Tag Archives: Pure Maths

Complex Analysis Assignment 1

My first complex analysis assignment has been marked and returned. I don't think I've ever felt the urge quite so much to learn from my mistakes.

Consequently there has been quite a lot of post-assignment learning... :/

This assignment featured a very brief introduction to complex numbers as a refresher, then broadly covered complex functions, the concept of continuity and complex differentiation.

So in no particular order, below are some notes on mistakes I made and how I could've avoided them! There's a lot to reflect on here...


Read questions carefully. One of the first very simple questions read "express z in polar form and determine all fourth roots". I did the second bit, but not the first.


I feel this is a bit "Complex Numbers 101", but the square root sign is defined as the principal square root (of a complex number), i.e. there's no need to calculate the second root.


If you're using the triangle inequality, state it specifically.


Again, this is fairly "Complex Numbers 101", but the polar form of a complex number isn't just a cosine function as the real part, and a sine function as the imaginary part. The arguments to both functions must be identical to qualify as "polar form". ie, you should be able to write the complex number as an exponential form.


Top tip: Be mindful about using identities. In complex analysis there are loads of them and they help a great deal.


When working out the inverse of a complex function, it's important to use your common sense. Part of one inverse I'd calculated had a square root in it. Just by looking at that, you know it could never produce a unique answer (it isn't a one-to-one function).


For another, I had to find the inverse of \text{Log}(3z) and the domain of that inverse. I got this spectacularly wrong. I'd written: given w=\text{Log}(3z), hence z=e^{3w}.

Trick here was to exponentiate each side, leading to e^{w}=3z. But the domain of the inverse isn't affected by the "3" above, the image set of the original function is still \{z: -\pi <\text{Im}z \leq \pi\}.


Some complex functions are very very different to their real equivalents. Case in point: \text{cosh}(x)\neq 0 , \forall x \in \mathbb{R}, but \exists\: z \in \mathbb{C}\: \text{s.t.}\: \text{cosh}(z)=0. Which leads to the next note:


If \text{cosh}(z) is the divisor in a complex quotient, you need to show that it's only 0 for values outside of the given range of the equation (eg |z|<1).


For one question, I had to prove that f(z)=z^{i},\:\: (\text{Re}\:z>0) was continuous. I thought this was easy.

z^{\alpha},\: \alpha \in \mathbb{C} is a basic continuous function on \mathbb{C}-\{x\in\mathbb{R} : x \leq 0\}. So if you let \alpha=i, then f(z) is continuous, right?

Not quite. I had entirely forgotten to state that the given set (\text{Re}\:z>0) is a subset of the set I gave: \mathbb{C}-\{x\in\mathbb{R} : x \leq 0\}.

The answer can appear obvious sometimes, but you have to keep your answer rigorous, otherwise you risk losing half marks or whole marks here and there.


Note:
z^{\alpha} = e^{\alpha Log(z)}
z^{\alpha} \neq e^{z Log(\alpha)}
🙁


For one question I had to prove whether a set was a region or not. For reference, a region is a non-empty, connected, open subset of \mathbb{C}. In the usual manner, if you can prove that any of those three properties don't hold then you've managed to prove that your set isn't a region. Easy.

I realised I could prove a set was closed, and hence not a region. Turns out this was incorrect. A set being "closed" and a set being "not open" hold two completely different definitions, and are seen as different things. I was meant to show it was "not open" as opposed to showing it was "closed".

In other words, mathematically:

Closed is not the same as not-open.
Closed is not the opposite of open.
Not-open is the opposite of open.


Again, here I needed to provide a proof based on the properties of various objects. Given a set that was compact (closed and bounded), I needed to prove that a function f was bounded on that set.

The Boundedness Theorem states that if a function is continuous on a compact set, then that function is bounded on that set.

The function was: f(z) = \frac{1}{7z^{7}-1}

I proved that the given function was continuous on it's domain, but I'd failed to prove it was continuous on the set. Here, I needed to show where the function was undefined, THEN show that those points at which it was undefined all lay outside of the set. So there was quite a lot of work I missed out from this answer.


My simultaneous requirement for the Cauchy-Riemann theorem, AND the Cauchy-Riemann Converse theorem within a proof ended up not flowing very well logically. Once again, I'd jumped ahead with my logic. As soon as I had seen something obvious, I felt the urge to state it immediately.

