$$\begin{aligned} \pi(x) \thicksim \frac{x}{\ln x} \quad \text{as } x \rightarrow \infty \\ \end{aligned}$$

where $\pi(x)$ is the number of primes less than or equal to $x$. A nice video.

# posted by rot13(Unafba Pune) @ 5:04 PM 0 comments

Let $f: U \rightarrow \mathbb{C}$ be analytic and let $\{|z - z_0| < r \subset U\}$. Then in this disk, $f$ has a power series representation:

$$\begin{aligned} f(z) = \sum_{k=0}^\infty a_k (z - z_0)^k, \quad |z - z_0| < r, \quad \text{where } a_k = \frac{f^{(k)}(z_0)}{k!} \\ \end{aligned}$$

The radius of convergence of this power series is $R \ge r$. Note this means an analytic function is determined entirely in a disk by all it's derivatives $f^{(k)}(z_0)$ at the center $z_0$ of the disk!

In other words, if $f$ and $g$ are analytic in $\{|z - z_0| < r\}$ and if $f^{(k)}(z_0) = g^{(k)}(z_0)$ for all k, then $f(z) = g(z)$ for all $z$ in $\{|z - z_0| < r\}$ !

More at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 4:34 PM 0 comments

For $s \in \mathbb{C}$ with $\textrm{Re } s > 1$, the zeta function is defined as

$$\begin{aligned} \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} \end{aligned}$$

We can readily see $\zeta(s)$ converges absolutely in $\{\textrm{Re } s > 1\}$. What's interesting is the convergence is uniform in $\{\textrm{Re } s > r\}$ for any $r > 1$, which can be used to show that $\zeta(s)$ is analytic in $\{\textrm{Re } s > 1\}$.

Riemann was able to show that $\zeta(s)$ has an analytic continuation into $\mathbb{C} \setminus \{1\}$, and this continuation satisfies that $\zeta(s) \rightarrow \infty$ as $s \rightarrow 1$.

Some interesting facts:

• The only zeros of the zeta function (ie when $\zeta(s) = 0$) outside the strip $\{0 \le \textrm{Re } s \le 1\}$ are at the negative even integers, $-2, -4, -6, \cdots$, which are called the "trivial zeros".
• Zeta has no zeros on the line $\{ \textrm{Re } s = 1\}$, nor the line $\{ \textrm{Re } s = 0\}$.

Riemann Hypothesis

    In the strip $\{0 \le \textrm{Re } s \le 1\}$, all zeros of $\zeta$ are on the line $\displaystyle \{ \textrm{Re } s = \frac{1}{2} \}$

which has significant implication on the distribution of prime numbers and the growth of many important arithmetic functions, but remains unproved.

Amazingly,

$$\begin{aligned} \zeta(s) = \prod_p \frac{1}{1 - p^{-s}} \end{aligned}$$

where the product is over all primes ! Why ? Observe

$$\begin{aligned} \zeta(s) &= \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots \\ &= \big(\frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{4^s} + \frac{1}{8^s} + \cdots\big)\big(\frac{1}{1^s} + \frac{1}{3^s} + \frac{1}{9^s} + \cdots\big)\big(\frac{1}{1^s} + \frac{1}{5^s} + \frac{1}{25^s} + \cdots\big) \cdots \\ \end{aligned}$$

More at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 10:23 AM 0 comments

Writing $z = r \,e^{i \,\theta}$, on one hand, we can easily split the series into real and imaginary parts

$$\begin{aligned} \sum_{k=0}^\infty z^k = \sum_{k=0}^\infty r^k \cos(k\,\theta) + i \sum_{k=0}^\infty r^k \sin(k\,\theta) \end{aligned}$$

On the other hand,

$$\begin{aligned} \frac{1}{1-z} = \frac{1}{1 - r\,e^{i \,\theta}} = \cdots = \frac{1 - r \cos \theta + i r \sin \theta}{1 - 2r \cos \theta + r^2} \\ \end{aligned}$$

Check it! This necessarily means, in the peculiar case when $|z| < 1$, ie when $r < 1$,

$$\begin{aligned} \sum_{k=0}^\infty r^k \cos(k \, \theta) = \frac{1 - r \cos \theta}{1 - 2r \cos \theta + r^2} \\ \sum_{k=0}^\infty r^k \sin(k \, \theta) = \frac{r \sin \theta}{1 - 2r \cos \theta + r^2} \\ \end{aligned}$$

!

