Complex Analysis - Maths KU

7.5 Applications 433
Step 2 From Step 1 we have w = T (z) maps the circle |z| = 1 onto the
horizontal line v = 0, and it maps the circle � �z − 1
�
�
2 = 1
2 onto the horizontal
line v = 1. Thus, the transformed boundaryconditions are Φ = −10 on the
line v = 0 and Φ = 20 on the line v = 1. See Figure 7.50(b).
Step 3 Modeled after Example 2 in Section 3.4 and Problem 12 in Exercises
3.4, a solution of the transformed Dirichlet problem is given by
Φ(u, v) =30v − 10.
Step 4 A solution of the original Dirichlet problem is now obtained bysubstituting
the real and imaginaryparts of T (z) defined in (5) for the variables
u and v in Φ(u, v). Byreplacing the symbol z with x + iy in T (z) and
simplifying we obtain:
Therefore,
x + iy − i
T (x + iy) =(1−i) x + iy − 1
= x2 + y 2 − 2x − 2y +1
(x − 1) 2 + y 2
is the desired electrostatic potential function.
+ i(y − 1) x − 1 − iy
=(1−i)x x − 1+yi x − 1 − iy
+ 1 − x2 − y2 (x − 1) 2 i.
+ y2 φ(x, y) =30 1 − x2 − y2 (x − 1) 2 − 10 (6)
+ y2 A complex potential function for the harmonic function φ(x, y) given by
(6) in Example 2 can be found as follows. If Ω(z) is a complex potential for
φ, then Ω(z) =φ(x, y) +iψ(x, y) and Ω(z) is analytic in D. From Step
4 of Example 2 we have that the complex function T (z) given by(5) has
real and imaginaryparts u = x2 + y2 − 2x − 2y +1
(x − 1) 2 + y2 and v = 1 − x2 − y2 (x − 1) 2 ,
+ y2 respectively. That is, T (z) =u + iv. We also have from Step 4 that φ(x, y) =
30v − 10. In order to obtain a function with 30v − 10 as its real part, we
multiply T (z) by−30i then subtract 10:
−30iT (z) − 10 = −30i(u + iv) − 10 = 30v − 10 − 30ui.
z − i
Since T (z) = (1 − i) is analytic in D, it follows that the function
z − 1
−30iT (z) − 10 is also analytic in D. Therefore,
Ω(z) =−30i (1 − i)
z − i
− 10 (7)
z − 1
is a complex potential function for φ(x, y). Since φ represents the electrostatic
potential, the level curves of the real and imaginaryparts of Ω represent