The Real And Complex Number Systems

D 2,2 f0, 0 lim
y0
D 2 f0, y D 2 f0, 0
y 0
lim
y0
0 y 0.
Remark: We do not give a detail computation, but here are answers. Leave to the
reader as a practice. For x, y 0, 0, wehave
D 1,1 fx, y 4y3 y 2 3x 2
x 2 y 2 3 ,
and
complex-valued functions
D 1,2 fx, y 4xy2 3x 2 y 2
x 2 y 2 3 ,
D 2,1 fx, y 4xy2 3x 2 y 2
x 2 y 2 3 ,
D 2,2 fx, y 4x2 yy 2 3x 2
x 2 y 2 3 .
5.35 Let S be an open set in C and let S be the set of complex conjugates z, where
z S. Iff is defined on S, defineg on S as follows: gz fz, the complex conjugate
of fz. Iff is differentiable at c, prove that g is differentiable at c and that g c f c.
Proof: Sincec S, we know that c is an interior point. Thus, it is clear that c is also an
interior point of S . Note that we have
the conjugate of
fz fc
z c
z
f fc
z c
gz gc
by gz fz.
z c
Note that z c z c, so we know that if f is differentiable at c, prove that g is
differentiable at c and that g c f c.
5.36 (i) In each of the following examples write f u iv and find explicit formulas
for ux, y and vx, y : ( These functions are to be defined as indicated in Charpter 1.)
(a) fz sin z,
Solution: Since e iz cosz i sin z, we know that
sin z 1 2 ey e y sin x ie y e y cosx
from sin z eiz e iz
2i
and
. So, we have
ux, y ey e y sin x
2
vx, y ey e y cosx
.
2
(b) fz cosz,
Solution: Sincee iz cosz i sin z, we know that
cosz 1 2 ey e y cosx e y e y sin x
from cos z eiz e iz
2
. So, we have