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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 132
\noindent{\bf 1.}
\noindent{\it (a)}
$$\int_0^\pi {2e^{i\theta}+2\over 2e^{i\theta}} 2ie^{i\theta} d\theta
= \int_0^\pi i(2e^{i\theta} + 2) d\theta
= \int_0^\pi 2ie^{i\theta} d\theta + \int_0^\pi 2i d\theta
= 2(-1 - 1) + 2\pi i.$$
\noindent{\it (b)}
$$\int_\pi^{2\pi} {2e^{i\theta} + 2\over 2e^{i\theta}} 2ie^{i\theta} d\theta
= \int_\pi^{2\pi} i(2e^{i\theta} + 2) d\theta
= \int_\pi^{2\pi} 2ie^{i\theta} d\theta + \int_\pi^{2\pi} 2 d\theta
= 4 + 2\pi i.
$$
\noindent{\it (c)}
This is equivalent to the sum of integrals over $(0, \pi)$ and $(\pi,
2\pi).$
\noindent{\bf 4.}
Parameterizing the curve as $x + ix^3,$ with $x\in (-1, 1),$
this integral can be taken as the two integrals (and the discontinuity
ignored because the function is defined at either side of it),
$$\int_{-1}^0 (1+3x^2) dx + \int_0^1 4x^3 (1+3x^2) dx
= (1 + 1) + (1 + 2) = 5.$$
\hrule
%page 138
\noindent{\bf 3.}
The length of the triangle is $3+4+5 = 12,$ and the maximum value of
$$e^z - \overline{z}$$ is $$|e^z - \overline{z}| \leq e^x + \sqrt{x^2 + y^2}
\leq \max{e^z} + \max{\sqrt{x^2 + y^2}} \leq 1 + 4,$$
so by the theorem, the integral will be less that $ML = 12*5 = 60.$
\noindent{\bf 4.}
The length of the top half of the circle will be $\pi R,$ and the
maximum value of the integrand will be less than the quotient of the
maximum value of the dividend and the minimum value of the divisor.
$|2z^2 - 1|$ has a maximum value of $2R^2 + 1$ at $z = iR,$ and
$$z^4 + 5z^2 + 4 = (z^2 + 1)(z^2 + 4)$$ has a minimum value of $(R^2 -
1)(R^2 - 4)$ (coincidentally at $z = iR,$ although this isn't necessary)
The upper bound for this integral is, therefore, $ML,$ which corresponds
to the given upper bound.
Dividing the numerator and the denominator by $R^4$ gives
$$\pi({1/R} + O(1/R^2)) \over 1 + O(1/R),$$ and because $1/R$ and
$1/R^2$ both approach $0$ as $R$ approaches $\infty,$ we get
$$\pi(1/R),$$ which also approaches $0.$
\bye
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