Articles of complex analysis

A boundary version of Cauchy's Integral Theorem

In Complex Analysis by Kodaira, a more powerful version of Cauchy’s Integral Theorem (and consequently formula) was proven. The result generalizes the theorem to the boundary of an open set as follows Let $D$ be a domain and $\overline{D}$ be it’s closure. Suppose that $f:\overline{D} \rightarrow \mathbb{C}$ is holomorphic in $D$ and continuous on $\overline{D}$. […]

Approximating a polynomial to a piece-wise function

I was going through my introductory calculus book(for high-school student) by a Russian author(N.Psikunov) where I encountered a theorem without proof named: Weierstrass Approximation Theorem So how can we apply this theorem and apply it to piece wise functions?(any general approach?)(say a simple function like |x|)

Compute $\int_\Gamma \frac{e^\frac{1}{z}}{z-1}dz$, where $\Gamma$ is the circle $|z-1|\le\frac{3}{2}$, positively oriented.

Compute $\int_\Gamma \frac{e^\frac{1}{z}}{z-1}dz$, where $\Gamma$ is the circle $|z-1|\le\frac{3}{2}$, positively oriented. The numerator is not analytic in $\Gamma$ so we can’t use Cauchy integral formula. I’m thinking maybe I shold use residue theorem. But then I have these two questions: Should I look for the Laurent series of $e^\frac{1}{z}$ around $z=0$? What if I look […]

Find all entire functions with $f(z) = f(\frac{1}{z})$

Find all entire functions with $f(z) = f(\frac{1}{z})$, for all $z\ne 0$. I tried to use the power series of $f$ but this did not help. I also tried to use Liouville’s Theorem on $\frac{f(z)}{f(\frac{1}{z})}$ but then I have to deal with possible singularities. Can someone help me with this problem?

Character space of $\mathcal P(K)$

Let $K$ be a compact subset of $\Bbb C$. Let $\mathcal P(K):=$ closure in $||\cdot||_{\infty}$ of all complex polynomials on $K$. Note that $\mathcal P(K) = C(K)$ do not generally hold since Stone-Weierstrass do not apply to complex polynomial. What is the character space $\Phi_{\mathcal P(K)}$ of $\mathcal P(K)$? It is commonly known that $\Phi_{C(K)}=\{ […]

Stitching two analytic functions?

Let $f$ be an analytic function on the open unit disc and let $g$ be an analytic function on the complement of its closure. Further assume that the two functions have a the same continuous limit on the common boundary of their domain. Is it possible to build an entire function that settles with $f$ […]

Evaluate $\int \cos^2\theta\space d\theta$ using complex numbers.

Evaluate $$\int \cos^2\theta \space d\theta$$using complex numbers. My attempt: $\displaystyle\int \cos^2\theta\space d\theta=\displaystyle\int\left(\dfrac{e^{i\theta}+e^{-i\theta}}{2}\right)^2\space d\theta$ Then I tried to simplify the integrand: $\displaystyle\int\dfrac{1}{4}(e^{i\theta}+e^{-i\theta})^2\space d\theta \\=\dfrac{1}{4}\displaystyle\int(e^{2i\theta}+2e^{i\theta}\cdot e^{-i\theta}+e^{-2i\theta}\space d\theta) \\=\dfrac{1}{4}\displaystyle\int(e^{2i\theta}+e^{-2i\theta}+2)\space d\theta$ At this point, I’m not sure how to proceed though. I haven’t learned how to integrate terms with $i$ in them yet.

How to solve a complex polynomial?

Solve: $$ z^3 – 3z^2 + 6z – 4 = 0$$ How do I solve this? Can I do it by basically letting $ z = x + iy$ such that $ i = \sqrt{-1}$ and $ x, y \in \mathbf R $ and then substitute that into the equation and get a crazy long […]

Fourier series without Fourier analysis techniques

It is known that one can sometimes derive certain Fourier series without alluding to the methods of Fourier analysis. It is often done using complex analysis. There is a way of deriving the formula $$\sum_{k = 1}^\infty \frac{\sin(kz)}{k} = \frac{\pi – z}{2}$$ using complex analysis for some $z$. In other words, it can be shown […]

Can someone please explain the following definition of $\ln(e^z)$

I noticed someone do this from one of the questions is asked on here i had: $$e^z = -0.5$$ $$e^z = 0.5e^{i\pi}$$ which magically became: $$z = \ln\left(\frac12\right) + iπ + 2ikπ$$ does this mean that if i have: $$e^z = -r = re^{i\pi} = \ln(r) + iπ + 2ikπ$$ Thanks for any help you […]