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\textbf{Math 425: Complex Variables}\hfill \textbf{Exam I Solutions}

\bigskip

\begin{enumerate}

\item
\begin{itemize}

\item[(a)] A rotation of the plane obviously preserves distances and
angles, and so takes circles to circles and lines to lines. Let $C$ be
a circle with center $z_0$, and let $\theta = \arg(z_0)$ so that
$e^{-i\theta}z_0 = |z_0|$. Then the rotation $R(z) = e^{-i\theta}z$
takes $C$ to a circle $R(C)$ with center $R(z_0) = |z_0|$ on the positive
real axis.

Now let $C$ be a line and let $z_0$ be the point on $C$ closest to the
origin.  Note that the line through 0 and $z_0$ is perpendicular to $C$,
so when we rotate the plane by $R(z) = e^{-i\theta}z$, where $\theta =
\arg(z_0)$, we see that the line through $R(0) = 0$ and $R(z_0) = |z_0|$,
i.e. the real axis, is perpendicular to the line $R(C)$. This proves
that $R(C)$ is a vertical line passing through the real axis at the
point $|z_0|$. (If $|z_0| = 0\in C$, then it is clear we may rotate $C$ so
that it becomes the vertical line $x=0$.)

Finally, for any rotation $R(z) = e^{i\theta}$, we find 
\[R(I(R(z))) = R(I(e^{i\theta}z)) = R\Big(\frac{1}{e^{i\theta}z}\Big) =
\frac{e^{i\theta}}{e^{i\theta}z} = \frac{1}{z} = I(z)\]

\item[(b)] Let $C$ be the circle $\{c + r e^{it}\,|\, 0\le t\le 2\pi\}$
with center $c>0$ and radius $r \ne c$. To see that $I(C)$ is a circle with
center $c_1 = c/(c^2-r^2)$  and radius $r_1 = r/|c^2 - r^2|$, we must show
$|I(z) - c_1| = r_1$ for all $z\in C$:
\[ |I(z) - c_1|
= \left| \frac{1}{c + re^{it}} - \frac{c}{c^2 - r^2}\right|
= \left| \frac{-r(r + ce^{it})}{(c + re^{it})(c^2 - r^2)}\right|
= \frac{r}{|c^2 - r^2|}\left| \frac{r + ce^{it}}{c + re^{it}}\right|
= \frac{r}{|c^2 - r^2|}
\]
The last step follows from
\[\left|\frac{r + ce^{it}}{c + re^{it}}\right|
= |e^{it}| \frac{|c + re^{-it}|}{|c + re^{it}|} = 1
\]
because $|e^{it}|=1$ and the numerator $c+re^{-it}$ is the conjugate of
the denominator $c+re^{it}$ ($|\overline{z}|=|z|$).

If $C$ is a vertical line, $x = c>0$, then $C = \{ c + iy \,|\, y\in\R\}$.
To see that $I(C)$ is a circle with center and radius $1/(2c)$ we must show
$|I(z) - 1/(2c)| = 1/(2c)$ for all $z\in C$:
%\begin{eqnarray*}
\[\left|I(z) - \frac{1}{2c}\right| = \left|\frac{1}{c+iy} - \frac{1}{2c}\right|
 = \left|\frac{c - iy}{(c+iy)2c}\right| 
 = \frac{1}{2c}\frac{|c-iy|}{|c+iy|}
 =\frac{1}{2c}
\]
because $|c-iy| = |c+iy|$.

Now, if $C$ is any circle or a line and $0\notin C$, then by (a) we may
rotate the plane by $R(z) = e^{-i\theta}z$ so that $R(C)$ is a circle
with center $c>0$ and radius $r \ne c$, or a vertical line $x = c>0$.
By what we have just proved, $I(R(C))$ is a circle. Rotating once more by
the same angle, $R(I(R(C))) = I(C)$ is still a circle.

\item[(c)] Let $C$ be the circle $\{ c + ce^{it}\,|\, 0\le t\le2\pi\}$
with center and radius~$c>0$ (note that $0\in C$).  To see that $I(C)$
is a vertical line $x = 1/(2c)$, we must show $\Re(I(z)) = 1/(2c)$
for all $z\in C$:
\[ \Re(I(z)) = \Re\Big(\frac{1}{c+ce^{it}}\Big) =
\Re\Big(\frac{1+e^{-it}}{c|1+e^{it}|^2}\Big) = \frac{1+\cos(t)}{c(2 +
2\cos(t))} = \frac{1}{2c} \]

If $C$ is the vertical line $x=0$, then $C = \{ iy \,|\, y\in\R\}$,
and $I(C) = \{ 1/(iy)\,|\, y\in\R\} = \{ i(-1/y)\,|\, y\in\R\} = C$.

