Use \(n\)-th roots of unity (i.e. solutions of \(z^n - 1 =0\)) to show that \[2^{n-1} \sin\frac{\pi}{n} \sin\frac{2\pi}{n} \cdots \sin\frac{(n-1)\pi}{n} = n \; .\]
> Hint: $1 - \cos 2 \theta = 2 \sin^2 \theta,\; \sin 2 \theta = 2 \sin \theta \cos \theta$.
(a) Show that in polar coordinates, the Cauchy-Riemann
equations take the form
$$\frac{\partial u}{\partial r} = \frac{1}{r} \frac{\partial v}{\partial \theta}
\; \; \; \text{and} \; \; \;
\frac{\partial v}{\partial r} = - \frac{1}{r} \frac{\partial u}{\partial \theta}$$
(b) Use these equations to show that the logarithm function
defined by $$\log z = \log r + i \theta \; \;
\mbox{where} \; z = r e^{i \theta } \; \mbox{with} \; - \pi < \theta < \pi$$
is a holomorphic function in the region
$r>0, \; - \pi < \theta < \pi$. Also show that $\log z$ defined
above is not continuous in $r>0$.Solution
Goal: 1. Prove using \(n\)-th roots of unity that \(2^{n-1} \prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right) = n\) for any integer \(n \geq 2\). 2. (a) Derive the polar form of the Cauchy-Riemann equations: \(\frac{\partial u}{\partial r} = \frac{1}{r}\frac{\partial v}{\partial \theta}\) and \(\frac{\partial v}{\partial r} = -\frac{1}{r}\frac{\partial u}{\partial \theta}\). 3. (b) Verify that the principal branch of \(\log z = \ln r + i\theta\) on \(U = \{r e^{i\theta} : r > 0, -\pi < \theta < \pi\}\) is holomorphic, and prove it cannot be extended continuously to the punctured plane \(\mathbb{C}^* = \{z \in \mathbb{C} : r > 0\}\).
Part 1: Product of Sines via Roots of Unity
1. Factorization of the cyclotomic polynomial.
1.1. The polynomial \(z^n - 1\) factors as \((z - 1)\sum_{k=0}^{n-1} z^k = (z-1)(z^{n-1} + z^{n-2} + \dots + z + 1)\).Proof: Standard algebraic identity for geometric series / difference of powers.
1.2. The roots of \(z^n - 1 = 0\) are \(\omega_k = e^{i 2\pi k/n}\) for \(k = 0, 1, \dots, n-1\).Proof: \((e^{i 2\pi k/n})^n = e^{i 2\pi k} = 1\).
1.3. Since \(\omega_0 = 1\), the remaining \(n-1\) roots of unity \(\omega_1, \dots, \omega_{n-1}\) are the roots of \(P(z) = z^{n-1} + z^{n-2} + \dots + z + 1 = \prod_{k=1}^{n-1} (z - \omega_k)\).Proof: A monic polynomial of degree \(n-1\) with \(n-1\) distinct roots is uniquely factored as the product of its linear factors.
1.4. Evaluating \(P(z)\) at \(z=1\):\[\prod_{k=1}^{n-1} (1 - e^{i 2\pi k/n}) = P(1) = \underbrace{1 + 1 + \dots + 1}_{n \text{ terms}} = n.\]Proof: Direct substitution of \(z=1\) into \(P(z)\).
1.5. Q.E.D.
2. Compute \(|1 - e^{i 2\pi k/n}|\).
2.1. For any real \(\phi\), \(|1 - e^{i\phi}|^2 = (1 - \cos\phi)^2 + (\sin\phi)^2 = 1 - 2\cos\phi + \cos^2\phi + \sin^2\phi = 2 - 2\cos\phi = 2(1 - \cos\phi)\).Proof: Expansion of complex modulus squared.
