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For subsets \(U\) and \(V\) of \(ℝ\), we define
\[U + V=\{u + v : u∈U, v∈V\}\](b)(ii) Give an example where \(U\) is not open but \(U + V\) is open for all open \(V\).
Answer. \(U=\{0 \}\)
(iii) Show that if \(V\) is open, then \(U + V\) is open.
Answer. \(∀x∈U, y∈V,\) since \(V\) is open, \(∃r : B (y, r)∈V→B (x + y, r)∈U + V\)
(c) Give an example of \(U, V\) closed and \(U + V\) is not closed.
Example 1. \(U=ℤ, V=\left\{\frac{1}{n}+ n : n=2, 3, … \right\}\)
\(\frac{1}{n}=(- n) + \left( \frac{1}{n}+ n \right)∈U + V\)
0 is a limit point of \(U + V\) but not in \(U + V\).
Example 2. \(U=ℤ, V=πℤ\)
\(π∉ℚ⇒∄p, q∈ℤ: p + qπ=0\) but \(p + q π\) can be arbitrarily close to 0.
0 is a limit point of \(U + V\) but not in \(U + V\).
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(b)(ii) \(M=\{(x, y)∈ℝ^2 : x^2 - y^2=1 \}\) has 2 connected components.
Answer. Consider \(M \cap \{(x, y) : x > 0 \}\) and \(M \cap \{(x, y) : x < 0 \}\)
The sets \(\left\{\left( \pm \sqrt{y^2 + 1}, y \right) : y∈ℝ\right\}\) are connected by continuity.
(iii) Let \(X_1, X_2, X_3, …\) be path-connected subsets of \(M\) such that \(X_i \cap X_{i + 1}≠\varnothing\) for all \(i\). Show that \(\bigcup X_i\) is path-connected.
Answer. \(∀x, y∈\bigcup X_i\)
If \(x, y\) are in same \(X_i\), ∃path from \(x\) to \(y\). If not, suppose \(x∈X_i, Y∈X_{i + k}, i≥1, k≥1\).
Let \(x_1∈X_i \cap X_{i + 1}, x_2∈X_{i + 1}\cap X_{i + 2}, …, x_k ∈X_{i + k - 1}\cap X_{i + k}\)
∃a path from \(x\) to \(x_1\) in \(X_i\)
∃a path from \(x_1\) to \(x_2\) in \(X_{i + 1}\)
⋮
∃a path from \(x_k\) to \(y\) in \(X_{i + k}\)
The concatenation is a path from \(x\) to \(y\) in \(\bigcup X_i\).
(c) Let \(γ(a, r)\) denote the circle with centre \(a∈ℂ\) and radius \(r > 0\). Show that the following spaces are not homeomorphic to one another.
\[A=γ(0, 1), \quad B=γ(0, 1) \cup γ(1 / 2, 1 / 2), \quad C=γ(0, 1) \cup γ(1 / 2, 1 / 2) \cup γ(- 1 / 2, 1 / 2)\]Answer. \(B∖\{0, - 1 \}\) is connected, \(C∖\{i, - i \}\) connected, but removing any 2 points from \(A\) disconnects \(A \Rightarrow\)\(A\) is not homeomorphic to \(B\) or \(C\)
\(B∖\{1 \}\) is disconnected, but cannot disconnect \(C\) by removing a point\(⇒B\) is not homeomorphic to \(C\)
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(c) Let \(f\) be a non-constant holomorphic function on \(ℂ\). Show that \(f (ℂ)\) is dense in \(ℂ\) – i.e. given \(ε > 0\) and \(w∈ℂ\) there is \(z∈ℂ\) such that \(| f (z) - w | < ε\).
Answer. Suppose \(∃w_0∈ℂ, ε_0 > 0\) such that \(| f (z) - w | > ε_0\) for all \(z∈ℂ\)
\(g (z)=\frac{1}{f (z) - w_0}\) is holomorphic in \(ℂ\) and bounded
By Liouville's Theorem, \(g\) is constant, so \(f\) is constant
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(c) Show that
\[\int_{γ(0, r)}(\operatorname{Re}z)^{2 n + 1}\mathrm{d}z=πi\binom{2 n + 1}{n}\frac{r^{2 n + 2}}{2^{2 n}}\]Proof. By binomial expansion
\begin{eqnarray*}(\operatorname{Re}z)^{2 n + 1}&=& \left( \frac{1}{2}\left( z + \frac{r^2}{z}\right) \right)^{2 n + 1}\\ &=& … + \frac{1}{2^{2 n + 1}}\binom{2 n + 1}{n}r^{2 n + 2}z^n z^{- n - 1}+ \cdots\end{eqnarray*} By Residue Theorem \(\Box\)
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(b) Let \(a > 0\). Use contour integration to evaluate
\[\int_0^{∞}\frac{\cos t \mathrm{d}t}{t^2 + a^2}\]Answer. Use a semicircle \(\Gamma_R\) in the upper half plane.