More at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 8:02 AM 0 comments

Let $f$ be analytic in a domain $D$ and suppose there exists a point $z_0 \in D$ such that $\left|\,f(z)\,\right| \le \left|\,f(z_0)\,\right| \, \forall z \in D$. Then $f$ is constant in $D$ !

As a consequence, if $D \subset \mathbb{C}$ is a bounded domain, and if $f: \overline D \mapsto \mathbb{C}$ is continuous on $\overline D$ and analytic in $D$, then $\left|\, f \,\right|$ reaches it's maximum on $\partial D$.

(If I understand the notation correctly, $\overline D$ refers to the union of both the open domain $D$ and it's boundary, whereas $\partial D$ refers to only the boundary.)

More at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 3:49 PM 0 comments

If $f$ is an entire function (ie analytic in the complex plane) and is bounded, then f must be a constant

! Not hard to prove using Cauchy's Estimate. For example, since $\sin z$ is not a constant, by Lioville's Theorem we know that there must be some points in the complex plane that $\sin z$ will go off to infinity.

A rather counter-intuitive example is when $f$ is an entire function with $u(z) \le 0 \, \forall z \in \mathbb{C}$, then $f$ must be constant ! The proof starts with considering the function $g(z) = e^{f(z)}$.

Furthermore, Liouville's Theorem can be used to prove (via contradiction) the Fundamental Theorem of Algebra, which states that any polynomial

$$\begin{aligned} p(z) = a_0 + a_1z + \cdots + a_n z^n \end{aligned}$$

with $a_0, \cdots, a_n \in \mathbb{C}$ with $n \ge 1$ and $a_n \not = 0$ has a zero in $\mathbb{C}$. In other words, there exists $z_0 \in \mathbb{C}$ such that $p(z_0) = 0$. As a consequence, polynomials can always be factored in $\mathbb{C}$ (but this is not the case in $\mathbb{R}$) ! So

$$\begin{aligned} p(z) = a_n(z - z_1)(z - z_2) \cdots (z - z_n) \end{aligned}$$

where $z_1, \cdots , z_n \in \mathbb{C}$ are the zeros of $p$. A typical example would be $p(z) = z^2 + 1$.

More at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 3:40 PM 0 comments

Suppose $f$ is analytic in an open set that contains $\overline{B_r(z_0)}$, and $\left|\, f(z) \,\right| \le m$ holds on $\partial B_r(z_0)$ for some constant m. Then for all $k \ge 0$,

$$\begin{aligned} \left|\, f^{(k)}(z_0) \,\right| \le \frac{k!\,m}{r^k} \\ \end{aligned}$$

which turns out to be remarkably not hard to prove based on the Cauchy's Integral Formula. If I understand the notation correctly, $\overline{B_r(z_0)}$ is the disk with radius $r$ centered at $z_0$, whereas $\partial B_r(z_0)$ is the boundary of that disk.

More at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 2:20 PM 0 comments

Let $D$ be a simply connected domain, bounded by a piecewise smooth curve $\gamma$, and let $\,f$ be analytic in a set $U$ that contains the closure of $D$ (ie $D$ and $\gamma$). Then for all $w \in D$,

$$\begin{aligned} f(w) = \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{z - w} \,dz \\ \end{aligned}$$

!

For example, what is the value of

$$\begin{aligned} \oint_{\left| z \right| = 2} \frac{z^2}{z - 1} \,dz \\ \end{aligned}$$

?

