# Cauchy’s Theorem

## Entire and $$O$$ of polynomial implies polynomial #complex/exercise/completed

Let $$f(z)$$ be entire and assume that $${\left\lvert {f(z)} \right\rvert} \leq M |z|^2$$ outside of some disk for some constant $$M$$. Show that $$f(z)$$ is a polynomial in $$z$$ of degree $$\leq 2$$.

• Prove a more general statement: if $${\left\lvert {f(z)} \right\rvert} \leq M{\left\lvert {z} \right\rvert}^n$$, then $$f$$ is a polynomial of degree at most $$n$$.

• Since $$f$$ is entire, it is analytic everywhere, so $$f(z) = \sum_{k\geq 0}c_k z^k$$ where $$c_k = f^{(k)}(0)/n!$$ is given by the coefficient of its Taylor expansion about $$z=0$$.

• Applying Cauchy’s estimate, on a circle of radius $$R$$, \begin{align*} {\left\lvert {f^{(k)}(0)} \right\rvert} \leq { \sup_{\gamma}{\left\lvert {f(z)} \right\rvert} n! \over R^k} \leq {M{\left\lvert {z} \right\rvert}^n n! \over R^k} = {M R^n n! \over R^k} .\end{align*}

• So for $$k \geq n+1$$, this goes to zero as $$R\to \infty$$, so $${\left\lvert {f^{k}(0)} \right\rvert} = 0$$ for all such $$k$$.

• But then $$f$$ is a power series annihilated by taking $$n+1$$ derivatives, so it is a polynomial of degree at most $$n$$.

## Uniform sequence implies uniform derivatives #complex/exercise/work

Let $$a_n(z)$$ be an analytic sequence in a domain $$D$$ such that $$\displaystyle \sum_{n=0}^\infty |a_n(z)|$$ converges uniformly on bounded and closed sub-regions of $$D$$. Show that $$\displaystyle \sum_{n=0}^\infty |a'_n(z)|$$ converges uniformly on bounded and closed sub-regions of $$D$$.

## Tie’s Extra Questions: Spring 2014 #complex/exercise/completed

The question provides some insight into Cauchy’s theorem. Solve the problem without using the Cauchy theorem.

• Evaluate the integral $$\displaystyle{\int_{\gamma} z^n dz}$$ for all integers $$n$$. Here $$\gamma$$ is any circle centered at the origin with the positive (counterclockwise) orientation.

• Same question as (a), but with $$\gamma$$ any circle not containing the origin.

• Show that if $$|a|<r<|b|$$, then $$\displaystyle{\int_{\gamma}\frac{dz}{(z-a)(z-b)} dz=\frac{2\pi i}{a-b}}$$. Here $$\gamma$$ denotes the circle centered at the origin, of radius $$r$$, with the positive orientation.

\begin{align*} \int_\gamma z^n\,dz= \int_0^{2\pi} R^n e^{itn} \cdot iRe^{it} \,dt = R^{n+1} \int_0^{2\pi} e^{i(t+1)n}\,dt = { i R^{n+1} \over i(n+1) } \delta_{n+1 = 0} .\end{align*}

About a point $$a$$ and $$R<{\left\lvert {a} \right\rvert}$$, \begin{align*} \int_{{\left\lvert {z-a} \right\rvert} = R} z^n\,dz &= \int_0^{2\pi} (a + re^{it})^n \cdot ire^{it}\,dt\\ &= \int_0^{2\pi} \sum_{1\leq k\leq n} {n \choose k} a_k R^{n-k+1} e^{it(n-k)} \cdot ire^{it}\,dt\\ &= i \int_0^{2\pi} \sum_{1\leq k\leq n} {n \choose k} a_k R^{n-k+1} e^{it(n-k+1)} \,dt\\ &= i \sum_{1\leq k\leq n} {n \choose k} a_k R^{n-k+1} \int_0^{2\pi} e^{it(n-k+1)} \,dt\\ &= i \sum_{1\leq k\leq n} {n \choose k} a_k R^{n-k+1} \cdot 0 \\ &= 0 ,\end{align*} provided $$n\neq 0$$, in which case $$\int_\gamma \,dz= 2\pi$$.

For the third computation, this follows from partial fraction decomposition.

