Complex Analysis - Maths KU

4
2
y
–4
–2 2 4
–2
–4
Figure 2.63 Streamlines in the planar
flow associated with f(z) =z 2
x
2.7 Applications 137
To find a function F for which ∂F/∂x = M and ∂F/∂y = N, we first partially
integrate the function M(x, y) =2xy with respect to the variable x:
�
F (x, y) = 2xy dx = x 2 y + g(y).
The function g(y) is then determined bytaking the partial derivative of F with
respect to the variable y and setting this expression equal to N(x, y) =x 2 −y 2 :
∂F
∂y = x2 + g ′ (y) =x 2 − y 2 .
This implies that g ′ (y) =−y 2 , and so we can take g(y) =− 1
3 y3 . In conclusion,
F (x, y) =x 2 y − 1
3 y3 = c is an implicit solution of the differential equation in
(5), and so the streamlines of the planar flow associated with f(z) =¯z 2 are
given by:
x 2 y − 1
3 y3 = c
where c is a real constant. In Figure 2.63, Mathematica has been used to plot
the streamlines corresponding to c = ± 2 16
3 , ± 3 , ±18. These streamlines are
shown in black superimposed over the plot of the normalized vector field for
the flow.
EXERCISES 2.7 Answers to selected odd-numbered problems begin on page ANS-11.
In Problems 1–8, (a) plot the images of the complex numbers z =1,1+i, 1−i, and
i under the given function f as position vectors, and (b) plot the images as vectors
in the vector field associated with f.
1. f(z) =2z− i 2. f(z) =z 3
3. f(z) =1 − z 2 4. f(z) = 1
z
5. f(z) =z − 1
z
6. f(z) = z 1/2 , the principal square root function
given by (7) of Section 2.4
7. f(z) = 1
8. f(z) = loge |z| + iArg(z)
¯z
In Problems 9–12, (a) find the streamlines of the planar flow associated with the
given complex function f and (b) sketch the streamlines.
9. f(z) =1− 2i 10. f(z) = 1
¯z
11. f(z) =iz 12. f(z) =(1+i)¯z
Focus on Concepts
13. Let f be a complex function. Explain the relationship between the vector
field associated with f(z) and the vector field associated with g(z) =f(z − 1).
Illustrate with sketches using a simple function for f.