1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

10.4 • MAPPING BY TRIGONOMETRIC FUNCTIONS 421

for u and v expressed as functions of x and y. To solve for u, we first equate the

real and imaginary parts of Equation (10- 25 ) and obtain the equations

coshv = ~ and sinhv = Y.
smu cosu

Then we eliminate v from these equations and obtain the single equation

x2 y2

-----=!.

sin^2 u cos^2 u

If we treat u as a constant, this equation represents a hyperbola in the xy plane,
the foci occur at the points (±1,0), and the transverse axis is given by 2sinu.
Therefore, a point (x, y) on the hyperbola must satisfy the equation

The quantity on t he right side of this equation is the difference of the distances

from (x, y) to (- 1, 0) and from (x, y) to (1, 0). We now solve the equation for u

to obtain the real part:

. [V(x+l)

2

+y2- J(x-1)

2

+y2]

u(x,y)=Arcsm

2

. (10-26)

The principal branch of the real function Arcsin tis used in Equation (10-26),

where the range values satisfy the inequality - 2 " < Arcsin t < ~.

Similarly, we can start with Equation (10-25) and obtain the equations

smu=. --x an d

coshv

y

cosu = sinhv ·

We then eliminate u from these equations and obtain the single equation

x2 y2

--+--=l.

cosh^2 v sinh^2 v

If we treat v as a constant, then this equation represents an ellipse in the xy

plane, the foci occur at the points (±1, 0), and the major axis has length 2 cosh v.

Therefore, a point (x, y) on this ellipse must satisfy the equation

2coshv = V(x + 1)^2 +y2 + Jcx-1)^2 +y2.

The quantity on the right side of this equation is the sum of the distances from

(x, y) to (-1, 0) and from (x, y) to (1, 0).

The function z = sin w maps points in the upper half (lower half) of the

vertical strip - 2 " < u < ~ onto the upper half-plane (lower half-plane), respec-

tively. Hence we can solve the preceding equation and obtain v as a function of