## Fall 2019.3, Spring 2020 HW 2.9 (Cauchy’s Formula for Exterior Regions) #complex/qual/completed

Let $$\gamma$$ be a piecewise smooth simple closed curve with interior $$\Omega_1$$ and exterior $$\Omega_2$$. Assume $$f'$$ exists in an open set containing $$\gamma$$ and $$\Omega_2$$ with $$\lim_{z\to \infty} f(z) = A$$. Show that \begin{align*} F(z) \coloneqq\frac{1}{2 \pi i} \int_{\gamma} \frac{f(\xi)}{\xi-z} d \xi=\left\{\begin{array}{ll} A, & \text { if } z \in \Omega_{1} \\ -f(z)+A, & \text { if } z \in \Omega_{2} \end{array}\right. .\end{align*}

NOTE (DZG): I think there is a typo in this question….probably this should equal $$f(z)$$ for $$z\in \Omega_1$$, which is Cauchy’s formula…

Note that $$G_z(\xi) \coloneqq{f(\xi) \over \xi - z}$$ has a pole of order one at $$\xi = z$$ and also a pole at $$\xi = \infty$$. If $$z\in \Omega_1$$, then $$\gamma$$ encloses just the pole $$\xi = z$$, so apply the residue theorem: \begin{align*} F(z) &\coloneqq{1\over 2\pi i}\oint_\gamma {f(\xi) \over \xi - z}\,d\xi\\ &= {1\over 2\pi i}\oint_\gamma G_z(\xi) \,d\xi\\ &= \mathop{\mathrm{Res}}_{\xi = z} G_z(\xi) \\ &= \lim_{\xi\to z} (\xi - z) G_z(\xi) \\ &= \lim_{\xi\to z} (\xi - z) {f(\xi) \over \xi-z} \\ &= \lim_{\xi\to z} f(\xi) \\ &= f(z) .\end{align*}

Now if $$z\in \Omega_2$$, then $$\gamma$$ encloses both $$\xi=z, \infty$$, and is oriented negatively,so \begin{align*} F(z) &= {1\over 2\pi i} \oint_\gamma G_z(\xi) \,d\xi\\ &= -\qty{\mathop{\mathrm{Res}}_{\xi = z} G_z(\xi) + \mathop{\mathrm{Res}}_{\xi = \infty} G_z(\xi)}\\ &= -\qty{f(z) + \mathop{\mathrm{Res}}_{\xi = \infty} G_z(\xi)}\\ ,\end{align*} where the last line proceeds by the same calculation as above. It remains to compute the unknown residue. Residues at $$\xi = \infty$$ are computed as residues at $$\xi =0$$, and the change of variables $$G_z(\xi)\,d\xi\mapsto G_z(w) \,dw$$ for $$w\coloneqq 1/\xi$$ yields $$G_z(\xi)\,d\xi\to G_z\qty{1\over \xi}(-1/\xi^2)\,d\xi$$. Thus \begin{align*} \mathop{\mathrm{Res}}_{\xi=\infty} G_z(\xi) &= -\mathop{\mathrm{Res}}_{\xi=0} G_z\qty{\xi^{-1}}\xi^{-2} \\ &= - \mathop{\mathrm{Res}}_{\xi=0} {f(\xi^{-1}) \over \xi^2(\xi^{-1}- z) } \\ &= - \mathop{\mathrm{Res}}_{\xi=0} {f(\xi^{-1}) \over \xi(1 - z\xi) } \\ &= -\lim_{\xi \to 0} {f(\xi^{-1}) \over 1-z\xi} \\ &= -\lim_{\xi \to 0}f(\xi ^{-1}) \\ &= -\lim_{\xi\to\infty} f(\xi) \\ &= -A .\end{align*} So combining this yields \begin{align*} F(z) = -\qty{f(z) - A} = -f(z) + A .\end{align*}

## Tie’s Extra Questions: Fall 2009 (Proving Cauchy using Green’s) #complex/exercise/completed

State and prove Green’s Theorem for rectangles. Use this to prove Cauchy’s Theorem for functions that are analytic in a rectangle.

Suppose $$f\in C_{\mathbf{C}}^1(\Omega)$$ and $$T\subset \Omega$$ is a triangle with $$T^\circ \subset \Omega$$.

• Apply Green’s theorem to show that $$\int_T f(z) ~dz = 0$$.
• Assume that $$f'$$ is continuous and prove Goursat’s theorem.

