Complex numbers can be though of as an ordered pair of real numbers.
If z is an ordered pair (x,y) such that x,y \in \mathbb{R}, then z is a Complex number.
Then we define the following relations
The Addition relation + \colon \mathbb{C} \to \mathbb{C} is defined
(x_1,y_1) + (x_2,y_2) \to (x_1 + x_2, y_1 + y_2).
The Subtraction relation - \colon \mathbb{C} \to \mathbb{C} is defined
(x_1,y_1) + (x_2,y_2) \to (x_1 - x_2, y_1 - y_2).
The Multiplication relation \times \colon \mathbb{C} \to \mathbb{C} is defined
(x_1,y_1) \times (x_2,y_2) \to (x_1x_2 - y_1y_2, x_1y_2 + x_2y_1).
The Division relation \ \colon \mathbb{C} \to \mathbb{C} is defined
(x_1,y_1) \times (x_2,y_2) \to (x_1x_2 - y_1y_2, x_1y_2 + x_2y_1).
The Quotient relation $ $ is defined $$
$$
Real and Imaginary functions
The Real part of a complex number is given by the function
\text{Re} \colon (x,y) \to x
The Imaginary part of a complex number is given by the function
\text{Im} \colon (x,y) \to y
TO-DO: Theorems
A Circle around an inmaginary number z_0 with radius r is
\{z \, \vert \, \text{abs}(z - z_0) = r\}
Given any cubic equation of the form
z^3 + az^2 + bz + c = 0
the substitution z = x - \frac{a}{3} gives the equivalent depressed cubic
x^3 + bx + c = 0.
\begin{align*}
z^3 + az^2 + bz + c &= 0\\
\left(x - \frac{a}{3}\right)^3 + a\left(x - \frac{a}{3}\right)^2 + b\left(x - \frac{a}{3}\right) + c &= 0\\
x^3 + bx + c &= 0
\end{align*}
A depressed cubic has the following solution
x = \sqrt[3]{-\frac{c}{2} + \sqrt{\frac{c^2}{4} + \frac{b^3}{27}}} + \sqrt[3]{-\frac{c}{2} - \sqrt{\frac{c^2}{4} + \frac{b^3}{27}}}.
Given a = 0, b=3, c =-4
x = \sqrt[3]{-\frac{c}{2} + \sqrt{\frac{c^2}{4} + \frac{b^3}{27}}} + \sqrt[3]{-\frac{c}{2} - \sqrt{\frac{c^2}{4} + \frac{b^3}{27}}}.
Gives x=1, allowing us to factor and obtain the solutions
x = \frac{1 \pm i\sqrt{11}}{2}.
- Create monic polynomial
- Preform change of variables
- Use Ferro-Tartaglia formula
- Factor to obtain a quadratic
- Find remaining solutions
TO-DO: Prove solutions of real cubic equations
Conjugates
Given a complex number z \in \mathbb{C}, we define its conjugate
\overline{z} = \text{Re}(z) - \text{Im}(z)
The following identities hold true
z + \overline{z} = 2\text{Re}(z)
\overline{\overline{z}} = z
\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}
\overline{z_1z_2} = \overline{z_1} \cdot \overline{z_2}
$ = $
Let z \in \mathbb{C}. Then
Given z_1,z_2 \in \mathbb{C}
Modulus
Let
r = |z| = \sqrt{x^2+y^2}
To-DO: Polar coordinates for complex numbers
Polar-coords
When |z| \not = 0, we can define a complex number by two quantities r for radius, and \theta for angle.
A complex number has an infinite amount of polar representations.
Let z \in \mathbb{C}, then exists r \in \mathbb{R} and \theta \in [0,2\pi] such that
z = r\cos\theta + r\sin\theta i.
The argument of a complex number is given by the function
\arg z = \{\theta \mid r\cos\theta + r\sin\theta i = z\}
where r = |z|.
The argument is any \theta \not = 0 such that for some r the polar representation is accurate.
\arg (1 + i) = \{\frac{\pi}{4} + 2\pi n \mid n \in \mathbb{N}\}
The principle argument of a complex number is given by the function
\text{Arg } \colon z \to (-\pi,\pi].
such that
\text{Arg } z = \{\theta \mid r\cos\theta + r\sin\theta i = z\}
where r = |z|.
De Moives formula is used when exponetiating complex numbers.
(e^{i\theta})^n = e^{in\theta}.