This is not an easy integral to evaluate directly, but with Cauchy's formula, the problem can be reduced into simple pattern matching ! Can you see why the value is $2 \pi i$ ? Now, how about

$$\begin{aligned} \oint_{\left| z \right| = 1} \frac{z^2}{z - 2} \,dz \\ \end{aligned}$$

?

Don't get tricked, and hopefully you can see the value is clearly zero.

An amazing consequence of this formula is that as long as $f$ is analytic in an open set $U$, then $f'$ is also analytic in $U$ ! Indeed,

$$\begin{aligned} f'(w) = \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{(z - w)^2} \,dz \\ \end{aligned}$$

And in general,

$$\begin{aligned} f^{(k)}(w) = \frac{k!}{2\pi i} \oint_\gamma \frac{f(z)}{(z - w)^{k+1}} \,dz \\ \end{aligned}$$

!

Doesn't this look déjà vu and perhaps somehow related to the coefficient of the $k$th term of a Taylor series expansion ?

For example, what is the value of

$$\begin{aligned} \oint_{\left| z \right| = 2\pi} \frac{z^2\sin z}{(z - \pi)^3} \,dz \\ \end{aligned}$$

? It becomes easy to realize it's exactly $-4\pi^2i$. The incredible Cauchy !

More at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 9:12 AM 0 comments

Suppose $D$ is a simply connected domain in $\mathbb{C}$, $\,f$ be analytic, and $\gamma: [a, b] \mapsto D$ be a piecewise smooth, closed curve in $D$ (so that $\gamma(b) = \gamma(a)$), then

$$\begin{aligned} \oint_\gamma f(z)\,dz = 0 \\ \end{aligned}$$

!

Corollary

Suppose $\gamma_1$ and $\gamma_2$ are two simple closed curves (ie neither of them intersects itself), oriented clockwise, where $\gamma_2$ is inside $\gamma_1$. Suppose $f$ is analytic in a domain $D$ that contains both curves as well as the region between them, then

$$\begin{aligned} \oint_{\gamma_1} f(z)\,dz = \oint_{\gamma_2} f(z)\,dz \\ \end{aligned}$$

!

For example, suppose $R$ is the rectangle with vertices $−2−i, 2−i, 2+i, −2+i$, the integral

$$\begin{aligned} \oint_{\partial R} \frac{1}{z - z_0} \,dz \\ \end{aligned}$$

can have zero vs. non-zero value depending $z_0$. Can you see why and how to compute the non-zero value ?

More at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 7:22 AM 0 comments

A primitive of $f$ on $D$ is an analytic function $F: D \mapsto \mathbb{C}$ such that $F' = f$ on $D$, where $D \subset \mathbb{C}$.

If $f$ is continuous on a domain $D$ and if $f$ has a primitive $F$ in $D$, then for any curve $\gamma: [a,b] \mapsto D$,

$$\begin{aligned} \int_\gamma f(z)\,dz = F(\gamma(b)) - F(\gamma(a)) \\ \end{aligned}$$

Observe that

• The integral depends only on the initial and terminal points of $\gamma$ !
• The critical assumption of $f$ having a primitive in $d$.

When does $f$ have a primitive ?

By Goursat Theorem, if $D$ is a simply connected domain in $\mathbb{C}$, and $f$ is analytic in $D$, then $f$ has a primitive in $D$.

More at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 7:28 AM 0 comments

Suppose $\gamma$ is a curve and $f$ is continuous on $\gamma$, then

$$\begin{aligned} \left| \int_\gamma f(z) \, dz \,\right|\, \le \int_\gamma \left|\, f(z) \,\right|\, \left| \, dz \,\right| \\ \end{aligned}$$

In particular, if $\left|\, f(z) \,\right| \le M$ on $\gamma$, then

$$\begin{aligned} \left| \int_\gamma f(z) \, dz \,\right|\, \le M \int_\gamma \, \left| \, dz \,\right| = M \cdot \text{length}(\gamma) \\ \end{aligned}$$

!

