homeworks and attendance quizzes for kang

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+\def\Re{\mathop{\rm Re}\nolimits}
+\def\Im{\mathop{\rm Im}\nolimits}
+\def\Log{\mathop{\rm Log}\nolimits}
+\let\rule\hrule
+\def\hrule{\medskip\rule\medskip}
+
+%page 95
+\noindent{\bf 8.}
+
+$$\log z = i\pi/2 \to z = e^{i\pi/2} = i,$$ by Euler's formula.
+
+\hrule
+%page 99
+\noindent{\bf 2.}
+
+\noindent{\it (a)}
+
+Where $z_n = r_ne^{i\theta_n},$ for all values of $\theta_n,$
+$$\log\left({z_1\over z_2}\right) = \ln|z_1/z_2| + i\arg(z_1/z_2)$$$$ =
+(\ln(r_1) - \ln(r_2)) + i(\theta_1-\theta_2) = (\ln(r_1) + i\theta_1) -
+(\ln(r_2) + i\theta_2) = \log z_1 - \log z_2,$$
+because $\ln(a-b) = \ln a - \ln b$ and $\arg(z_1/z_2) = \arg(z_1) -
+\arg(z_2).$
+
+\noindent{\it (b)}
+
+$1/z = \overline z/|z|^2.$ $$\log(1/z) = \ln|1/z| + i\arg(1/z) =
+- \ln|z| - i\arg z = - (\ln|z| + i\arg z) = -\log(z),$$ so
+$$\log(z*(1/z)) = \log(z) + \log(1/z) = \log(z) - \log(z) = 0.$$
+
+\hrule
+%page 103
+\noindent{\bf 2.}
+
+\noindent{\it (a)}
+
+The principal value of $(-i)^i$ is $e^{i\Log(i)} = e^{i\pi/2}.$
+
+\noindent{\it (b)}
+
+The base of this power function is $e^(1-2\pi/3) = b \to \Log(b) = 1-2\pi i/3.$
+This gives a principal value of $e^{3\pi i(1-2\pi/3)} = e^{3\pi i - 2\pi^2} = 
+-e^{2\pi^2}.$
+
+\noindent{\it (c)}
+
+The principal value of $(1-i)^{4i}$ is $e^{\Log(1-i)4i} =
+e^{(4i)(\ln(2)/2 - \pi/4)} = e^{(2\ln(2)i)-\pi} = e^{-\pi}(\cos(2\ln(2))
++ i\sin(2\ln(2))).$
+
+\noindent{\bf 8.}
+
+\noindent{\it (a)}
+
+$$z^{c_1}z^{c_2} = e^{c_1\Log(z)}e^{c_2\Log(z)} = e^{\Log(z)(c_1+c_2)} =
+z^{c_1+c_2}.$$
+
+\noindent{\it (b)}
+
+$${z^{c_1}\over z^{c_2}} = {e^{\Log(z)c_1}\over e^{\Log(z)c_2}} =
+e^{c_1\Log z - c_2\Log z} = e^{\Log(z)(c_1-c_2)}.$$
+
+\noindent{\it (c)}
+
+$$(z^c)^n = (e^{c\Log z})^n = e^{nc\Log z} = z^{cn}.$$
+
+\hrule
+%page 107
+\noindent{\bf 3.}
+
+$$\sin(z+z_2) = \sin z \cos z_2 + \cos z \sin z_2 \to
+\cos(z+z_2)   = \cos z \cos z_2 - \sin z \sin z_2 \to
+\cos(z_1+z_2) = \cos z_1\cos z_2-\sin z_1\sin z_2.$$
+
+\noindent{\bf 7.}
+% need to do more here
+
+$\sin z = \sin x \cosh y + i\cos x \sinh y.$
+$\cos z = \cos x \cosh y - i\sin x \sinh y.$
+
+$$|\sin z|^2 = (\sin x\cosh y)^2 + (\cos x\sinh y)^2
+= \sin^2 x(1+\sinh^2 y) + \cos^2 x\sinh^2 y =
+\sin^2 x + \sinh^2 y.$$
+
+$$|\cos z|^2 = \cos^2 x \cosh^2 y + \sin^2 x \sinh^2 y
+= \cos^2 x \cosh^2 y + (1-\cos^2 x)\sinh^2 y
+= \sinh^2 y + \cos^2 x.$$
+
+
+\noindent{\bf 12.}
+
+$\sin z$ and $\cos z$ are real with $z$ on the real axis.
