- The triangle with vertices - 5 - 2i, -6, and 3 + 2i.
15a. w = f (z) =^3 tJi z +^7 if.
15 c. w = f (z) = !z+^7 ~^4 i.
ANSWERS 587- Let f (z) = Az + B and g ( z) = C z + E be two linear transformations. Then
h(z) = j (g(z)) = f (Cz+E) = A(Cz +E) +B = ACz+ (E+B), which
is the required form for a linear transformation.Sectio n 2.2. The Mappings w = zn and w = z~: page 69
la. Using Equation (2-9) we see that, if A = {(x, y): y = 1}, then f (A)
{ ( u, v) : u = x^2 - 1, v = 2x} = { ( u, v) : u = ~ -1}.l e. The region in the upper half-plane Im (w) > 0 that lies between the parabolas
4
.,2 .,2
u = - 16 and u = 4 - 1.l e. The point (x, y) in the xy-plane is mapped to the point (u, v) = (x^2 - y^2 ,
2 2
2xy). For any x, u = x^2 - ~· Ifx = 1, then u = 1 - " 4. Ifx = 2, then u =
24-~ 6 • Your only remaining task is to show that the strip {(x, y): 1 < x < 2}
indeed is mapped between these two parabolas.lg. The infinite strip {(u, v): 1 < v < 2}, which is the region in the uv plane
between v = i and v = 2i. Show the details in a manner similar to the answer
for part a.3a. The points that lie to the ext reme right or left of the branches of the hyper-bola x^2 - y^2 = 4.
2- The region in thew plane that lies to the right of the parabola u = 4 - r 6 •
7a. The set {re;e: r > 8, and^3 .;' < 8 < 7T}.7c. The set {re'^9 : r > 64 , and^3 ; < 8 < 27T}.
- See also problem 2. T he fallacy lies in the assumption implicit in the second
equality that y1ziZ2 = Fi ..fZ2 for all complex numbers z1 and z2. Assuming
~ .Arg(.i1.i2)
the principal square root is used, then ylziZ2 = jz 1 z2I e• 2. This will
equal ./Zi.../Z2 = lzd! ei~ lz2I! ei~ precisely when Arg(z1z2) =
Arg (zi} + Arg (z2)-explain! The latter equality is plainly false when z 1 =
z 2 = -1. (Again, explain.) To give a very thorough answer to this problem,
you should state precisely when the last equality is true, and justify your
assertion.