Hint: Green’s theorem states \begin{align*} \int_{T} F d x+G d y=\int_{T^\circ}\left(\frac{\partial G}{\partial x}-\frac{\partial F}{\partial y}\right) d x d y .\end{align*}

Green’s theorem: if $$\Omega$$ is a domain with positively oriented boundary with $$u, v$$ continuously differentiable in $$\overline{\Omega}$$, then \begin{align*} \int_{{{\partial}}\Omega} u\,dx+ v\,dy= \iint_{\Omega}\qty{v_x - u_y}\,dx\,dy .\end{align*} Now use that if $$f = u+iv$$ is analytic in a region, it satisfies Cauchy-Riemann: \begin{align*} u_x = v_y \qquad u_y = -v_x .\end{align*}

Now integrating $$f$$: \begin{align*} \oint_{{{\partial}}\Omega} f(z) \,dz &= \oint_{{{\partial}}\Omega} (u+iv)(\,dx+ i\,dy)\\ &= \oint_{{{\partial}}\Omega} \qty{u\,dx- v\,dy} + i\oint_{{{\partial}}\Omega} \qty{v\,dx+ u\,dy} \\ &= \iint_\Omega\qty{v_x + u_y}\,dx\,dy+ \iint_\Omega\qty{u_x - v_y}\,dx\,dy\\ &= \iint_\Omega\qty{v_x -v_x }\,dx\,dy+ \iint_\Omega\qty{u_x - u_x}\,dx\,dy\\ &= 0 .\end{align*}

## No polynomials converging uniformly to $$1/z$$#complex/exercise/completed

Prove that there is no sequence of polynomials that uniformly converge to $$f(z) = {1\over z}$$ on $$S^1$$.

• By Cauchy’s integral formula, $$\int_{S^1} f = 2\pi i$$
• If $$p_j$$ is any polynomial, then $$p_j$$ is holomorphic in $${\mathbb{D}}$$, so $$\int_{S^1} p_j = 0$$.
• Contradiction: compact sets in $${\mathbf{C}}$$ are bounded, so \begin{align*} {\left\lvert {\int f - \int p_j} \right\rvert} &\leq \int {\left\lvert {p_j - f} \right\rvert} \\ &\leq \int {\left\lVert {p_j - f} \right\rVert}_\infty \\ &= {\left\lVert {p_j - f} \right\rVert}_\infty \int_{S^1} 1 \,dz \\ &= {\left\lVert {p_j-f} \right\rVert}_\infty \cdot 2\pi \\ &\to 0 \end{align*} which forces $$\int f = \int p_j = 0$$.

## Eventually sublinear implies constant #complex/exercise/completed

Suppose $$f: {\mathbf{C}}\to {\mathbf{C}}$$ is entire and \begin{align*} {\left\lvert {f(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}^{1\over 2} \quad\text{ when } {\left\lvert {z} \right\rvert} > 10 .\end{align*}

Prove that $$f$$ is constant.

Let $$R> 10$$, then by Cauchy: \begin{align*} 2\pi {\left\lvert {f'(z)} \right\rvert} &\leq \oint_{{\left\lvert {\xi} \right\rvert} = R} { {\left\lvert { f(\xi)} \right\rvert} \over {\left\lvert {\xi} \right\rvert}^2 } \,d\xi\\ &\leq \oint_{{\left\lvert {\xi} \right\rvert} = R} R^{-2} {\left\lvert {\xi} \right\rvert}^{1\over 2} \,d\xi\\ &= R^{-{3\over 2}} \cdot 2\pi R \\ &\sim R^{-{1\over 2}} \\ &\overset{R\to\infty}\longrightarrow 0 .\end{align*}

## The Cauchy pole function is holomorphic #complex/exercise/completed

Let $$\gamma$$ be a smooth curve joining two distinct points $$a, b\in {\mathbf{C}}$$.

Prove that the function \begin{align*} f(z) \coloneqq\int_\gamma {g(w) \over w-z} \,dw \end{align*} is analytic in $${\mathbf{C}}\setminus\gamma$$.