+Therefore, $\overline{\sin z} = \sin\overline z$ and $\overline{\cos z}
+= \cos\overline z.$
+
+\hrule
+%page 111
+\noindent{\bf 6.}
+
+\noindent{\it (a)}
+$$|\cosh z|^2 = \sinh^2 x + \cos^2 y \to |\sinh x|^2 \leq |\cosh z|^2
+\to |\sinh x|\leq |\cosh z|.$$
+$$|\cosh z|^2 = \cosh^2 x + cos^2 y \to |\cosh z|^2 \leq \cosh^2 x \to
+|\cosh z| \leq \cosh x,$$
+because $\cosh x > 0.$
+
+\noindent{\it (b)}
+%9b of sec 38 would probably be good
+
+$$|\sinh y| \leq |\cos z| \leq \cosh y.$$
+We change $z$ to $z' = zi,$ so $y$ becomes $x.$
+$$|\sinh x| \leq |\cosh z| \leq \cosh x.$$
+
+\noindent{\bf 16.}
+$$\sinh z = \sinh x \cos y + i\cosh x \sin y.$$
+$$\cosh z = \cosh x \cos y + i\sinh x \sin y.$$
+
+\noindent{\it (a)}
+$$\sinh z = i = \sinh x\cos y + i\cosh x\sin y \to 1 = \cosh x\sin y =
+\sin y,$$
+because all zeroes of $\cosh x$ are on imaginary axis, so
+$x = 0 \to \cosh x = 1.$
+
+$\sin y = 1$ on $y = 2n\pi + \pi/2 \to z = (2n\pi + \pi/2)i.$
+
+\noindent{\it (b)}
+$$\cosh z = \cosh(yi) = 1/2 = \cos y \to y = (2n\pi \pm \pi/3)i.$$
+
+\hrule
+%page 113
+\noindent{\bf 1.}
+%learn the derivation of inverse sin
+
+\noindent{\it (c)}
+
+$$\cosh^{-1}(-1) = \log\left[-1+0^{1/2}\right] = \log(1)
+= 2n\pi i+\pi.$$
+
+\hrule
+%page 119
+\noindent{\bf 2.}
+
+\noindent{\it (a)}
+
+$\int_0^1 (1+it)^2 dt = \int_0^1 1 + 2it - t^2 dt =
+[t + it^2 - t^3/3]_0^1 = 1 + i - 1/3 = 2/3 + i.$
+
+\noindent{\it (b)}
+
+$\int_1^2 \left({1\over t} -i\right)^2 dt = \int_1^2 {1\over t^2} -
+{2i\over t} - 1 dt = [-{1\over t} - 2i\ln t - t]_1^2 =
+1-{1\over 2} - 2i\ln2 + 2 - 1 = 3/2 - 2i\ln2.$
+
+\noindent{\it (c)}
+
+$\int_0^{\pi/6} e^{i2t} dt = \int_0^{\pi/6} \cos(2t) + i\sin(2t) dt =
+(1/2)[\sin(2t) - i\cos(2t)]_0^{\pi/6} (1/2)[\sqrt3/2 - i/2].$
+
+\noindent{\it (d)}
+
+$\int_0^\infty e^{-zt} dt = \lim_n\to\infty [-e^{-zt}/z]_0^n = 1/z -
+\lim_t\to\infty e^{-zt}/z = 1/z.$
+
+\hrule
+%page 123
+\noindent{\bf 1.}
+%do no 2 to confirm knowledge of jordan curve
+
+\noindent{\it (a)}
+
+This is true for a real-valued function, so 
+$$\int_{-b}^{-a} u(-t) dt + i\int_{-b}^{-a} v(-t) = \int_a^b u(t) dt +
+i \int_a^b v(t) dt,$$
+because $u$ and $v$ are real-valued, so the corresponding imaginary and
+real components are equal.
+
+\noindent{\it (b)}
+
+This is true of a real function on $(a,b),$ so $$\int_a^b u(t) dt =
+\int_\alpha^\beta u(\phi(\tau))\phi'(\tau) d\tau,$$
+and similarly for $v(t),$ so
+$$\int_a^b u(t)+iv(t) dt =
+\int_\alpha^\beta (u(\phi(\tau)) + iv(\phi(\tau))\phi'(\tau) d\tau.$$
+
+\bye