2.2. Using the half-angle identity \(1 - \cos\phi = 2\sin^2(\phi/2)\), \(|1 - e^{i\phi}|^2 = 4\sin^2(\phi/2)\), so \(|1 - e^{i\phi}| = 2\left|\sin\left(\frac{\phi}{2}\right)\right|\).Proof: Taking the principal square root.
2.3. For \(\phi = \frac{2\pi k}{n}\) with \(1 \leq k \leq n-1\), \(\frac{\phi}{2} = \frac{\pi k}{n} \in (0, \pi)\), so \(\sin\left(\frac{\pi k}{n}\right) > 0\).Proof: The sine function is strictly positive on \((0, \pi)\).
2.4. Thus, \(|1 - e^{i 2\pi k/n}| = 2\sin\left(\frac{\pi k}{n}\right)\) for every \(1 \leq k \leq n-1\).Proof: Follows from 2 and
2.3.
2.5. Q.E.D.
3. Take the modulus of the product.
3.1. Taking the absolute value of both sides of <1>1.<2>4:\[\left|\prod_{k=1}^{n-1} (1 - e^{i 2\pi k/n})\right| = |n| = n \implies \prod_{k=1}^{n-1} |1 - e^{i 2\pi k/n}| = n.\]Proof: Multiplicativity of the complex modulus.
3.2. Substituting <1>2.<2>4 into the product:\[\prod_{k=1}^{n-1} \left( 2\sin\left(\frac{k\pi}{n}\right) \right) = 2^{n-1} \prod_{k=1}^{n-1} \sin\left(\frac{k\pi}{n}\right) = n.\]Proof: Factoring out the \(n-1\) factors of \(2\).
3.3. Q.E.D.
Part 2: (a) Cauchy-Riemann Equations in Polar Coordinates
4. Express partial derivatives with respect to \(r\) and \(\theta\) in terms of \(x\) and \(y\).
4.1. The coordinate transformation is \(x(r,\theta) = r\cos\theta\) and \(y(r,\theta) = r\sin\theta\).Proof: Definition of polar coordinates.
4.2. The partial derivatives are:\[\frac{\partial x}{\partial r} = \cos\theta, \quad \frac{\partial y}{\partial r} = \sin\theta, \quad \frac{\partial x}{\partial \theta} = -r\sin\theta, \quad \frac{\partial y}{\partial \theta} = r\cos\theta.\]Proof: Differentiating \(x\) and \(y\) with respect to \(r\) and \(\theta\).
4.3. By the multivariable chain rule for \(u(r,\theta) = u(x(r,\theta), y(r,\theta))\):\[\frac{\partial u}{\partial r} = u_x \cos\theta + u_y \sin\theta, \qquad \frac{\partial u}{\partial \theta} = -u_x r\sin\theta + u_y r\cos\theta.\]\[\frac{\partial v}{\partial r} = v_x \cos\theta + v_y \sin\theta, \qquad \frac{\partial v}{\partial \theta} = -v_x r\sin\theta + v_y r\cos\theta.\]Proof: Chain rule for compositions of \(C^1\) functions.
4.4. Q.E.D.
5. Substitute Cartesian Cauchy-Riemann equations \(u_x = v_y\) and \(u_y = -v_x\).
5.1. Substitute \(u_x = v_y\) and \(u_y = -v_x\) into the expression for \(\frac{\partial v}{\partial \theta}\):\[\frac{\partial v}{\partial \theta} = -v_x r\sin\theta + v_y r\cos\theta = u_y r\sin\theta + u_x r\cos\theta = r(u_x \cos\theta + u_y \sin\theta) = r \frac{\partial u}{\partial r}.\]Proof: Direct substitution and grouping by <1>4.<2>3.
5.2. Dividing by \(r > 0\) yields \(\frac{\partial u}{\partial r} = \frac{1}{r}\frac{\partial v}{\partial \theta}\).Proof: Division by \(r\).