In polar form
(\cos\theta + \sin\theta i)^n = (\cos(n\theta) + \sin(n\theta) i)
Using De Moivres formulas, prove
\cos(5\theta) = ???
We have
(\cos\theta + \sin \theta i)^5 = \cos(5\theta) + \sin(5\theta)i
Which using binomial theorem
(a + b)^5 = a^5 + \binom{5}{1}a^4b + \binom{5}{2}a^3b^2 + \binom{5}{3}a^2b^3 + \binom{5}{4}ab^4 + b^5.
Which evaluates into
(a + b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5.
Plugging this back into our equation
a^5 + 5\cos^4\theta \sin \theta i + 10\cos^3 \theta \sin^2 \theta i^2 + 10\cos^2\theta \sin^3\theta i^3 \\+ 5\cos\theta \sin^4\theta i^4 + \sin^5\theta i^5 = \cos(5\theta) + \sin(5\theta)i
Matching the real parts we have
cos(5\theta) = ??
Solving equations
The equation z^n = 1 can easily be solved with z = 1, but there should exist n many solutions.
We can solve it with polar representation.
The Roots of unity are complez solutions to the polynomial z^n - 1 = 0, where n is an integer.
Notice that
|z^n| = 1 \implies |z|^n = 1
by our modulus operations. Using our polar representation, we have
z = re^{i\theta}
where z is not zero. Then
(e^{i\theta})^n = e^{ni\theta} = 1
Thus the solutions to z^n = 1 are
z_k = e^{2ki\pi/n}
for k as an integer. Thus we have obtained all n solutions to the polynomial.
The primitive nth root of unity is given by z_1 = \omega_n = e^{2i\pi/n}
Find all 5th roots of 9i.
Let z = re^{i\theta}. Converting to polar form
r^5e^{i5\theta} = 9e^{i\pi/2}.
Then r^5 = 9 and e^{i5\theta} = e^{i\pi/2}.
Then \theta = \pi/10 + 2\pi k /5, and r = \sqrt[5]{9}.
We say z = \zeta \omega_n^k is a solution to the equation z^n = \zeta^n.
Topology of the Complex numbers
A parametrized curve in \mathbb{C} is
C \colon z(t) = x(t) + y(t)i
for a \leq t \leq b.
A curve is smooth if x(t) and y(t) are differentiable.
A curve is closed if z(a) = z(b).
A curve is non-intersecting if t_1 \not = t_2 implies z(t_1) \not= z(t_2).
A positive orientation is counter clockwise, while a negative orientation is clockwise.
A circle is denoted C_\epsilon(z_0) for a circle of radius \epsilon and center z_0.
A disk is denoted D_\epsilon(z_0) and is all the stuff inside the circle.
The closed disk is denoted \overline{D}_\varepsilon(z_0).
Let S \subset \mathbb{C} be a set. A point p in S is an interior point of S if there exists some \varepsilon > 0 such that some \varepsilon neighborhood of p is completely contained in S.
Let S \subset \mathbb{C} be a set. A point p in S is a boundary point of S if for all \varepsilon > 0, there exist ptext
A set S is called open if every point in S is open.
A set S is called open if every point in S is open.
A set S is called a domain if it is open and pathwise-connected.
A region is a domain together with an arbitrary amount of boundary points.
Every point in D_1(0) is an interior point.
Let z_0 \in D_1(0). Then |z_0| < 1, so \varepsilon = 1 - |z_0| is positive.
We want to show D_\varepsilon(z_0) \subset S. Any point z \in D_\varepsilon(z_0) had the property
\begin{align*}
|z - z_0| &< \varepsilon\\
&<1 - |z_0|
\end{align*}
And by triangle inequality
|z| = |(z - z_0) + z_0| \leq |z - z_0| + |z_0| < 1.
Prove every point in C_1(0) is on the boundary.
Let z_0 \in C_1(0). Then |z_0| = 1. Write z_0 = e^{i\theta_0}. Then we have z_1 = (1 + \varepsilon/2)z_0.
|z_1 - z_0| < \varepsilon.
Thus z_0 is in the boundary of S.
Thus we have.
Functions and linear mappings
A complex-valued function is a function f : \mathbb{C} \to \mathbb{C}. Also called a map.
Let f(z) = z^2. The domain and range of the function is \mathbb{C}.
Let g(z) = |z|^2. The domain is the complex numbers, while the range is strictly the positive real numbers with zero.
Let h(z) = 1/z. Then the domain and range are \mathbb{C}^*.
Functions can be in polar, cartesian, or complex form.
The image of A is the output of f(A).