Toward applying Morera, let $$T \subseteq {\mathbf{C}}\setminus\gamma$$ be a triangle, so that $$z\in T$$ and $$w\in \gamma$$ implies $$z-w\neq 0$$. Then \begin{align*} \oint_T f(z) \,dz &= \oint_T \int_\gamma {g(w)\over w-z}\,dw\,dz\\ &= \int_\gamma \oint_T {g(w)\over w-z}\,dz\,dw\\ &= \int_\gamma g(w) \qty{ \oint_T {1 \over w-z}\,dz} \,dw\\ &= \int_\gamma g(w) \cdot 0 \,dw\\ &= 0 ,\end{align*} where the exchange of integrals is justified by compactness of $$\gamma, T$$, and the inner integral vanishes because for a fixed $$w\in \gamma$$, the function $$z\mapsto {1\over w-z}$$ has a simple pole at $$w$$, and so is holomorphic in $$\gamma^c$$ and vanishes by Goursat.

## Schwarz reflection proof #complex/exercise/completed

Suppose that $$f: {\mathbf{C}}\to{\mathbf{C}}$$ is continuous everywhere and analytic on $${\mathbf{C}}\setminus {\mathbf{R}}$$ and prove that $$f$$ is entire.

Just reproducing the proof of holomorphicity in the Schwarz reflection theorem.

• Note $$f$$ is continuous on $${\mathbf{C}}$$ since analytic implies continuous ($$f$$ equals its power series, where the partials sums uniformly converge to it, and uniform limit of continuous is continuous).
• Strategy: take $$D$$ a disc centered at a point $$x\in {\mathbf{R}}$$, show $$f$$ is holomorphic in $$D$$ by Morera’s theorem.
• Let $$\Delta \subset D$$ be a triangle in $$D$$.
• Case 1: If $$\Delta \cap{\mathbf{R}}= 0$$, then $$f$$ is holomorphic on $$\Delta$$ and $$\int_\Delta f = 0$$.
• Case 2: one side or vertex of $$\Delta$$ intersects $${\mathbf{R}}$$, and wlog the rest of $$\Delta$$ is in $${\mathbb{H}}^+$$.
• Then let $$\Delta_{\varepsilon}$$ be the perturbation $$\Delta + i{\varepsilon}= \left\{{z+ i{\varepsilon}{~\mathrel{\Big\vert}~}z\in \Delta}\right\}$$; then $$\Delta_{\varepsilon}\cap{\mathbf{R}}= 0$$ and $$\int_{\Delta_{\varepsilon}} f = 0$$.
• Now let $${\varepsilon}\to 0$$ and conclude by continuity of $$f$$ (???)
• We want \begin{align*} \int_{\Delta_{\varepsilon}} f = \int_a^b f(\gamma_{\varepsilon}(t)) \gamma_{\varepsilon}'(t)\,dt \overset{{\varepsilon}\to 0}\to \int_a^b f(\gamma(t)) \gamma_{\varepsilon}'(t)\,dt =\int_{\Delta} f \end{align*} where $$\gamma_{\varepsilon}, \gamma$$ are curves parametrizing $$\Delta_{\varepsilon}, \Delta$$ respectively.
• Since $$\gamma, \gamma_{\varepsilon}$$ are closed and bounded in $${\mathbf{C}}$$, they are compact subsets. Thus it suffices to show that $$f(\gamma_{\varepsilon}(t)) \gamma_{\varepsilon}'(t)$$ converges uniformly to $$f(\gamma(t))\gamma'(t)$$.
• ??
• Case 3: $$\Delta$$ intersects both $${\mathbb{H}}^+$$ and $${\mathbb{H}}^-$$.
• Break into smaller triangles, each of which falls into one of the previous two cases.

## Prove Liouville #complex/exercise/completed

Prove Liouville’s theorem: suppose $$f:{\mathbf{C}}\to{\mathbf{C}}$$ is entire and bounded. Use Cauchy’s formula to prove that $$f'\equiv 0$$ and hence $$f$$ is constant.

The main idea: \begin{align*} {\left\lvert {f'(z)} \right\rvert} &\leq {1\over 2\pi }\oint_R {{\left\lvert {f(\xi)} \right\rvert} \over {\left\lvert {\xi} \right\rvert}^2 } \,d\xi\\ &= {1\over 2\pi }\oint_R {{\left\lvert {f(\xi)} \right\rvert} } R^{-2} \,d\xi\\ &\leq {1\over 2\pi }\oint_R M R^{-2} \,d\xi\\ &= {1\over 2\pi} MR^{-2}\cdot 2\pi R \\ &= MR^{-1} \\ &\overset{R\to\infty}\longrightarrow 0 .\end{align*} So $$f'\equiv 0$$.