5.3. Substitute \(u_x = v_y\) and \(u_y = -v_x\) into the expression for \(\frac{\partial u}{\partial \theta}\):\[\frac{\partial u}{\partial \theta} = -u_x r\sin\theta + u_y r\cos\theta = -v_y r\sin\theta - v_x r\cos\theta = -r(v_x \cos\theta + v_y \sin\theta) = -r \frac{\partial v}{\partial r}.\]Proof: Direct substitution and grouping by <1>4.<2>3.
5.4. Dividing by \(-r\) yields \(\frac{\partial v}{\partial r} = -\frac{1}{r}\frac{\partial u}{\partial \theta}\).Proof: Division by \(-r\).
5.5. Q.E.D.
Part 3: (b) Holomorphicity and Discontinuity of \(\log z\)
6. \(\log z = \ln r + i\theta\) is holomorphic on \(U = \{r e^{i\theta} : r > 0, -\pi < \theta < \pi\}\).
6.1. For \(\log z\), the real and imaginary parts are \(u(r,\theta) = \ln r\) and \(v(r,\theta) = \theta\).Proof: By definition of \(\log z\).
6.2. Compute the polar partial derivatives:\[\frac{\partial u}{\partial r} = \frac{1}{r}, \quad \frac{\partial u}{\partial \theta} = 0, \quad \frac{\partial v}{\partial r} = 0, \quad \frac{\partial v}{\partial \theta} = 1.\]Proof: Differentiation of \(\ln r\) and \(\theta\).
6.3. Check the polar Cauchy-Riemann equations on \(U\):\[\frac{\partial u}{\partial r} = \frac{1}{r} = \frac{1}{r}\cdot 1 = \frac{1}{r}\frac{\partial v}{\partial \theta}, \qquad \frac{\partial v}{\partial r} = 0 = -\frac{1}{r}\cdot 0 = -\frac{1}{r}\frac{\partial u}{\partial \theta}.\]Proof: Direct comparison of 2 with the polar CR equations.
6.4. Since \(u, v \in C^1(U)\) and satisfy the Cauchy-Riemann equations, \(\log z\) is holomorphic on \(U\).Proof: Equivalence of real differentiability + CR equations with complex differentiability (holomorphicity).
6.5. Q.E.D.
7. \(\log z\) cannot be continuous on the punctured plane \(\mathbb{C}^* = \{z : r > 0\}\).
7.1. Consider the point \(z_0 = -1 \in \mathbb{C}^*\) (which has \(r = 1\), \(\theta = \pm \pi\)).Proof: Test point on the branch cut \((-\infty, 0)\).
7.2. Approach \(z_0 = -1\) from the upper half-plane along the unit circle \(z(\theta) = e^{i\theta}\) as \(\theta \to \pi^-\):\[\lim_{\theta \to \pi^-} \log(e^{i\theta}) = \lim_{\theta \to \pi^-} (\ln 1 + i\theta) = i\pi.\]Proof: Since \(-\pi < \theta < \pi\), \(\log(e^{i\theta}) = i\theta\).
7.3. Approach \(z_0 = -1\) from the lower half-plane along the unit circle \(z(\theta) = e^{i\theta}\) as \(\theta \to (-\pi)^+\):\[\lim_{\theta \to (-\pi)^+} \log(e^{i\theta}) = \lim_{\theta \to (-\pi)^+} (\ln 1 + i\theta) = -i\pi.\]Proof: In the lower half-plane, \(-\pi < \theta < 0\), so \(\log(e^{i\theta}) = i\theta\).
7.4. Since the two directional limits \(i\pi\) and \(-i\pi\) do not agree (\(i\pi \neq -i\pi\)), the limit \(\lim_{z \to -1} \log z\) does not exist.Proof: Failure of two path limits to coincide.
7.5. Therefore, \(\log z\) is discontinuous at every point on the negative real axis \((-\infty, 0]\), and is not continuous on \(r > 0\).Proof: A function lacking a limit at a point is not continuous at that point.
7.6. Q.E.D.