Properties
The most important properties of a function depend on if they said function is injective or surjective.
A function f \colon A \to B is injective if for all a,a^\prime \in A
f(a) = f(a^\prime) \implies a = a^\prime.
Let f(z) = z^2. This is not an injective function as 1 and -1 map to the same number.
Let g(z) = iz. This is an injective function.
A function f \colon A \to B is surjective if for all b \in B, there exists a \in A such that
f(a) = b.
And the most appealing type of function,
A function f \colon A \to B is bijection if f is both injective and surjective.
If a function is injective, there exists an inverse.
Let f(z) = iz + 2. Then by setting w = iz+2 and solving for w, we have
f^{-1}(w) = -iz + 2i.
Inverse exponents
Let f(z) = z^2. In exponential form we have f(re^{i\theta}) = r^2e^{2i\theta}. We know that this function is very much not injective.
We want to find some restriction for our function such that f \colon D \to \mathbb{C} is injective. This gives us
D = \{re^{i\theta} \colon r > 0, \frac{-\pi}{2} < \theta \leq \frac{\pi}{2}\}.
Let A = \{z \mid \Im(z) = 2\} maps to a parabola.
This allows us to create a principle root function. This is a well defined function that gives us an inverse for any squared complex numbers.
Let g be the principle root function. Then g(z) = \sqrt{|z|}e^{i\frac{\arg(z)}{2}}.
The principal nth root is given by
g(w) = \sqrt[n]{|w|} e^{i\frac{\text{Arg}(w)}{n}}.
Limits and continuity
Let z,w \in \mathbb{C}, and f \colon \mathbb{C} \to \mathbb{C} be a function. The limit of f at z is w if for all D_\varepsilon(w), there exists D_\delta(z) such that f(D_\delta(z)) \subset D_\varepsilon(w).
Equivalently, if |z - z_0| < \delta then |f(z) - w_0| < \varepsilon.
Prove \lim_{z \to 0}z^2 = 0
Let \varepsilon > 0, and choose \delta = \sqrt{\varepsilon}. If z \in D_\delta(0), then |z| < \sqrt{\varepsilon}. Thus z^2 = f(z) < \varepsilon and f(z) \in D_\varepsilon(0).
Prove \lim_{z \to 0} \overline{z}/z does not exist.
Let \Im(z) = 0, then as z approaches 0, we have that f(z) = 1. But if \Re(z) = 0, as z approaches 0 we have that f(z) -1.
Thus, the limit does not exist.
Lots of theorems exist.
$$
$$
Limit breaks up, limit follows limit properties.
A complex function is continuous if for all z \in D, we have
\lim_{z \to z_0} = f(z_0).
Prove f(z)=z^{-1} is continuous on it’s domain, \mathbb{C}^*.
A branch of a multivalued function f is a single values function.
The function f(z)=z^{1/2} has the branches
- f_1(z) = \sqrt{r}e^{i\frac{\text{Arg}(z)}{2}}
- f_2(z) = -\sqrt{r}e^{i\frac{\text{Arg}(z)}{2}}
- f_3(z) = -\sqrt{r}e^{i\frac{\text{arg}(z)}{2}}
Not continuous on negative x-axis.
A branch cut of a multivalued function f is the set of discontinuties for all branches.
Depressed cubic Products and algebra Modulus, argument, conjugate formulas Triangle equality reverse and direct Limits Exponential/Polar/Cartesian Find all roots of numbers Prove topological results (Disk,Circle,Etc.) Openness/Pathconnectedness Find images under mappings Find limits and epsilon delta proofs
Differentiable and holomorphic functions
Let z_0 \in \mathbb{C}, and f be a complex valued function. Then the derivative of f is defined
f^\prime(z_0) = \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0}
if limit exists. If f^\prime(z_0) exists, we say f is differentiable at z_0.
Example of finding a functions derivative
Let f(z) = z^3. Find the derivative of f.
\begin{align*}
f^\prime(z_0) &= \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0}\\
&= \lim_{z \to z_0} \frac{z^3 - z^3_0}{z - z_0}\\
&= \lim_{z \to z_0} \frac{(z - z_0)(z^2 + zz_0 + z^2_0)}{z - z_0}\\
&= \lim_{z \to z_0} z^2 + zz_0 + z^2_0\\
&= 3z^2_0.
\end{align*}
Do all functions have a derivative at some point? No.
