\def\arg{\mathop{\rm arg}\nolimits}
\let\rule\hrule
\def\hrule{\goodbreak\medskip\rule\medskip\goodbreak}

%page 317
\noindent{\bf 1.}

$$w = {i-x\over i+x} = {(i-x)^2\over -1 - x^2} = -{-1 - 2ix + x^2 \over
1+x^2} = {1-x^2\over 1+x^2} + i{2x\over 1+x^2}.$$
$y=0,$ the boundary of the preimage in Figure 13 transforms into 

$$|w| = {\sqrt{(1-x^2)^2+4x^2}\over 1+x^2} =
{\sqrt{1-2x^2+x^4+4x^2}\over 1+x^2} = 1.$$

\noindent{\bf 7.}

Take $$\rho = \ln|Z|, \quad 0< \Theta = \arg(Z) \leq 2\pi.$$
$$w = \rho + i\Theta.$$

\hrule
%page 330
\noindent{\bf 1.}

With constant $y_1,$
$$u = x^2 - y_1^2, \quad v = 2xy_1.$$
$$u = (v/2y_1)^2 - y_1^2 \to 4y_1^2(u+y_1^2) = v^2,$$
assuming $y_1>0.$

\noindent{\bf 4.}

$y=0$ gives $u = x_1^2, v = 0,$ and $v(y) = -v(-y), u(y) = u(-y),$
telling us that $x_1 = 1$ transforms into a complete parabolic region
with form $v^2 = -4(u-1).$

$x=\pm y$ similarly gives
$$u = x^2 - x^2, \quad v = \pm 2x^2 \to u = 0, \quad v\in {\bf R,}$$
which at the given boundary points has $A = 0 \mapsto A' = 0,$ $C =
1\mapsto C' = 1,$ $D = 1+i \mapsto D' = 2,$ and $B' = 1-i\mapsto -2,$
under the $w = z^2$ transformation.
This completes the boundary of the region, which is interior to the
boundary.

\hrule
%page 352
\noindent{\bf 1.}

$${dw\over dz} = 2z = 2(2+i)$$
at $z_0 = 2+i,$ giving an angle of rotation $\arg 2+i = \tan^{-1}(1/2)$
and scale factor $|2(2+i)| = 2\sqrt{5}.$
This can be demonstrated with $y=1,$ passing through $z_0$ at angle
$\theta = 0,$ transformed into
$$(x+iy)^2 = (x+i)^2 = (t+2+i)^2 = t^2 + 2(1+i)t + 2,$$
which, since $x(0) = z_0,$ gives an angle equal to the argument of
$2(1+i),$ which is $\pi/4.$

\noindent{\bf 4.}

The transformation $w=z^n$ gives
$${dw\over dz} = nz^{n-1},$$
meaning that $\rho_0\exp(i\theta_0)$ gives angle of rotation
$\arg(n\rho_0^{n-1}\exp(i(n-1)\theta_0)) = (n-1)\theta_0$ and scale
factor $|n\rho_0^{n-1}\exp(i(n-1)\theta_0)| = n\rho_0^{n-1}.$

\hrule
%page 357
\noindent{\bf 1.}

\noindent{\it (a)}

$$u(x, y) = 2x - x^3 + 3xy^2.$$

This is harmonic because $$u_{xx} + u_{yy} = - 6x + 6x = 0$$ on all
$x\in{\bf R}.$

$$v_y = u_x = 2 + 3y^2 \to v(x,y) = 2y + y^3 + g(x).$$

$$g'(x) = v_x = -u_y = -6xy = \to v(x,y) = 2y + y^3 - 3x^2y+C.$$

\noindent{\it (b)}

$$u(x,y) = \sinh x \sin y$$

This is harmonic because $$u_{xx} + u_{yy} =
\sinh x\sin y - \sinh x\sin y = 0.$$

$$v_y = u_x = \cosh x\sin y \to v(x,y) = -\cosh x\cos y + g(x).$$
$$g'(x) = v_x = -u_y = -\sinh x\cos y \to v(x,y) = -\cosh x\cos y -
\cosh x\cos y + C = C - 2\cosh x\cos y.$$

\noindent{\it (c)}

$$u(x,y) = {y\over x^2+y^2}$$

This is harmonic because $$u_{xx} + u_{yy} = 0.$$

$$v_y = u_x = {-2yx \over (x^2+y^2)^2} \to v = {x\over x^2+y^2} + g(x).$$

$$g'(x) = v_x = -u_y = -{x^2+y^2 - 2y^2\over (x^2+y^2)^2} \to g(x) =
{x\over x^2+y^2} + C.$$

% also read sec 127-128, 129 (question unnecessary)

\bye