ACIT4330-Page/content/Lectures/Lecture 18 - Complex Analysis.md

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18	2025-03-20

Overview of Complex Analysis of the Course

• Will do analysis using complex numbers.
• Concrete Goal: compute integrals such as \int_{-\infty}^{+\infty} \frac{e^{i \omega t}}{t^2 + a^2} \, dt = \frac{\pi}{a}e^{-a\mid \omega \mid} (a \gt 0) using complex techniques.

Complex Numbers

The definition of a Complex Numbers as defined in the lecture.

All algebraic properties follow from the definition. For instance

(x_{1}+iy_{1})(x_{2}+iy_{2}) = (x_{1}x_{2} -y_{1}y_{2}) + i(x_{1}y_{2} + y_{1}x_{2}).

This operation is commutative, that is z_{1}z_{2} = z_{2}z_{1}.

Proposition

Let z = x+iy, and it should be non-zero. Then there exists another complex number z^{-1} \in \mathbb{C} such that zz^{-1} = 1. It is given by

z^{-1} = \frac{x-iy}{x^2 + y^2}

Proof

We compute

(x + iy)(x-iy) = x^2 + y^2.

This is non-zero. Then

zz^{-1} = (x+iy) \frac{x-iy}{x^2 + y^2} = 1

[!info] Remark This means that \mathbb{C} is a field like \mathbb{R}.

Definition - Absolute Value

Let z = x+iy. Its absolute value (or Norm) is defined by

\mid z \mid \sqrt{ x^2 + y^2 }.

An argument for z is a real number \phi \in \mathbb{R} such that

x = \mid z \mid \cos \phi, \; y = \mid z \mid \sin \phi.

This allows us to identify complex numbers with points in the plane \mathbb{R}^2:

! %%Drawing 2025-03-20 10.50.15.excalidraw.md, and the Drawing 2025-03-20 10.50.15.excalidraw.light.svg%% (note that angles are only defined up to multiples of 2\pi).

You could multiply pairs in \mathbb{R}^2 by

(x_{1},y_{1}) \times (x_{2}, y_{2}) = (x_{1}x_{2}, y_{1}y_{2}).

But this does not correspond to the multiplication of complex numbers.

Polar Form

• Using the previous definitions, we can write any z \in \mathbb{C} as:

z = x + iy = \mid z \mid (\cos \phi + i \sin \phi).

Definition

We introduce the following notations:

r := \mid z \mid, e^{i\phi} := \cos \phi + i \sin \phi.

Then we have that z = r e^{i\phi}, which we call the polar form of z.

[!info] Remark For now, e^{i\phi} is just a symbol. Later we will identify it with the Complex Exponential Function

We have that e^{i\phi} shares many properties with e^{x}.

Proposition

Let \phi, \; \theta \in \mathbb{R}. Then:

1. e^{i \phi} e^{i \theta} = e^{i (\phi + \theta)},
2. e^{i 0} = 1,
3. \frac{1}{e^{i \phi}} = e^{-i \phi},
4. e^{i(\phi + 2 \pi k)} = e^{i \phi}, for k \in \mathbb{Z},
5. \mid e^{i \phi} \mid = 1,
6. \frac{d e^{i \phi}}{d \phi} = i e^{i \phi}.

Proof (only for 3.)

According to a previous result, we have:

(\cos \phi + i \sin \phi)^{-1} = \frac{\cos \phi - i \sin \phi}{\overbrace{(\cos \phi)^2 + (\sin \phi)^2}^{=1}} = \cos \phi - i \sin \phi = e^{- i \phi},

using \cos(-\phi) = \cos \phi and \sin(-\phi) = -\sin \phi.

Complex Conjugation

Definition of Complex Conjugation, as defined in the lecture.

Triangle Inequality

Definition of Triangle Inequality, as defined in the lecture.

But we want to show a similar result for \mathbb{C}.