For example, suppose $\gamma(t) = t + it, 0 \le t \le 1$, what is the upper bound of $\displaystyle \int_\gamma z^2\,dz$ ? It should be easy to check that it's $2\sqrt{2}$. Turns out a better and indeed exact estimate is $\displaystyle \frac{2}{3}\sqrt{2}$, using the first part of the theorem !

More at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 7:36 PM 0 comments

Given a curve $\gamma: [a, b] \mapsto \mathbb{C}$, how do we find it's length ?

$$\begin{aligned} \text{length}(\gamma) &\approx \sum_{j=0}^n \,\left|\, \gamma(t_{j+1}) - \gamma(t_j) \,\right|\, \\ &\approx \sum_{j=0}^n \frac{\,\left|\, \gamma(t_{j+1}) - \gamma(t_j) \,\right|\,}{t_{j+1} - t_j} (t_{j+1} - t_j) \\ \text{length}(\gamma) &= \int_a^b \,\left|\, \gamma'(t) \,\right|\, dt \qquad\\ \end{aligned}$$

!

Here is the definition of Integration with respect to Arc Length:

$$\begin{aligned} \int_\gamma f(z)\,\left|\, dz \,\right| = \int_a^b f(\gamma(t)) \,\left|\, \gamma'(t) \,\right|\,dt \end{aligned}$$

Note when $f(z) = 1$,

$$\begin{aligned} \int_\gamma f(z)\,\left|\, dz \,\right| = \int_\gamma \,\left|\, dz \,\right| = \int_a^b \,\left|\, \gamma'(t) \,\right|\,dt = \text{length}(\gamma) \end{aligned}$$

!

More at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 8:28 AM 0 comments

Let $f: [a, b] \rightarrow \mathbb{R}$ be continuous. Then

$$\begin{aligned} \int_a^b f(t)\,dt &= \lim_{n \rightarrow \infty} \sum_{j=0}^{n-1} f(t_j)(t_{j+1} - t_j) \\ F(x) &= \int_a^x f(t)\,dt \\ F'(x) &= f(x) \,\,\, \text{ for } x \in [a, b] \\ \end{aligned}$$

Let $\gamma$ be a curve ie a smooth or piecewise smooth function in $\mathbb{C}$:

$$\begin{aligned} &\gamma: [a, b] \rightarrow \mathbb{C} \\ &\gamma(t) = x(t) + iy(t) \\ \end{aligned}$$

If $f$ is complex-valued on $\gamma$,

$$\begin{aligned} \int_\gamma f(z)\,dz &= \lim_{n \rightarrow \infty} \sum_{j=0}^{n-1} f(z_j)(z_{j+1} - z_j) \\ \end{aligned}$$

where $z_j = \gamma(t_j)$.

But

$$\begin{aligned} \int_\gamma f(z)\,dz &= \int_a^b f(\gamma(t)) \, \gamma'(t)\,dt \\ \end{aligned}$$

Why ? This is easy to see, as

$$\begin{aligned} z &= \gamma(t) \\ dz &= \gamma'(t)\,dt \\ \end{aligned}$$

Such integration over the curve is called the Path Integral. Furthermore, if $g(t) = u(t) + iv(t)$, then

$$\begin{aligned} \int_a^b g(t)\,dt &= \int_a^b u(t) \,dt + i\int_a^b v(t) \,dt \\ \end{aligned}$$

What is $\displaystyle \int_{\left| z \right| = 1} \frac{1}{z}\,dz$ ? Kind of cute.

Turns out, in general, $\displaystyle \int_{\left| z \right| = 1} z^m \,dz = 2\pi i$ when $m = -1$ but zero otherwise !

