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\noindent {\sc Complex Variables I
			\hfill Practice Midterm 2  \vspace{0.3cm} \\
			Name:\underline{\hspace*{1.1in}} Id No.:\underline{\hspace*{1.9in}}  Class: 	\underline{\hspace*{0.9in}}	} \\


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{\bf Problem 1: (15 points)}
Show that $f(z) = \frac{\operatorname{Log}(z+5i)}{z^2+3z+2}$ is analytic everywhere except at the points $-1, -2,$ and on the ray $\{(x, y)| x \le 0, y=-5 \}$.
\begin{proof}
First, the function $\frac{1}{z^2+3z+2}$ is analytic everywhere except at $-1$, and $-2$. Second, the function
$\operatorname{Log}(z+5i)$ is analytic except at those points where $z+5i = 0$, or where $z+5i$
lies on the negative real axis, i.e. $\{(x; y) | x \le 0, y = -5\}.$ Hence the function $f(z) = \frac{\operatorname{Log}(z+5i)}{z^2+3z+2}$ is analytic everywhere except at the points $-1, -2,$ and on the ray $\{(x, y)| x \le 0, y=-5 \}$.
\end{proof}

{\bf Problem 2:}
\begin{enumerate}
\item {\bf (10 points)}
Show that $ \operatorname{tanh}^{-1} z = \frac{1}{2} \log \left( \frac{1+z}{1-z} \right).$
\begin{proof}[Hint]
Consider $z= \operatorname{tanh} w = \frac{(e^{w}-e^{-w})}{(e^{w}+e^{-w})}.$ Then write $w$ as a function of $z$.
\end{proof}
\item {\bf (5 points)}
Compute $\operatorname{tanh}^{-1} (1+2i)$.
\begin{proof}[Solution] 
$ \frac{1}{4} \operatorname{ln} 2 + i (\frac{3}{8} +n) \pi$, where $n$ is an integer.
\end{proof}
\end{enumerate}

{\bf Problem 3:(10 points)}
Determine whether $\log (i^2)$ and $2 \log i$ are equal or not when the branch $\log z = \operatorname{ln} r + i \theta (r>0, \frac{3 \pi}{4} < \theta < \frac{11\pi}{4}) $ is used. Show your work. 
\begin{proof}[Solution]
$\log (i^2) = \log (-1) = i \pi$ and $2 \log i = 2 (i \frac{10 \pi}{4}) = i \frac{5 \pi}{2}$ when the stated branch of the logarithmic function is used. Hence $\log (i^2) \not= 2 \log i.$
\end{proof}

{\bf Problem 3: (15 points)}
Consider $I=\int_C \frac{1}{z^3(z+4)}dz.$
\begin{enumerate}
\item Evaluate the integral $I$ when the contour $C$ is the positively oriented circle $|z|=2$.
   \begin{proof}[Hint]
  First we notice that $\frac{1}{z+4}$ is analytic inside and on $C$. Then we use the Extension of Cauchy Integral Formula.
$\cdots$ The solution is $\frac{\pi i}{32}$.
   \end{proof}
\item Evaluate the integral $I$ when the contour $C$ is the positively oriented circle $|z+2| =3.$
   \begin{proof}[Hint]
   The singularities $0$ and $-4$ of $\frac{1}{z^3(z+4)}$ are inside the contour $C$. We consider the contour $C_1$ the positively oriented circle $|z|=\frac{1}{2}$ and the contour $C_2$ the positively oriented circle $|z+4|=\frac{1}{2}$. Then $C_1$ and $C_2$ are inside $C$. So we can apply the Cauchy-Goursat theorem for multiply connected domain. $\cdots$ The solution is $0$
   \end{proof}
\end{enumerate}

{\bf Problem 4:(10 points)}
Consider the function $f(z) = (z-2)^4$ and the closed rectangular region $R$ with vertices at $0, -1, -1+4i, 4i.$ Find the points on or inside $R$ at which $|f(z)|$ attains its maximum and minimum values.
\begin{proof}[Solution]
Observe that $|z-2| = d$ is the distance from $z$ to 2 and $|(z-2)^4| = |z-2|^4 = d^4$. The maximum (minimum) modulus principle implies that the maximum and minimum values occur at the boundary. From the geometry, we can see that the maximum and minimum values of $d$, and therefore $|f(z)|$, occur at the boundary points, namely $-1+4i$ and $0$. Hence $\operatorname{max} |f(z)|$ occurs at $z=-1+4i$ and $\operatorname{min}|f(z)|$ occurs at $z=0$.
\end{proof}

%{\bf Problem 4:(10 points)}
%Let $R$ region $0 \le x \le \pi, 0 \le y \le 1$. Show that the modulus of the entire function $f(z)= \sin z$ has a maximum value in $R$ at the boundary point $z=(\pi/2) +i$.
%\begin{proof}[Hint]
%Write $|f(z)|^2=\sin^2 x + \sinh^2 y$ and locate points in $R$ at which $\sin^2 x$ and $\sinh^2 y$ are the largest.
%\end{proof}

{\bf Problem 5:(10 points)}
Let $f$ be an entire function with the property that $|f(z)| \ge 1$ for all $z$. Show that $f$ is constant.
\begin{proof}[Hint]
Liouville's theorem.
\end{proof}



{\bf Problem 6: (15 points)}
Compute $\int_C \frac{1}{z^2-z}dz$, where $C$ is a line segment from $2$ to $2+i$.
\begin{proof}
$\frac{1}{z^2-z} = -\frac{1}{z} + \frac{1}{z-1}$ is analytic everywhere except at $0$ and $1$.
Let $D=\mathbb{C} \backslash \{ (x,y) | x \le 1, y=0 \}$. Then $\frac{1}{z^2-z} = -\frac{1}{z} + \frac{1}{z-1}$ is analytic in $D$ and $-\operatorname{Log} z + \operatorname{Log}(z-1)$ is an antiderivative of $\frac{1}{z^2-z} = -\frac{1}{z} + \frac{1}{z-1}$. By the extension of fundamental theorem of calculus, $\int_C \frac{1}{z^2-z}da = \int_2^{2+i} \frac{1}{z^2-z}dz =(-\operatorname{Log}z + \operatorname{Log}(z-1))|_{2}^{2+i} = -\frac{1}{2} \ln 5+ \frac{3}{2} \ln 2 + i(\operatorname{arc tan} \frac{1}{2} + \frac{1}{4} \pi).$
\end{proof}

{\bf Problem 7: (10 points)}

Let $C_R$ denote the upper half of the circle $|z|=R (R>2)$, taken in the counterclockwise direction. Use ML bounds to find an upper bound on the modulus of the following contour integral $\int_{C_R} \frac{2z^2-1}{z^4+5z^2+4} dz$. 
\begin{proof}[Hint]
$M = \frac{ (2R^2+1)}{(R^2-1)(R^2-4)}$ and $L=\pi R$. The upper bound is
$\frac{\pi R (2R^2+1)}{(R^2-1)(R^2-4)}$
%$|\int_{C_R} \frac{\operatorname{Log} z}{z^2} dz| \le ML= \operatorname{max}_{z \in C_R} |\frac{\operatorname{Log}z}{z^2}|(2 \pi R) = \frac{\operatorname{max}_{z \in C_R}|\operatorname{ln}|z|+i\theta|}{R^2}(2 \pi R) = 2 \pi \frac{\sqrt{(\operatorname{ln}R)^2+\pi^2}}{R}.$
\end{proof}
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