10.4  Mapping Involving Trigonometric Functions

    The trigonometric functions can be expressed with compositions that involve the exponential function followed by a bilinear function. We can find images of certain regions by following the shapes of successive images in the composite mapping.

Example 10.12.  The transformation    is a one-to-one conformal mapping of the vertical strip    onto the disk  .  

Solution.  Recall Equations (5-32) and (5-34), which are complex trigonometric identities:

                 and     .

Then we can write  ,  and we use the formula  

            .  

Then, mapping can be considered to be the composition  

                 and       .  

The transformation    maps the vertical strip    one-to-one and onto the right half-plane  .  Then the bilinear transformation    maps the right half-plane one-to-one and onto the disk, as shown in Figure 10.16.

Figure 10.16  The composite transformation  .

Explore Solution 10.12.

Example 10.13.  Show that the transformation    is a one-to-one conformal mapping of the vertical  strip    onto the w-plane slit along the rays    and  .  

Solution.  Because    for values of z satisfying the inequality  ,  it follows that    is a conformal mapping.  Using Equation (5-33), we write  

            .  

If  ,  then the image of the vertical line    is the curve in the w plane given by the parametric equations  

            ,

for  .  Next, we rewrite these equations as  

            .  

We now eliminate y from these equations by squaring and using the hyperbolic identity  .  The result is the single equation  

(10-23)        .  

The curve given by Equation (10-23) is identified as a hyperbola in the uv plane that has foci at the points .  Therefore the vertical line    is mapped one-to-one onto the branch of the hyperbola given by Equation (10-23) that passes through the point .  If  ,  then it is the right branch;  if  ,  it is the left branch.  The image of the y axis, which is the line  ,  is the v axis.  The images of several vertical lines are shown in Figure 10.17(a).

Figure 10.17  The transformation  .

The image of the horizontal segment    is the curve in the w plane given by the parametric equations  

            ,

for  .  We rewrite them as  

            .  

We now eliminate x from the equations by squaring and using the trigonometric identity  .  The result is the single equation

(10-24)        .  

The curve given by Equation (10-24) is identified as an ellipse in the uv plane that passes through the points    and   and has foci at the points .  Therefore, if  ,  then  ,  and the image of the horizontal segment is the portion of the ellipse given by Equation (10-24) that lies in the upper half-plane  .  If  ,  then  ,  and the image of the horizontal segment is the portion of the ellipse given by Equation (10-24) that lies in the lower half-plane  .  The images of several segments are shown in Figure 10.17(b).

Explore Solution 10.13.

10.4.1  The Complex Arcsine Function

    We now develop explicit formulas for the real and imaginary parts of the principal value of the arcsine function  .  We use this mapping to solve problems involving steady temperatures and ideal fluid flow in Section 10.7. The mapping is found by solving the equation  

(10-25)        ,

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  

            .  

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

            .  

If we treat u as a constant, this equation represents a hyperbola in the xy plane, the foci occur at the points  ,  and the transverse axis is given by  .  Therefore a point on the hyperbola must satisfy the equation  

            .  

The quantity on the right side of this equation is the difference of the distances from   to   and from   to .  We now solve the equation for u to obtain the real part:
            
(10-26)        .  

The principal branch of the real function    is used in Equation (10-26), where the range values satisfy the inequality  .

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

              

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

            .  

If we treat v as a constant, then this equation represents an ellipse in the xy plane, the foci occur at the points  ,  and the major axis has length .  Therefore a point on this ellipse must satisfy the equation  

            .  

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

    The function    maps points in the upper half (lower half) of the vertical strip     onto the upper half-plane (lower half-plane), respectively.  Hence we can solve the preceding equation and obtain v as a function of x and y:  

(10-27)        ,  

where , and .  The real function given by    with t is used in Equation (10-27).

    Therefore the mapping    is a one-to-one conformal mapping of the z plane cut along the rays  ,  and  ,  onto the vertical strip    in the w plane, which can be construed from Figure 10.17 if we interchange the roles of the z and w planes.  The image of the square  ,  under  , is shown in Figure 10.18.  We obtained it by plotting the two families of curves    and  ,  where  .  

Figure 10.18  The mapping   .  

Extra Example 1.  The transformation    is a one-to-one conformal mapping of the first quadrant    onto the semi-infinite strip  .  

Explore Solution for Extra Example 1.

The formulas in Equations (10-26) and (10-27) are also convenient for evaluating  , as shown in Example 10.14.

Example 10.14.  Find the principal value of  .  

Solution.  Using Formulas (10-26) and (10-27), we get  

              

Evaluating these equations at the point  ,  we get  

               

Explore Solution 10.14.

    Is there any reason to assume that there exists a conformal mapping for some specified domain D onto another domain G?  The theorem concerning the existence of conformal mappings is attributed to Riemann and can be found in the book by Lars V. Ahlfors Complex Analysis (New York: McGraw-Hill Book Co.) Chapter 6, 1966.

Theorem 10.4 (Riemann Mapping Theorem).  If D is any simply connected domain in the plane (other than the entire plane itself), then there exists a one-to-one conformal mapping    that maps D onto the unit disk  .

Proof.