Lemma

Let z \in \mathbb{C}, then we can take a look at the \mathrm{Re} and \mathrm{Im}:

- \mid z \mid \leq Re z \leq \mid z \mid - \mid z \mid \leq Im z \leq \mid z \mid.

Proof

Recall that for z = x + iy we have \mid z \mid = \sqrt{ x^2 + y^2 }. Then the results follow from the following inequalities (a, b \in \mathbb{R}):

-\sqrt{ a^2 + b^2 } \leq - \sqrt{ a^2 } \leq \; \mid a \mid \; \leq \sqrt{ e^2 } \leq \sqrt{ a^2 + b^2 }.

This gives the result.

We use this to prove the triangle inequality for \mathbb{C}.

Proposition

Let z_{1}, z_{2} \in \mathbb{C} we have

\mid z_{1} + z_{2} \mid \; \leq \; \mid z_{1} \mid + \mid z_{2} \mid.

Proof

We compute:

\mid z_{1} + z_{2} \mid^2 \; = (z_{1} + z_{2})(\overline{z_{1}} + \overline{z_{2}}) = z_{1}\overline{z_{1}} + z_{1} \overline{z_{2}} + z_{2} \overline{z_{1}} + z_{2} \overline{z_{2}} = \; \mid z_{1} \mid^{2} + 2 \mathrm{Re} (z_{1} \overline{z_{2}}) \mid z_{2} \mid^2

(Noting that z_{2}\overline{z_{1}} = \overline{z_{1}\overline{z_{2}}}).

Next we use the property that \mathrm{Re} z \leq \; \mid z \mid. Then

\mid z_{1} + z_{2} \mid^2 \; \leq \; \mid z_{1} \mid^2 + 2 \mid z_{1} \overline{z_{2}} \mid + \mid z_{2} \mid^2 = \; \mid z_{1} \mid^2 + \; 2 \mid z_{1} \mid \times \mid \overline{z_{2}} \mid + \mid z_{2} \mid^2 = \; \mid z_{1} \mid^2 + \; 2 \mid z_{1} \mid \mid z_{2} \mid + \mid z_{2} \mid^2 = (\mid z_{1} \mid + \mid z_{2} \mid)^2.

Taking square roots of both (non-negative) sides, we get the result.

This means that \mathbb{C} is a Norm space.

Limits

Before being able to define a limit, we need to define a Complex Functions, from the lecture.

Definition

Complex Limits as defined in the lecture

Proposition

Let f and g be Complex Functions.

Suppose that \lim_{ z \to z_{0} } f(z) and \lim_{ z \to z_{0} } g(z) exist. Then:

1. For a, b \in \mathbb{C} we have \lim_{ z \to z_{0} } (af(z) + bg(z)) = a \lim_{ z \to z_{0} } f(z) + b \lim_{ z \to z_{0} } g(z),
2. We have \lim_{ z \to z_{0} } (f(z) \times g(z)) = \left(\lim_{ z \to z_{0} } f(z) \right) \times \left(\lim_{ z \to z_{0} } g(z) \right).

Example

(Computing the Complex Limits, but will not happen again, just for understanding)

Let f(z) = z^2. We want to show that \lim_{ z \to 2 } f(z) = 4. Fix \varepsilon \gt 0 and observe that

f(z) - 4 = z^2 - 4 = (z - 2)^2 + 4(z - 2).

Write \mid z - 2 \mid \; \lt \delta for some \delta. Then using the Triangle Inequality, we get

\mid f(z) - 4 \mid \; = \; \mid(z-2)^2 + 4 (z-2) \mid \leq \; \mid z - 2 \mid^2 + \; 4 \mid z - 2 \mid \lt \delta^2 + 4 \delta.

It suffices to pick \delta such that \delta^2 + 4 \delta \leq \varepsilon (which is possible).

A much easier way to compute the following: First \lim_{ z \to z_{0} } z = z_{0}, then

\lim_{ z \to 2 } z^2 = \left( \lim_{ z \to 2 } z \right) \times \left( \lim_{ z \to 2 } z \right) = 2 \times 2 = 4