by (ii).
5.37 Write f u iv and assume that f has a derivative at each point of an open disk D
centered at 0, 0. Ifau 2 bv 2 is constant on D for some real a and b, not both 0. Prove
that f is constant on D.
Proof: Letau 2 bv 2 be constant on D. We consider three cases as follows.
1. As a 0, b 0, then we have
v 2 is constant on D
which implies that
vv x 0.
If v 0onD, it is clear that f is constant.
If v 0onD, that is v x 0onD. So, we still have f is contant.
2. As a 0, b 0, then it is similar. We omit it.
3. As a 0, b 0, Taking partial derivatives we find
auu x bvv x 0onD. 1
and
auu y bvv y 0onD.
By Cauchy-Riemann equations the second equation can be written as we have
auv x bvu x 0onD. 2
We consider 1v x 2u x and 1u x 2v x , then we have
bvv x2 u x2 0 3
and
auv x2 u x2 0 4
which imply that
au 2 bv 2 v x2 u x2 0. 5
If au 2 bv 2 c, constant on D, wherec 0, then v x2 u x2 0. So, f is constant.
If au 2 bv 2 c, constant on D, wherec 0, then if there exists x, y such that
v x2 u x2 0, then by (3) and (4), ux, y vx, y 0. By continuity of v x2 u x2 , we know
that there exists an open region S D such that u v 0onS. Hence, by Uniqueness
Theorem, we know that f is constant.
Remark: In complex theory, the Uniqueness theorem is fundamental and important.
The reader can see this from the book named Complex Analysis by Joseph Bak and
Donald J. Newman.