\def\Re{\mathop{\rm Re}\nolimits}
\def\Im{\mathop{\rm Im}\nolimits}
\let\rule\hrule
\def\hrule{\medskip\rule\medskip}

%page 70
\noindent{\bf 1.}

\noindent{\it (a)}

If $f(z) = \overline z = x - yi,$ $U_x = 1$ and $V_y = -1.$ $U_x\neq V_y,$
so because the Cauchy-Riemann equations are dissatisfied, the derivative
does not exist.

\noindent{\it (b)}

If $f(z) = z-\overline z = x + yi - (x - yi) = 2yi,$ $U_x = 0,$ and $V_y
= 2i.$ Again, $U_x\neq V_y.$

\noindent{\it (c)}

If $f(z) = 2x+ixy^2,$ 

$U_x = 2.$ $V_y = 2ix.$
$U_x = V_y \to 2 = 2ix \to x = 1/i = -i.$ $x\in R,$ so this is false,
and the function is not analytic.

\noindent{\it (d)}

If $f(z) = e^xe^{-iy},$ 

$$U_x = e^xe^{-iy}.$$
$$V_y = -ie^xe^{-iy} \neq U_x.$$

\noindent{\bf 3.}
% only info from secs 21 and 23

\noindent{\it (a)}

$f(z) = 1/z = \overline z/|z|^2 = (a-bi)/(a^2+b^2).$

The partial derivatives are $U_x = (y^2-x^2)/(x^2+y^2)^2,$
$V_y = (y^2-x^2)/(x^2+y^2)^2,$ $U_y = (-2xy)/(x^2+y^2)^2,$ and $V_x =
(2xy)/(x^2+y^2)^2.$ These satisfy the Cauchy-Riemann equations except
where they are undefined, $x^2+y^2 = 0 \to z = 0.$ Using the power rule
on $f(z) = z^{-1},$ $f'(z) = -z^{-2}.$

\noindent{\it (b)}

$f(z) = x^2+iy^2.$

The partial derivatives are $U_x = 2x,$ $U_y = 0,$ $V_y = 2y,$ and $V_x
= 0.$ $0 = 0$ at all points, but $2y = 2x$ only for $x = y$ or
$z = x+ix.$

Its derivative is, from these partial derivatives, $2x.$

\noindent{\it (c)}

$f(z) = z\Im z = (x+iy)y = xy + iy^2.$

$U_x = y = V_y = 2y \to y = 0.$ $U_y = x = -V_x = 0 \to x = 0.$
Therefore, $z = 0 + i0 = 0.$ From these partial derivatives, its
derivative on this domain is $0.$

\hrule
%page 76
\noindent{\bf 4.}

\noindent{\it (a)}

$$f(z) = {2z+1\over z(z^2+1)}$$

$f(z)$ is not analytic on $z=0$ or $z^2+1=0 \to z = \pm i.$ It is,
however, analytic on the remainder of the plane because, using the chain
rule and the product rule, its derivative can be constructed from its
component polynomials (which are entire functions).

\noindent{\it (b)}

$$f(z) = {z^3+i\over z^2-3z+2}$$

This is not analytic on $z^2-3z+2 = 0 = (z-1)(z-2) \to z = 1,2.$ It is
analytic on the remainder of the plane for the same reason as (a).

\noindent{\it (c)}

$$f(z) = {z^2+1\over (z+2)(z^2+2z+2)}.$$

This is not analytic on $z+2 = 0 \to z = -2$ or $z^2+2z+2 = 0 =
(z+(1-i))(z+(1+i)) \to z = 1\pm i.$ It is analytic for the same reason
as (a).

\hrule
%page 79
\noindent{\bf 4.}

See addendum.

\hrule
%page 89
\noindent{\bf 2.}

$$f(z) = 2z^2 - 3 - ze^z + e^{-z}.$$

If the component functions $f$ and $g$ are analytic on the plane (have a
defined derivative), $f+g$ is also analytic on the plane.
A similar rule exists for the product of entire functions.
$2z^2$ follows the power rule, and has a derivative over the entire
plane, and $-3$ follows the constant rule, having a derivative of 0.
$-z$ is entire, also from the power rule, and $e^z$ has been shown to
have a derivative in chapter 3. Using the product rule, $-ze^z$ is
entire.
$e^{-z}$ is a composition of $e^z$ and $-z,$ which each are analytic
over the plane, so their composition is also analytic over the plane.

This means $f(z)$ is entire.

\noindent{\bf 8.}

\noindent{\it (a)}

$e^z = -2.$ $e^x = |z| = |-2| = 2 \to x = \ln 2,$ and
$\arg z = \pi + 2n\pi,$ so $z = \ln2 + i(\pi+2n\pi).$

\noindent{\it (b)}

$e^z = 1 + i.$ $e^x = |z| = |1+i| = \sqrt 2 \to x = (1/2)\ln2,$ and
$\arg z = \pi/4 + 2n\pi.$ Therefore, $z = (1/2)\ln2 + i(\pi/4+2n\pi).$

\noindent{\it (c)}

If $\exp(2z-1) = 1,$ $\exp(2z)e^{-1} = 1 \to e^{2z} = e.$ This gives
$e^{2x} = |e| = e \to 2x = 1 \to x = 1/2,$ and $\arg(2z) = 2n\pi \to
\arg z = n\pi,$ so $z = 1/2 + in\pi.$

\hrule
%page 95
\noindent{\bf 2.}

\noindent{\it (a)}

$$\log e = 1 + 2n\pi i$$

$\ln e = 1,$ and $\arg e = 2n\pi.$ This proves the above identity (with
$n = 0,\pm1,\pm2,\ldots$)

\noindent{\it (b)}

$$\log i = \left(2n + {1\over2}\right)\pi i$$

$\ln|i| = \ln1 = 0,$ and $\arg i = \pi/2 + 2n\pi.$ This makes $\log i =
(2n+1/2)\pi i.$

\noindent{\it (c)}

$$\log(-1+\sqrt3i) = \ln2 + 2\left(n + {1\over3}\right)\pi i$$

$\ln|-1+\sqrt3i| = \ln2,$ and $\arg(-1+\sqrt3i) = 2\pi/3 + 2n\pi,$ which
corresponds to the given identity ($\log(-1+\sqrt3i) = (2n\pi+2\pi/3)i +
\ln2.$)

\bye