More at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 8:11 AM 0 comments

Linear Induction

Tree Induction

Structural Induction

The most general form of induction !

$$\begin{aligned} &\phi[a] \\ &\phi[b] \\ &\forall \lambda. \forall \mu.((\phi[\lambda] \land \phi[\mu]) \implies \phi[c(\lambda, \mu)]) \\ \hline \\ &\forall \nu.\phi[\nu] \\ \end{aligned}$$

More at Introduction to Logic.

# posted by rot13(Unafba Pune) @ 8:11 AM 0 comments

Induction for finite languages is trivial. We simply use the Domain Closure rule of inference. For a language with object constants $\sigma_1, \cdots, \sigma_n$,

$$\begin{aligned} \phi[\sigma_1] \\ \cdots \,\,\, \\ \phi[\sigma_n] \\ \hline \\ \forall \nu.\phi[\nu] \end{aligned}$$

If we believe a schema is true for every instance, then we can infer a universally quantified version of that schema.

More at Introduction to Logic.

# posted by rot13(Unafba Pune) @ 8:42 PM 0 comments

What conformal mappings are there of the form $f: \mathbb{D} \mapsto D$, where $\mathbb{D} = B_1(0)$ is the unit disk and $D \in \mathbb{C}$ ?

If $D$ is a simply connected domain (ie open, connected, no holes) in the complex plane, but not the entire complex plane, then there is a conformal map (ie analytic, one-to-one, onto) of $D$ onto the open unit disk $\mathbb{D}$.

We say that "$D$ is conformally equivalent to $\mathbb{D}$".

To find a unique conformal mapping $\, f$ from $D$ to $\mathbb{D}$, we need to specify "3 real parameters".

For example,

• We can use Möbius transformation that maps the upper half plane $\mathbb{D}^+$ to $\mathbb{D}$, by examining the mappings from $0, 1, \infty$ to $1, i, -1$. This would lead to

  $$\begin{aligned} f(z) = \frac{-z + i}{z + i} \end{aligned}$$

  Don't forget to check.

• $f$ would map the first quadrant $Q$ of the complex plane to $\mathbb{D}^+$. Again, check the mappings from $0, i, \infty$ (ie. the imaginary axis) to the line through $1, 0, -1$ (ie. the real axis).
• $g(z) = z^2$ is injective and analytic in $Q$, and $g$ maps $Q$ conformally onto $D$. Should check.

# posted by rot13(Unafba Pune) @ 5:35 PM 0 comments

Every Möbius transformation is the composition of mappings of the type:

$$\begin{aligned} z &\mapsto az \qquad \text{(rotation & dilation)} \\ z &\mapsto z + b \quad \text{(translation)} \\ z &\mapsto \frac{1}{z} \qquad \text{(inversion)} \\ \end{aligned}$$

which all map circles and lines to circles or lines.

More at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 8:43 AM 0 comments

1. The composition of two Möbius transformations is a Möbius transformation, and so is the inverse.
2. Given three distinct points $z_1, z_2, z_3$ and three distinct points $w_1, w_2, w_3$, there exists a unique Möbius transformation $f : \hat{\mathbb{C}} \mapsto \hat{\mathbb{C}}$ that maps $z_j$ to $w_j , j = 1, 2, 3$:

$$\begin{aligned} f_2^{-1} \circ f_1 \end{aligned}$$

where $f_1$ maps $z_1, z_2, z_3$ to $0, 1, \infty$, and $f_2$ maps $w_1, w_2, w_3$ to $0, 1, \infty$.

More at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 12:47 AM 0 comments

$$\begin{aligned} \therefore \text{Re } w & = \frac{1}{2} \\ \end{aligned}$$

!

which happens to be the image of the circle $K = \{z: \left| z - 1 \right| = 1 \}$.

# posted by rot13(Unafba Pune) @ 8:20 AM 0 comments

What does $\left| z - 3 \right| = 1$ look like ? A circle of radius 1, centered at 3.