## Tie’s Extra Questions Fall 2009 (Fractional residue formula) #complex/exercise/completed

Assume $$f$$ is continuous in the region: \begin{align*} 0 < {\left\lvert {z-a} \right\rvert} \leq R,\quad 0 \leq \operatorname{Arg}(z-a) \leq \beta_0 \qquad \beta_0\in (0, 2\pi] .\end{align*}

and the following limit exists: \begin{align*} \lim_{z\to a}(z-a)f(z) = A .\end{align*} Show that \begin{align*}\lim_{r \rightarrow 0} \int_{\gamma_r} f(z) dz = i A \beta_0 \; , \; \;\end{align*} where \begin{align*} \gamma_r : = \{ z \; | \; z = a + r e^{it}, \; 0 \leq t \leq \beta_0 \}. .\end{align*}

Let $$f$$ be a continuous function in the region \begin{align*} D=\{z {~\mathrel{\Big\vert}~}{\left\lvert {z} \right\rvert}>R, 0\leq \arg z\leq \theta\}\quad\text{where}\quad 1\leq \theta \leq 2\pi .\end{align*} If there exists $$k$$ such that $$\displaystyle{\lim_{z\to\infty} zf(z)=k}$$ for $$z$$ in the region $$D$$. Show that \begin{align*} \lim_{R'\to\infty} \int_{L} f(z) dz=i\theta k ,\end{align*} where $$L$$ is the part of the circle $$|z|=R'$$ which lies in the region $$D$$.

Without loss of generality take $$a=0$$. Since $$zf(z) \to A$$ as $$z\to 0$$, $$z=0$$ is a simple pole of $$f$$ and we can write $$f(z) = c_{-1}z^{-1}+ c_0 + c_1z + \cdots$$. Then \begin{align*} \int_{\gamma_r} f(z)\,dz &= \int_{\gamma_r} \sum_{k\geq -1} c_k z^k \,dz\\ &= \sum_{k\geq -1} c_k \int_{\gamma_r} z^k \,dz\\ &= c_{-1}\int_{\gamma_r}{1\over z}\,dz\\ &= c_{-1}\int_{0}^{\beta_0} {1\over re^{i t}} ire^{it} \,dt\qquad z= re^{it}, \,dz= ire^{it} \,dt\\ &= i c_{-1}\int_{0}^{\beta_0} \,dt\\ &= i c_{-1}\beta_0 .\end{align*} Now use that \begin{align*} zf(z) = c_{-1} + c_0z + \cdots \overset{z\to 0}\longrightarrow c_{-1} = A ,\end{align*} so the integral is $$iA\beta_0$$.

## Spring 2020 HW 2, 2.6.7 #complex/exercise/work

Suppose $$f: {\mathbb{D}}\to {\mathbf{C}}$$ is holomorphic and let $$d \coloneqq\sup_{z, w\in {\mathbb{D}}}{\left\lvert {f(z) - f(w)} \right\rvert}$$ be the diameter of the image of $$f$$. Show that $$2 {\left\lvert {f'(0)} \right\rvert} \leq d$$, and that equality holds iff $$f$$ is linear, so $$f(z) = a_1 z + a_2$$.

Hint: \begin{align*} 2f'(0) = \frac{1}{2\pi i} \int_{{\left\lvert {\xi } \right\rvert}= r} \frac{ f(\xi) - f(-\xi) }{\xi^2} ~d\xi \end{align*} whenever $$0<r<1$$.

## Spring 2020 HW 2, 2.6.8 #complex/exercise/work

Suppose that $$f$$ is holomorphic on the strip $$S = \left\{{x+iy {~\mathrel{\Big\vert}~}x\in {\mathbf{R}},~ -1<y<1}\right\}$$ with $${\left\lvert {f(z)} \right\rvert} \leq A \qty{1 + {\left\lvert {z} \right\rvert}}^\nu$$ for $$\nu$$ some fixed real number. Show that for all $$z\in S$$, for each integer $$n\geq 0$$ there exists an $$A_n \geq 0$$ such that $${\left\lvert {f^{(n)}(x)} \right\rvert} \leq A_n (1 + {\left\lvert {x} \right\rvert})^\nu$$ for all $$x\in {\mathbf{R}}$$.

