Notes - Complex Analysis MT23, Misc

Created: January 31, 2024

Flashcards

Suppose $z \in \mathbb C$ with $\text{Im }z > 0$. Quickly deduce that $\text{Im } \frac 1 z < 0$.

Let $z = a+bi$ with $b > 0$. Then

\[\frac 1 z = \frac{a-bi}{|a|^2 + |b|^2}\]

which clearly has imaginary part less than $0$.

Rewrite $1 - e^{i\theta}$ in a way that expresses it purely in the form $r e^{i\theta}$.
\[\begin{aligned} 1 - e^{i\theta} &= 1 - \cos \theta - i\sin\theta \\\\ &= 1 - \left(1 - 2\sin^2 \frac \theta 2\right) - 2i\sin \frac \theta 2 \cos \frac \theta 2\\\\ &= 2\sin^2 \frac \theta 2 - 2i\sin \frac \theta 2 \cos \frac \theta 2 \\\\ &= 2\sin \frac \theta 2 (-i)\left( \cos \frac \theta 2 + i \sin \frac \theta 2 \right) \\\\ &= 2 \sin \frac \theta 2 e^{i (\theta / 2 - \pi / 4)} \end{aligned}\]

Quickly use the binomial theorem to show that

\[\frac{1}{\sqrt{1-z} } = \sum^{\infty}_{n = 0} {2n \choose n} \left(\frac{z}{4}\right)^n\]
\[\begin{aligned} \frac{1}{\sqrt{1 - z} } &= \sum^\infty_{n = 0} {-\frac 1 2 \choose n} (-z)^n \\\\ &= \sum^\infty_{n = 0} \frac{\left(-\frac 1 2\right) \times \left(-\frac 3 2\right) \times \cdots \times \left(-\frac 1 2 - n + 1\right)}{n!} (-z)^k \\\\ &= \sum^\infty_{n = 0} \frac{\left(-\frac 1 2\right) \times \left(-\frac 3 2\right) \times \cdots \times \left(-\frac{2n-1}{2}\right)}{n!} (-z)^k \\\\ &= \sum^\infty_{n = 0} \frac{1 \times 3 \times \cdots \times (2k -1)}{n! 2^n} z^k \\\\ &= \sum^\infty_{n = 0} \frac{(2n)!}{(2 \times 4 \cdots \times 2n) n! 2^n} z^k \\\\ &= \sum^\infty_{n = 0} \frac{(2n)!}{(n! 2^n) n! 2^n} z^k \\\\ &= \sum^\infty_{n = 0} \frac{(2n)!}{(n!)(2n - n)!} \frac{1}{4^n} z^k \\\\ &= \sum^\infty_{n = 0} {2n \choose n} \left(\frac z 4\right)^n\\\\ \end{aligned}\]

Suppose that $k^{z} := \exp(z \log k)$. How could you deduce that

\[\sum^\infty_{k = 1} \frac{1}{k^z}\]

converges to a holomorphic function on $\{z \in \mathbb C \mid \text{Re}(z) > 1\}$?

This is an application of the $M$-test, with the slight issue that we need each $\left \vert \frac{1}{k^z}\right \vert \le M _ k$ to be independent of $z$. Consider

\[f_ n(z) := \sum_ {k = 1}^n \frac{1}{k^z}\]

and fix some $t > 1$. Let $H _ t = \{x + iy \mid x > t, y \in \mathbb R\}$. Then define

\[M_k := \frac{1}{k^t}\]

so that

\[\left|\frac{1}{k^z}\right| \le M_k\]

Since

\[\sum M_k < \infty\]

it follows that $f _ n \to f$ uniformly for some $f$ on $H _ t$, and that $f$ is holomorphic on $H _ t$. Since $t > 1$ is arbitrary, it follows the sum converges to a holomorphic function on

\[\bigcup_{t > 1} H_t = \\{z \in \mathbb C \mid \text{Re}(z) > 1\\}\]

The Bessel functions $J _ n(w)$ satisfy the identity

\[\exp\left( \frac w 2 \left( z - \frac 1 z \right) \right) = \sum^\infty_{n = 0} J_n(w) z^n\]

for $z, w \ne 0$. How could you use this to deduce that

\[J_0(w) = \sum^\infty_{k = 0} (-1)^k \frac{w^{2k} }{(k!)^2 4^k}\]

?
  • Expand out the LHS to become a product of two power series
  • Collect $z^0$ terms to get the equality.

The Bessel functions $J _ n(w)$ satisfy the identity

\[\exp\left( \frac w 2 \left( z - \frac 1 z \right) \right) = \sum^\infty_{n = 0} J_n(w) z^n\]

for $z, w \ne 0$. How could you use this to show that

\[J_n(w) = \frac{1}{2\pi} \int^\pi_{-\pi} \cos(w \sin \theta - n \theta) \text d\theta\]

?

Note that the function on the RHS is holomorphic on an annulus about $0$. Then use Laurent’s theorem to get an expression for the coefficients of the power series in terms of an integral, and this integral ends up working.

Proofs



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