ANS\VERS 615Sect ion 10. 1. Basic Properties of Conformal Mappings: page 401
la. All z.
le. All z except z = - 1.l e. All z except z = 0.3. f' (1) = l,a = Argf' (1) = 0, If' (l)I = 1;
f' (1 + i) =! -~.a= Argf' (1 + i) = 7 , If' (1 + i)I = Yj.;
!' (i) = -i,a = Argf' (i) =^3 ;, If' (i)I = 1.- f' (~ + i) = -isinh 1, a = Argf' (~ + i) = -; , I f' (j + i) I = sinh 1;
f' (j + i) = -isinhl,a = Argf' (j +i) = - 2 " .If'(~ +i) I = sinh 1;
f' (O) = l,a = Argf' (0) = O,I !' (O)I = 1.
7. I!' (a+ ib)I = 12 v'~+ibl = 2 (a>+t>it•/4l =F 0, hence f (z) is conformal at
z = a+ ib. The lines z 1 (t) = a+ (b + t)i and, z 2 (t) = (a+ t) + ib intersect
orthogonally at the point z 1 (0) = z2(0) = a + ib, therefore, their image
curves will intersect orthogonally at the point v'ii'+ii).9. lf'(a+ib)I = lcos(a+ib)I = Jcos^2 acosh^2 b+sin^2 asinh^2 b =F 0, hence
f (z) is conformal at z =a+ ib. The lines z 1 (t) =a+ ti and z2(t) =a+ t
intersect orthogonally at t he point z 1 (0) = z2(0) =a; therefore, their image
curves will intersect orthogonally at the point sin (a + ib).- First show that the mapping W = Z preserves the magnitude, but reverses
the sense, of angles at Z 0 • Then consider the mapping w = f (z) as a
composition.
Sectio n 10.2. Bilinear Transformations : page 409(^1) • Z -- s- '( W ) -- (l+i)w-2w:'f2 - 1-i) -- (1-i)(l-w) i+w ·
T he disk lwl < 1.
The region lwl > 1.
w = S(z) =-:~ii·
w = S(z) = ;7;.
The disk lwl < 1.
The portion of the disk lwl < 1 that lies in t he upper half-plane Imw > 0.
The region that lies exterior to both the circles lw -! I =! and lw - ~I =!.