Hint: Use the Cauchy inequalities.

## Spring 2020 HW 2, 2.6.9 #complex/exercise/work

Let $$\Omega \subset {\mathbf{C}}$$ be open and bounded and $$\phi: \Omega \to \Omega$$ holomorphic. Prove that if there exists a point $$z_0 \in \Omega$$ such that $$\phi(z_0) = z_0$$ and $$\phi'(z_0) = 1$$, then $$\phi$$ is linear.

Hint: assume $$z_0 = 0$$ (explain why this can be done) and write $$\phi(z) = z + a_n z^n + O(z^{n+1})$$ near $$0$$. Let $$\phi_k = \phi \circ \phi \circ \cdots \circ \phi$$ and prove that $$\phi_k(z) = z + ka_nz^n + O(z^{n+1})$$. Apply Cauchy’s inequalities and let $$k\to \infty$$ to conclude.

## Spring 2020 HW 2, 6 #complex/exercise/work

Show by example that there exists a function $$f(z)$$ that is holomorphic on $$\left\{{z\in {\mathbf{C}}{~\mathrel{\Big\vert}~}0 < {\left\lvert {z} \right\rvert} < 1}\right\}$$ and for all $$r<1$$, \begin{align*} \int_{{\left\lvert {z} \right\rvert} = r} f(z) \, dz = 0 ,\end{align*} but $$f$$ is not holomorphic at $$z=0$$.

## Spring 2020 HW 2, 7 #complex/exercise/work

Let $$f$$ be analytic on a region $$R$$ and suppose $$f'(z_0) \neq 0$$ for some $$z_0 \in R$$. Show that if $$C$$ is a circle of sufficiently small radius centered at $$z_0$$, then \begin{align*} \frac{2 \pi i}{f^{\prime}\left(z_{0}\right)}=\int_{C} \frac{d z}{f(z)-f\left(z_{0}\right)} .\end{align*}

Hint: use the inverse function theorem.

## Spring 2020 HW 2, 8 #complex/exercise/work

Assume two functions $$u, b: {\mathbf{R}}^2 \to {\mathbf{R}}$$ have continuous partial derivatives at $$(x_0 ,y_0)$$. Show that $$f \coloneqq u + iv$$ has derivative $$f'(z_0)$$ at $$z_0 = x_0 + iy_0$$ if and only if \begin{align*} \lim _{r \rightarrow 0} \frac{1}{\pi r^{2}} \int_{\left|z-z_{0}\right|=r} f(z) d z=0 .\end{align*}

## Spring 2020 HW 2, 10 #complex/exercise/work

Let $$f(z)$$ be bounded and analytic in $${\mathbf{C}}$$. Let $$a\neq b$$ be any fixed complex numbers. Show that the following limit exists: \begin{align*} \lim_{R\to \infty} \int_{{\left\lvert {z} \right\rvert} = R} {f(z) \over (z-a)(z-b)} \,dz .\end{align*}

Use this to show that $$f(z)$$ must be constant.

## Spring 2020 HW 2, 11 #complex/exercise/work

Suppose $$f(z)$$ is entire and \begin{align*} \lim_{z\to\infty} {f(z) \over z} = 0 .\end{align*}

Show that $$f(z)$$ is a constant.

## Spring 2020 HW 2, 12 #complex/exercise/work

Let $$f$$ be analytic in a domain $$D$$ and $$\gamma$$ be a closed curve in $$D$$. For any $$z_0\in D$$ not on $$\gamma$$, show that \begin{align*} \int_{\gamma} \frac{f^{\prime}(z)}{\left(z-z_{0}\right)} d z=\int_{\gamma} \frac{f(z)}{\left(z-z_{0}\right)^{2}} d z .\end{align*} Give a generalization of this result.

## Spring 2020 HW 2, 13 #complex/exercise/work

Compute \begin{align*} \int_{{\left\lvert {z} \right\rvert} = 1} \qty{z + {1\over z}}^{2n} {dz \over z} \end{align*} and use it to show that \begin{align*} \int_0^{2\pi} \cos^{2n}(\theta) \, d\theta = 2\pi \qty{1\cdot 3 \cdot 5 \cdots (2n-1) \over 2 \cdot 4 \cdot 6 \cdots (2n)} .\end{align*}

#complex/exercise/completed #complex/exercise/work #complex/qual/completed