What does $\displaystyle \left| \frac{1}{z} - 3 \right| = 1$ look like ?

$$\begin{aligned} \left| \frac{1}{z} - 3 \right| &= 1 \\ \left| \frac{1-3z}{z} \right|^2 &= 1 \\ \left| 1-3z \right|^2 &= \left| z \right|^2 \\ (1-3z)\overline{(1-3z)} &= z \overline z \\ (1-3z)(1-3\overline z) &= z \overline z \\ 1-3z-3\overline z + 9z\overline z &= z \overline z \\ 8z\overline z - 3z - 3\overline z + 1 &= 0 \\ z\overline z - \frac{3}{8}z - \frac{3}{8} \overline z + \frac{1}{8} &= 0 \\ (z - \frac{3}{8})(\overline z - \frac{3}{8}) - \frac{3^2}{8^2} + \frac{1}{8} &= 0 \\ \left| z - \frac{3}{8} \right|^2 &= \frac{1}{64} \\ \left| z - \frac{3}{8} \right| &= \frac{1}{8} \\ \end{aligned}$$

$\therefore$ A circle of radius $\displaystyle \frac{1}{8}$, centered at $\displaystyle \frac{3}{8}$.

More at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 3:26 AM 0 comments

Möbius Transformation is a function of the form:

$$\begin{aligned} f(z) = \frac{az + b}{cz + d} \end{aligned}$$

where $a, b, c, d \in \mathbb{C}$ such that $ad - bc \not= 0$. (This looks surprisingly familiar to the determinant of a 2 by 2 invertible matrix !)

Interestingly,

• $f$ is a mapping from the extended complex plane $\hat{\mathbb{C}} = \mathbb{C} \cup \{\infty\}$ to $\hat{\mathbb{C}}$.
• $f$ is one-to-one and onto from $\hat{\mathbb{C}}$ to $\hat{\mathbb{C}}$ !
• $f$ does not uniquely determine the parameters $a, b, c, d$, for if we multiply each parameter by a constant $k \not= 0$, we obtain the same mapping.
• $\displaystyle f'(z) = \frac{ad - bc}{(cz + d)^2}$. Therefore, $f$ is a conformal mapping from $\hat{\mathbb{C}}$ to $\hat{\mathbb{C}}$.
• Möbius Transformation are the only conformal mappings from $\hat{\mathbb{C}}$ to $\hat{\mathbb{C}}$ !
• Every Möbius Transformation maps circles and lines to circles or lines !
• Given three distinct points $z_1,\, z_2,\, z_3 \in \hat{\mathbb{C}}$, there exists a unique Möbius Transformation $f$ such that $f(z_1) = 0,\, f(z_2) = 1,$ and $\, f(z_3) = \infty$ !

  Here it is:

  $$\begin{aligned} f(z) = \frac{z - z_1}{z - z_3} \cdot \frac{z_2 - z_3}{z_2 - z_1} \end{aligned}$$

Möbius Transformation is called Affine transformation when $c = 0$, and $d=1$. Interestingly,

• Affine transformations map $\infty$ to $\infty$ and therefore map $\mathbb{C}$ to $\mathbb{C}$.
• Hence Affine transformations are conformal mappings from $\mathbb{C}$ to $\mathbb{C}$.
• Affine transformations are the only conformal mappings from $\mathbb{C}$ to $\mathbb{C}$ !
• When $b = 0,\, f(z) = az$ which is a rotation and dilation.
• When $a = 1,\, f(z) = z + b$ which is a translation.
• When $\displaystyle a = 0, b = 1, c = 1, d = 0, f(z) = \frac{1}{z}$ which is an inversion.
  • $f$ interchanges outside and inside of the unit circle
  • a circle centered at $0$ is clearly mapped to a circle, centered at $0$, of reciprocal radius.

More at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 8:44 AM 0 comments

Intuitively, a conformal mapping is a “mapping that preserves angles between curves”.

