homeworks and attendance quizzes for kang

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+\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