Now, if $C$ is any circle or a line with $0\in C$, then by (a) we may
rotate the plane by $R(z) = e^{-i\theta}z$ so that $R(C)$ is a circle
with center $c>0$ and radius $r=c$, or the vertical line $x = 0$.
By what we have just proved, $I(R(C))$ is a line. Rotating once more by
the same angle, $R(I(R(C))) = I(C)$ is still a line.


\end{itemize}

\item

\begin{itemize}

\item[(a)] Let $D_1$ and $D_2$ be domains with a non-empty
intersection. To show that $D = D_1 \cup D_2$ is a domain, we must show
that $D$ is open and connected.

\emph{$D$ is open:} Let $p\in D$. Then $p\in D_1$ or $p\in D_2$. If
$p\in D_1$, then there is an $\epsilon>0$ such that $D(p,\epsilon)\subset
D_1\subset D$, since $D_1$ is open. Similarly, if $p\in D_2$, there is an
$\epsilon>0$ such that $D(p,\epsilon)\subset D_2\subset D$, since $D_2$ is
open.  This shows that every point of $D$ is an interior point and hence
$D$ is open.

\emph{$D$ is connected:} Let $p,q\in D$. If $p$ and $q$ are both in $D_1$
or $D_2$, then there is a polygonal path in $D$ connecting them because
$D_1$ and $D_2$ are connected.  So let us assume $p$ and $q$ are not both
in $D_1$ or $D_2$, say $p\in D_1$ and $q\in D_2$.  Let $r\in D_1\cap
D_2 \ne \emptyset$. Since $p$ and $r$ are both in $D_1$, there is a
polygonal path $\gamma_1$ in $D_1$ from $p$ to~$r$. Likewise, since both
$q$ and $r$ are in $D_2$ there is a polygonal path $\gamma_2$ in $D_2$
from $q$ to~$r$. Then $\gamma_1 - \gamma_2$ is a polygonal path in $D$
from $p$ to~$q$.

\item[(b)] Let $D$ be an arbitrary subset of the complex plane. We must
prove that every point of $D$ is either an interior point of $D$ or
a boundary point of~$D$. This is equivalent to showing that if $p\in D$
is not an interior point of~$D$, then $p$ is a boundary point of~$D$.
If $p\in D$ is not an interior point, then for all $r>0$, $D(p,r)
\not\subset D$. Therefore, $p\in D(p,r)\cap D \ne \emptyset$ and
$D(p,r)\cap(\C\setminus D) \ne \emptyset$  for all $r>0$ (i.e., $D(p,
r)$ contains points of $D$ and its complement for all $r>0$) and this is
the definition for $p$ to be a boundary point of~$D$.

\end{itemize}

\item 

\begin{itemize}

\item[(a)] Let $f(z) = \frac{z^2 - \sqrt{2} z + 1}{z^4+1}$ and let
$z_0 = (1+i)/\sqrt{2}$. For $f(z_0)$ to be continuous at $z = z_0$, we
must define
\[f(z_0) = \lim_{z\to z_0}\frac{z^2 - \sqrt{2} z + 1}{z^4+1}
= \lim_{z\to z_0}\frac{z^2 - \sqrt{2} z + 1}
{(z^2 - \sqrt{2} z + 1)(z^2 + \sqrt{2} z + 1)}
=\frac{1}{z_0^2 + \sqrt{2} z_0 + 1} = \frac{1}{2(1+i)} = \frac{1-i}{4}
\]
(The roots of $z^4+1$ are $z_0=(1+i)/\sqrt{2}$, $\overline{z}_0$,
$w_0=(-1+i)/\sqrt{2}$ and $\overline{w}_0$, so $(z - z_0)(z - \overline{z}_0)
= z^2 - \sqrt{2} z + 1$ and $(z - w_0)(z - \overline{w}_0) = z^2 +
\sqrt{2} z + 1$.)

\item[(b)] Assume that the series $\sum_{n=0}^\infty a_n$ converges. Then
$a_n \to 0$ as $n\to \infty$. It follows that there is a constant $M>0$
such that $|a_n| \le M$ for all $n\ge 0$. Therefore,
$\sum_{n=0}^\infty |a_nz^n| \le \sum_{n=0}^\infty M |z|^n =
M \frac{1}{1-z}$, if $|z| < 1$. We conclude that $\sum_{n=0}^\infty
a_nz^n$ is absolutely convergent for all $|z|<1$.