The Cauchy-Riemann theorem proves that a function is not differentiable at certain points. The Converse theorem then proves that a function IS differentiable on certain points. After using the Cauchy-Riemann theorem, it was extremely obvious where the function was differentiable, so I stated it. Then, as a matter of course, plodded through the Converse theorem to prove it. Complete lack of discipline! 🙂

Complex Functions: Domains, Image Sets and Inverses

I can imagine having to refer to these notes regularly, so I'm putting them here!

Image Sets

  1. State the domain A of f(z)=w.
  2. Rearrange so w is a function of z (to discover the condition under which w remains valid.

e.g., for f(z)=\frac{1}{z-1}:

    \begin{align*} f(A) =&\: \left\{ \frac{1}{z-1}\::\:z\in\mathbb{C}-\{1\}\right\} \\ f(A) =&\: \left\{w=\frac{1}{z-1}\::\:z\neq 1\right\} \\ f(A) =&\: \left\{w\::\: z=\frac{1}{w}+1\::\:z\neq 1\right\} \\ f(A) =&\: \{w\::\: w\neq 0\} \\ f(A) =&\: \mathbb{C}-\{0\} \\ \end{align*}


Domain of Combined Functions

Domain of combined functions are the intersection (A\cap B) of the domains of all component functions and that of the combined function. e.g:

f(z)=\frac{z-1}{z}\:,\:\: z\in\mathbb{Z}-\{0\}

g(z)=\frac{z}{z-1}\:,\:\: z\in\mathbb{Z}-\{1\}

\frac{f\left(z\right)}{g\left(z\right)} = \frac{z^{2}-2z+1}{z^{2}} \:,\:\: z\in\mathbb{Z}-\{0,\:1\}


Domain of Composite Functions

For f and g with domains A and B respectively, the domain of g\circ f is:

A-\{z\::\: f(z)\: \notin\: B\}

e.g., for:

f(z)=\frac{z-1}{z}\:,\:\: z\in\mathbb{C}-\{0\}

g(z)=\frac{z}{z-1}\:,\:\: z\in\mathbb{C}-\{1\}

\text{domain of }f\circ g = \text{domain of }g - \{z\::\:\frac{z}{z-1} \:\notin\: \mathbb{C}-\{0\}\}

\text{domain of }f\circ g = (\mathbb{C}-\{1\} ) - \{z\::\:\frac{z}{z-1} =0\}

\text{domain of }f\circ g = (\mathbb{C}-\{1\} ) - \{0\}

\text{domain of }f\circ g = (\mathbb{C}-\{0,\:1\} )


Inverses

  1. Determine image set of f(z)=w.
  2. Invert f(z) to find a unique z in the domain of f.

For f(z)=\frac{1}{z-1}

f(A) = \{\frac{1}{z-1}\::\:z\in\mathbb{C}-\{1\}\}

f(A) = \{w=\frac{1}{z-1}\::\:z\:\neq\: 1\}

f(A) = \{w\::\: z=\frac{1}{w}+1\:\neq\: 1\}

f(A) = \{w\::\: w\:\neq\: 0\}

f(A) = \mathbb{C}-\{0\}

(all same as above for finding an image set)

z=\frac{1}{w}+1 gives a unique soluition in \mathbb{C}-\{0\}, hence f has a unique inverse rule:

f^{-1}(z)=\frac{1}{w}+1\:,\:\:\:z\in\mathbb{C}-\{0\}

A New Year of Study Begins!

September has come around already!

My virology work placement came and went so quickly. Great experience. I was certainly not expecting it to be such a creative process. Or should I say "necessarily creative process". It seems with research like this, you really do need to be mindful of other results popping out of your work. If the tangent appears to be more important/meaningful than the original work, then it's best to follow it!

Of course this creative approach gives rise to a bit of a problem. The project has the potential to meander. I suppose this would be fine on a longer time scale, but for my all-too-brief 8 weeks it meant I wasn't able to neatly draw a line under it by the end of the placement.

Though having said that, the work that I was looking at wasn't time-dependant so I still have however long I want to complete the project in my spare time. After this next year of study I'd love to return to it.

Speaking which... my next year of study has started! I won't be informed of my assigned tutor for about another month, but I have my learning materials. So learning has begun! This year it's complex analysis. Very excited to be looking at this subject, and to be writing up the areas I have difficulty with on here.

(Side note: Wow, I've been writing on this blog for 5 years?!)