If $f:U \rightarrow \mathbb{C}$ is analytic and if $z_0 \in U$ such that $f′(z_0) \not = 0$, then $f$ is conformal at $z_0$ !

Observe the contrapositive: if a function $f$ is not conformal, then either $f$ is not analytic or $f′(z_0) = 0$. This means it would fail the Cauchy-Riemann equations ! The function $\overline z$ is a good example.

# posted by rot13(Unafba Pune) @ 11:21 PM 0 comments

What does the graph look like ?

In Desmos, choose polar form and enter:

$$\begin{aligned} r = \theta \end{aligned}$$

# posted by rot13(Unafba Pune) @ 8:29 AM 0 comments

$$\begin{aligned} z &= \left| z \right| \cdot e^{i \cdot \text{Arg}(z)}, \qquad\qquad\qquad\text{in polar form} \\ \text{Log}(z) &= \ln{\left| z \right|} + i \cdot \text{Arg}(z), \qquad\quad\quad\text{the principal branch of logarithm} \\ \text{log}(z) &= \ln{\left| z \right|} + i\left(\text{Arg}(z) + 2k\pi\right), \quad\,\,\text{where } k \in \mathbb{Z} \\ \text{log}(z) &= \ln{\left| z \right|} + i\cdot\text{arg}(z), \qquad\quad\quad\, \text{a multi-valued function} \\ \end{aligned}$$

Observe Log($z$):

$$\begin{aligned} z &\mapsto \left| z \right|, \qquad\quad\,\,\,\text{continuous in }\mathbb{C} \\ z &\mapsto \ln{\left| z \right|}, \qquad\,\,\;\text{continuous in }\mathbb{C} \setminus \{0\} \\ z &\mapsto \text{Arg}(z), \qquad\text{continuous in }\mathbb{C} \setminus (-\infty,0] \\ \end{aligned}$$

Recall $-\pi \lt \text{Arg}(z) \le \pi$. So as $z$ approaches the x-axis to the left of the origin, it would result in $\pi$ for Arg($z$) if from above, but $-\pi$ if from below.

$$\begin{aligned} \therefore \text{Log}(z) \text{ is continuous in } \mathbb{C} \setminus (-\infty,0]. \end{aligned}$$

Similar line of arguments can be used to reason that the domain of $z^2$ is continuous on $\mathbb{C}$, but the domain of $\sqrt{z}$ is only continuous on $\mathbb{C} \setminus (-\infty,0]$. Do you see how ?

Quiz: what is $\displaystyle\text{Log}(-e^xi)$ ?

Observe $\displaystyle z = -e^xi \,$ lies on the imaginary axis and therefore $\displaystyle\text{Arg}(z) = -\frac{\pi}{2}$

$$\begin{aligned} -e^x i &= e^x(0 - i) = e^x e^{-\frac{\pi}{2}i} \\ \text{Log}(-e^xi) &= x -\frac{\pi}{2}i \\ \end{aligned}$$

$\Box$

More at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 8:16 AM 0 comments

$$\begin{aligned} \cos x &= \frac{e^{ix} + e^{-ix}}{2} \\ \sin x &= \frac{e^{ix} - e^{-ix}}{2i} \\ \end{aligned}$$

This works whether $x$ is $\theta \in \mathbb{R}$ or $z \in \mathbb{C}$. Do you see why ?

Furthermore, it's not difficult to show their relationships with hyperbolic functions:

$$\begin{aligned} \sin z &= \sin(x + iy) = \sin x\cosh y + i\cos x\sinh y \\ \cos z &= \cos(x + iy) = \cos x\cosh y\, – i\sin x\sinh y \\ \end{aligned}$$

Why would this be useful ? It's particularly handy when there is the need to evaluate some form of conjugate or absolute value, where it's helpful to break down the function into real and imaginary parts.

Did I mention already ? Euler, the truly incredible.