\end{itemize}

\item 

\begin{itemize}

\item[(a)] Let $\gamma(t) = e^{it}$, $0\le t \le \pi$. Then
\[\int_\gamma \sqrt{z}\,dz = \int_0^\pi (e^{it})^{1/2}ie^{it}\,dt
= \int_0^\pi ie^{3it/2}\,dt = \frac{2}{3}e^{3it/2}\Big|_0^\pi
= \frac{2}{3}\big(e^{3\pi i/2} - 1\big) = \frac{2}{3}\big(-i-1\big)
\]

\item[(b)] Let $\gamma(t) = x(t) + iy(t)$, $a\le t \le b$ be a
smooth simple closed curve bounding a domain~$D$ and let $f = u + iv$.
Then
\begin{eqnarray*}
 \int_\gamma f(z)\,dz
&=& \int_a^b [u(\gamma(t)) + iv(\gamma(t))] [x'(t) + iy'(t)]\,dt\\
&=&\int_a^b u(\gamma(t))x'(t) - v(\gamma(t))y'(t)\,dt
+i \int_a^b u(\gamma(t))y'(t) + v(\gamma(t)) x'(t)\,dt\\
&=&\int_\gamma u\,dx - v\,dy + i\int_\gamma v\,dx + u\,dy\\
&=&\int\!\!\!\int_D -\frac{\partial v}{\partial x} -
\frac{\partial u}{\partial y} \,dx\,dy +
i\int\!\!\!\int_D \frac{\partial u}{\partial x}  -
\frac{\partial v}{\partial y} \,dx\,dy
\quad\mbox{[Green's Theorem]}\\
&=&\int\!\!\!\int_D i\frac{\partial u}{\partial x}
-\frac{\partial v}{\partial x}
-\frac{\partial u}{\partial y}
-i\frac{\partial v}{\partial y}\,dx\,dy\\
&=&i\int\!\!\!\int_D \frac{\partial u}{\partial x}
+i\frac{\partial v}{\partial x}
+ i\Big(\frac{\partial u}{\partial y}
+i\frac{\partial v}{\partial y}\Big)\,dx\,dy\\
&=&i \int\!\!\!\int_D \frac{\partial f}{\partial x}
+i\frac{\partial f}{\partial y}\,dx\,dy\\
\end{eqnarray*}

\end{itemize}

\item

\begin{itemize}

\item[(a)] Let $f$ be analytic on a domain $D$ and suppose $f'(z) =
0$ for all $z\in D$. Then $0 \equiv f'(z) = u_x(x,y) + iv_x(x,y) =
-i u_y(x,y) + v_y(x,y)$ so $u_x(x,y) = u_y(x,y) = v_x(x,y) = v_y(x,y)
\equiv 0$. Therefore, $u(x,y) = a$ and $v(x,y) = b$ are constant, and
$f(z) = a + ib$ is constant.

\item[(b)]

\begin{itemize}
\item[(i)] $\log z = u+iv$ with $u = (1/2)\ln(x^2+y^2)$ and $v=\arctan(y/x)$.
Thus, $(\log z)' = u_x + iv_x = x/(x^2+y^2) - iy/(x^2+y^2) = \overline{z}/|z|^2 = 1/z$.

\item[(ii)] $z^\alpha = e^{\alpha\log(z)}$, so
$(z^\alpha)' = e^{\alpha\log(z)}\alpha/z = \alpha z^{\alpha-1}$.

\item[(iii)] $\sin z = (e^{iz}-e^{-iz})/(2i)$, so
$(\sin z)' =(ie^{iz}+ie^{-iz})/(2i)
=(e^{iz}+e^{-iz})/2 =\cos z$.

\item[(iv)] $\cos z = (e^{iz}+e^{-iz})/2$, so
$(\cos z)' =(ie^{iz}-ie^{-iz})/2
=-(e^{iz}-e^{-iz})/(2i) =-\sin z$.

\item[(v)] $\tan z = \sin z/\cos z$ so
$(\tan z)' = (\cos z \cos z- (-\sin z)\sin z)/\cos^2 z = 1/\cos^2 z =
\sec^2 z$.

\end{itemize}

\item[(c)]

\begin{itemize}
\item[(i)] $u = \frac{x}{x^2+y^2}$, $u_x = \frac{y^2-x^2}{(x^2+y^2)^2}$,
$u_{xx} = \frac{2x^3 - 6xy^2}{(x^2+y^2)^3}$, $u_y = -\frac{2xy}{(x^2+y^2)^2}$,
$u_{yy} = \frac{6xy^2 - 2x^3}{(x^2+y^2)^3}$, so\newline
$u_{xx} + u_{yy} = 0$ and $u$ is harmonic.

Harmonic conjugate: $v_x = -u_y = \frac{2xy}{(x^2+y^2)^2}$. Integrate
with respect to~$x$ to get $v = -\frac{y}{x^2+y^2} + g(y)$. Then
$v_y = \frac{y^2 - x^2}{(x^2+y^2)^2} + g'(y) = u_x = \frac{y^2 -
x^2}{(x^2+y^2)^2}$ so $g'(y) = 0$ and $g(y) = c$ is constant.  Therefore
$v = -\frac{y}{x^2+y^2} + c$ and $f(z) = u + iv = \frac{x}{x^2+y^2} -
i\frac{y}{x^2+y^2} + ic = \frac{\overline{z}}{|z|^2} + ic = \frac{1}{z} +
ic$.