\(\lim_{z→∞}\frac{1}{z^2 + a^2}=0\), by Jordan's lemma, the integral on the arc→0 as \(R → ∞\).
There is one simple pole \(ai\) inside \(\Gamma_R\)
\[\operatorname{Res}(f ; ai)=\frac{e^{iz}}{2 z}|_{z=ai}=\frac{e^{- a}}{2ai}\]By residue theorem
\[\int_{- ∞}^{∞}\frac{\cos t \mathrm{d}t}{t^2 + a^2}=\lim_{R→∞}\operatorname{Re}\int_{\Gamma_R}\frac{e^{iz}\mathrm{d}z}{t^2 + a^2}=\operatorname{Re}\left( 2 πi \frac{e^{- a}}{2 ai}\right)=\frac{πe^{- a}}{a}\]Since the integrand is even
\[\int_0^{∞}\frac{\cos t \mathrm{d}t}{t^2 + a^2}=\frac{πe^{- a}}{2a}\]
(c) Use a keyhole contour to evaluate
\[\int_0^{∞}\frac{\sqrt{t}}{1 + t^2}\mathrm{d}t\]Answer. \(f (z)=\frac{\sqrt{z}}{1 + z^2}\) has 2 simple poles \(\pm i\) inside the contour.
\begin{align*}\operatorname{Res}(f ; i) &=\frac{\sqrt{z}}{2 z}|_{z=i}=- \frac{1}{2}i e^{\frac{π}{4}i}\\ \operatorname{Res}(f ; - i) &=\frac{\sqrt{z}}{2 z}|_{z=- i}=\frac{1}{2}ie^{\frac{3 π}{4}i}\end{align*}By estimation lemma,
\[\left| \int_{\Gamma_R}\frac{\sqrt{z}}{1 + z^2}\mathrm{d}z \right| \leqslant2 πR \frac{\sqrt{R}}{R^2 - 1}→0 \text{ as }R \rightarrow∞\]By residue theorem,
\begin{align*}2 πi (\operatorname{Res}(f ; i) +\operatorname{Res}(f ; - i)) &=\int_{\Gamma_R}- \int_{\Gamma_r}+ \int_0^R f (x) \mathrm{d}x + \int_R^0 f (z) \mathrm{d}z \text{ where }z=e^{2 πi}x, x∈[0, R]\\ &→2 \int_0^{∞}f (x) \mathrm{d}x\end{align*}Finally
\[\int_0^{∞}f (x) \mathrm{d}x=πi \left( - \frac{1}{2}ie^{\frac{π}{4}i}+ \frac{1}{2}ie^{\frac{3 π}{4}i}\right)=\frac{π}{\sqrt{2}}\]
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Let \(Q=\{z :\operatorname{Im}z≥0,\operatorname{Re}z≥0\}\). State a harmonic function \(\phi (x, y)\) on \(Q\) which satisfies the boundary conditions
\[\phi (x, 0)=0, \quad \phi (0, y)=1 \quad \text{for}x, y > 0.\]Find a holomorphic function \(f\) on \(Q\) such that \(\phi=\operatorname{Im}f\).
Find a conformal bijection from \(Q\) to the unit disc \(D (0, 1)\) which sends \(1 + i\) to the origin.
Answer. On real axis \(\arg z=0\). On imaginary axis \(\arg z=\frac{π}{2}\).
\[\phi (z)=\frac{2}{π}\arg z, f (z)=\frac{2}{π}\log z\]\[Q \xmapsto{\frac{z^2}{2}}\text{upper half plane}\xmapsto{\frac{z - i}{z +i}}D (0, 1)\]The composition map \(z↦\frac{z^2 - 2 i}{z^2 + 2 i}\) is a conformal bijection \(Q→D (0, 1)\) which sends \(1 + i\) to 0.