the two functions
u(x, Y) =
x2 - y2
and v ( x , y ) =
(x2 y 2 p
are harmonic throughout any domain in the xy plane that does not contain the origin.
If two given functions u and v are harmonic in a domain L) and their first-order
where grad u is the gradient vector
grad u = u,i
+ u, j.
Because u, and u, are zero everywhere in D, then, grad u is the zero vector at all
points on L. Hence it follows from equation (1) that the derivative d u l d s is zero along
L; and this
and 0 exist everywhere in that neighborhood. Ifthose partial derivatives are continuous
at (ro, 6,) and satisfy the polar form
o the Cauchy-Riemann equations at (ro, go), then f'(zo) exists.
The derivative f '(zo) here can b
( A X Ay)+(O,O)
v(x0. yo+ AY) - ~
( ~ YO)
= vy (xo, YO)
I m p -(Ar,Ay)+(O,O) A2
~ ( ~ Yo 7
~ -)U ( ~ 0YO)
= -uy ( ~ 0 9
Hence it follows from expression (3) that
respectively, it also shows that the real and imaginary components of a function of a
complex variable can have continuous partial derivatives of all orders at a point and
yet the function may not be differentiable there.
The function f ( 2 ) = 1z12 is co
0 < I (x
+ i y ) - (xo + YO) 1 < 6.
That is, limit (1) holds.
Let us now start with the assumption that limit (1) holds. With that assumption,
we know that, for each positive number E ,there is a positive number 6 such that
(See Fig. 27.) But the continuity of f at z0 ensures that the neighborhood lz - zol < 6
can be made small enough that the second of these inequalities holds. The continuity
of the composition g [ f (z)] is, therefore, established.
Our final example here uses the images of horizontal lines to find the image of a
EXAMPLE 3. When w = eZ,the image of the infinite strip 0 5 y 5 rr is the upper
half v 2 0 of the w plane (Fig. 22). This is s
EXAMPLE 1. According to Example 2 in Sec. 11, the mapping w = z2 can be
thought of as the transformation
from the xy plane to the u v plane. This form of the mapping is especially useful in
finding the images of certain hyperbolas.
5. Let S be the open set consisting of all points z such that lzl < 1 or lz - 21 < 1. State why
S is not connected.
6. Show that a set S is open if and only if each point in S is an interior point.
7. Determine the accumulation
EXAMPLE 3. The two values q (k = 0, 1) of
roots of & i, are found by writing
(J?+ i)1/2, which are the square
and (see Fig. 14)
Euler's formula (Sec. 6) tells us that
GO = h e x p ( i k )
= A (cos
- + i sin-
PRODUCTS QUOTIENTS EXPONENTIAL
geometrically obvious that
and e -i4n - 1.
e -in/2 - -i,
Note, too, that the equation
is a parametric representation of the circle )zl = R, centered at the origin with radius
R. As the para
and [see Exercise 2(b)]
10. Establish the identity
and then use it to derive Lagrange's trigonometric identity:
Suggestion: As for the first identity, write S = 1 + z z2 + .
zn and consider
the difference S - zS. To derive the second identity,
So the conjugate of the sum is the sum of the conjugates:
In like manner, it is easy to show that
The sum z
of a complex number z = x iy and its conjugate Z = x - iy is
the real number 2x, and the difference z - z is the