\item[(ii)] $v = x^2 - 2x + 1 - y^2$, $v_x = 2x - 2$, $v_{xx} = 2$,
$v_y = -2y$, $v_{yy} = -2$, so $v_{xx} + v_{yy} = 0$ and $v$ is harmonic.

Harmonic conjugate: $u_x = v_y = -2y$. Integrate with respect to~$x$ to get
$u = -2xy + g(y)$. Then $u_y = -2x + g'(y) = -v_x = -2x + 2$, so $g'(y) =
2$ and $g(y) = 2y + c$. Therefore, $u = -2xy + 2y + c$ and
$f(z) = u + iv = -2xy + 2y + c + i(x^2 - 2x + 1 - y^2) =
i[(x^2 - y^2 + i 2xy)  - 2(x + iy) + 1 ] + c = i[z^2 - 2z + 1] + c =
i(z-1)^2 + c$.

\item[(iii)] $u = \cosh(x)\cos(y)$, $u_x = \sinh(x)\cos(y)$, $u_{xx} =
\cosh(x)\cos(y)$, $u_y = -\cosh(x)\sin(y)$,\newline $u_{yy} =
-\cosh(x)\cos(y)$, so $u_{xx} + u_{yy} = 0$ and $u$ is harmonic.

Harmonic conjugate: $v_x = -u_y = \cosh(x)\sin(y)$. Integrate with respect
to~$x$ to get $v = \sinh(x)\sin(y) + g(y)$. Then $v_y = \sinh(x)\cos(y) +
g'(y) = u_x = \sinh(x)\cos(y)$, so $g'(y) = 0$ and $g(y) = c$ is constant.
Therefore, $v = \sinh(x)\sin(y) + c$, and $f(z) = u + iv = \cosh(x)\cos(y)
+ i\sinh(x)\sin(y) +ic= (e^x + e^{-x})\cos(y)/2 + i(e^x - e^{-x})\sin(y)/2
+ic =[ e^x(\cos(y) + i\sin(y)) + e^{-x}(\cos(-y) +i\sin(-y)) ]/2 +ic
=[e^z + e^{-z}]/2 +ic = \cosh(z) + ic$.

\item[(iv)] $v = \sinh(y)\cos(x)$, $v_x = -\sinh(y)\sin(x)$, $v_{xx} =
-\sinh(y)\cos(x)$, $v_y = \cosh(y)\cos(x)$,\newline
$v_{yy} = \sinh(y)\cos(x)$, so $v_{xx} + v_{yy} = 0$ and $v$ is harmonic.

Harmonic conjugate: $u_x = v_y = \cosh(y)\cos(x)$. Integrate with respect
to~$x$ to get $u = \cosh(y)\sin(x) + g(y)$. Then $u_y = \sinh(y)\sin(x)
+ g'(y) = -v_x = \sinh(y)\sin(x)$ so $g'(y) = 0$ and $g(y) = c$ is
constant. Therefore, $u = \cosh(y)\sin(x) + c$ and $f(z) = u + iv =
\cosh(y)\sin(x) + i\sinh(y)\cos(x) + c = (e^y + e^{-y})\sin(x)/2 + i(e^y
- e^{-y})\cos(x)/2 +c = ie^y(\cos(-x) + i\sin(-x))/2 - ie^{-y}(\cos(x) +
i\sin(x))/2 +c = i(e^{y-ix} - e^{-y+ix})/2 +c = i(e^{-iz} - e^{iz})/2 =
(e^{iz}-e^{-iz})/(2i) +c = \sin(z)+c$.

\item[(v)] $u = \arctan(y/x)$, $u_x = -y/(x^2+y^2)$, $u_{xx} =
2xy/(x^2+y^2)^2$, $u_y = x/(x^2+y^2)$, $u_{yy} = -2xy/(x^2+y^2)^2$, so
$u_{xx} + u_{yy} = 0$ and $u$ is harmonic.

Harmonic conjugate: $v_x = -u_y = -x/(x^2+y^2)$. Integrate with respect
to~$x$ to get $v = -(1/2)\ln(x^2+y^2) + g(y)$. Then $v_y = -y/(x^2+y^2) +
g'(y) = u_x = -y/(x^2+y^2)$, so $g'(y) = 0$ and $g(y) = c$ is constant.
Therefore, $v = -(1/2)\ln(x^2+y^2) + c$ and $f(z) = \arctan(y/x) -
i(1/2)\ln(x^2+y^2) +ic = \arg(z) - i\ln|z| + ic = -i\log(z) + ic$.

\end{itemize}

\end{itemize}

\end{enumerate}

\end{document}