# posted by rot13(Unafba Pune) @ 8:20 AM 0 comments

It's rather easy to show that:

$$\begin{aligned} \left|z\right|^2 = z \cdot \overline z \,\,\,\,\,\text{ where } z = x + iy \end{aligned}$$

How about:

$$\begin{aligned} \left | z + 1 \right |^2 = (z + 1) \cdot (\,\overline z + 1) \end{aligned}$$

?

As a challenge, consider:

$$\begin{aligned} Re(e^{\frac{z}{z+1}}) = \left( e^{\frac{\left|z\right|^2+x}{\left|z+1\right|^2}}\right) \cdot \cos\left(\frac{y}{\left| z+1 \right| ^2}\right) \end{aligned}$$

where the left hand side refers to the real part of $\displaystyle e^{\frac{z}{z+1}}$. Does the equation hold ? Note in particular the right hand side is a real number, but in terms of numbers in the complex plane !

More at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 11:44 PM 0 comments

In other words, $e$ raised to the power of the conjugate of $z$ is equal to the conjugate of $e^z$, where $z \in \mathbb{C}$. Not immediately obvious but not hard to show.

# posted by rot13(Unafba Pune) @ 8:18 AM 0 comments

The Cauchy-Riemann Equations are useful for checking differentiability in the complex plane, and computing them.

Suppose

$$\begin{aligned} f(z) = u(x,y) + i \cdot v(x,y) \end{aligned}$$

is differentiable at $z_0$. Then

1. the partial derivatives $u_x, u_y, v_x, v_y$ exist at $z_0$,
2. $u_x = v_y$, and
3. $u_y = -v_x$.

Conversely, suppose

$$\begin{aligned} f = u + i \cdot v \end{aligned}$$

is defined on a domain $D \in \mathbb{C}$. Then $f$ is analytic in $D$ iff $u(x,y)$ and $v(x,y)$ have continuous first partial derivatives on $D$ that satisfies the Cauchy-Riemann equations. The key point is continuity is necessary for the guarantee to work.

More info at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 10:48 PM 0 comments

What's peculiar about the function

$$\begin{aligned} f(x) = x \cdot \lvert x \rvert^2 \end{aligned}$$

in terms of differentiability ? More on this later ...

# posted by rot13(Unafba Pune) @ 8:10 AM 0 comments

A complex function $f$ is analytic if $f$ is differentiable at each point $z \in U$ where $U$ is an open set in $\mathbb{C}$. A function which is analytic in all of $\mathbb{C}$ is called an entire function.

Consider the complex functions:

$$\begin{aligned} f(z) &= \text{Re }(z) \\ f(z) &= \bar z \\ f(z) &= \lvert z \rvert^2 \end{aligned}$$

Can you see why they are not analytic ?

# posted by rot13(Unafba Pune) @ 7:57 PM 0 comments

Is this a correct Fitch proof ?

	1.	$p(x)$		Assumption
	2.	$p(x) \lor \lnot p(x)$		Or Introduction 1
	3.	$\forall x. p(x) \lor \lnot p(x)$		Universal Introduction 2
	4.	$p(x) \implies \forall x. p(x) \lor \lnot p(x)$		Implication Introduction 1-3

More information at Introduction to Logic.

# posted by rot13(Unafba Pune) @ 4:15 PM 0 comments

Is this a correct single step inference in the Fitch system for Relational Logic ?

$$\begin{aligned} \frac{\forall x . \forall y . (p(x) \wedge q(y))}{\forall y.(p(y) \wedge q(y))} \end{aligned}$$

More information at Introduction to Logic.

# posted by rot13(Unafba Pune) @ 4:00 PM 0 comments

It can be shown that if

$$\begin{aligned} \forall x .(\exists y. p(x,y) \implies q(x)) \end{aligned}$$

holds, then

$$\begin{aligned} \forall x .\forall y.(p(x,y) \implies q(x)) \end{aligned}$$

holds. Do you see why ?

# posted by rot13(Unafba Pune) @ 11:53 PM 0 comments