Let f(z) = \overline{z}. Show that f is differentiable nowhere.
\begin{align*}
f^\prime(z_0) &= \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0}\\
&= \lim_{z \to z_0} \frac{\overline{z} - \overline{z_0}}{z - z_0}
\end{align*}
We previously showed that this limit does not exist. Therefore the function is not differentiable at any point.
Let z_0 \in \mathbb{C} and f be a complex valued function. The function f is holomorphic at a point z_0 if it is differentiable at at every point in some neighborhood D_\varepsilon(z_0).
Let R \subset \mathbb{C} and f be a complex valued function. The function f is holomorphic on the region R if f is holomorphic at every point in R.
Let f be a complex valued function. The function f is entire if is differentiable at every point in the complex plane.
The above definition is eequivalent to being holomorphic at every point on the complex plane.
Let f(z) = z^3. Since f is differentiable at every point z_0 \in \mathbb{C}, it is entire.
Let f(z) = |z|. The function f is differentiable only at z_0 = 0, and nowhere else. Therefore the function is holomorphic nowhere, since there exists no neighborhood where every point is differentiable.
Now we can move on to some properties
Let z_0 \in \mathbb{C} and let f be a complex valued function.
If f is differentiable at z_0, then f is continuous at z_0.
We can convert L’Hopital’s rule into the complex plane.
Let z_0 \in \mathbb{C} and f,g be complex valued functions.
If f and g are holomorphic at z_0, and f(z_0) = 0 and g(z_0) = 0, then
\lim{z \to z_0} \frac{f(z)}{g(z)} = \frac{f^\prime(z)}{g^\prime(z)}
as long as g^\prime(z_0) \not= 0.
Cauchy-Riemann equations
Let f be a complex function differentiable at z_0 \in \mathbb{C} written in it’s real and imaginary parts, that is
f(x +iy) = u(x,y) + iv(x,y)
If f = u + vi is holomorphic, then both u and v are harmonic.
A function u(x,y) is harmonic if it satisfies
u_{xx} + u_{yy} = 0
Given a harmonic function u(x,y), we can choose v(x,y) such that f(z) = u(x,y) + iv(x,y) is holomorphic.
Let u(x,y) be a harmonic function. A harmonic conjugate is a function v such that f(z) = u(x,y) + iv(x,y) is holomorphic.
The following are functions
\text{curl}(\overline{V}) = \frac{\partial q}{\partial x} - \frac{\partial p}{\partial y}.
The curl detects rotation.
\text{div}(\overline{V}) = \frac{\partial p}{\partial x} - \frac{\partial q}{\partial y}.
The div detects divergence.
\text{grad}(\phi(x,y)) = \frac{\partial \phi}{\partial x} - i\frac{\partial \phi}{\partial y}.
If continuity holds, then V is holomorphic.
F(z) = \phi(x,y) + \psi(x,y)i
Sequences and Series
Many properties of sequences from real numbers carry over to complex numbers.
We have the same definition of sequence from real analysis, that is
Let A \subseteq \mathbb{N} and S be a set.
If f \colon A \to S is a function, then f is a sequence.
except now each term may be a complex number.
Let z_n = e^{2\pi n i/5} be a complex sequence. Then
z_n = (e^{2\pi i/5},e^{4\pi i/5},e^{6\pi i/5},\dots)
A sequence (z_n) of complex numbers is said to converge to a complex number w if, for every \varepsilon > 0, there exists some natural number N such that n > N implies z_n \in D_\varepsilon(w).
This creates are ability to make a limit.
Let z_n be a sequence of complex numbers and let w be a fixed complex number. Then
\lim_{n\to\infty} z_n = w
if and only if
\lim_{n\to\infty} \Re(z) = \Re(w) \text{ and } \lim_{n\to\infty} \Im(z) = \Im(w).
(\Rightarrow) Let z_n = x_n + y_ni and w = u + vi.
Since z_n converges to w, we have that for all \varepsilon > 0 there exist N such that n > N implies |z_n - w| < \varepsilon. Then
|x_n - u| = |\Re(z_n) - \Re(w)|.
(\Leftarrow) Let \lim_{n \to \infty} x_n = u and \lim_{n \to \infty} y_n = v.
When n > \max(N_{\varepsilon/2},M_{\varepsilon/2}), we have that
|x_n - u| + |y_n - v| < \varepsilon/2 + \varepsilon/2 = \varepsilon
And since |x_n - u| + |y_n - v| \geq |z_n - w|, we have that |z_n - w| < \varepsilon so we have our limit.
Let z_n = \frac{1}{n} + \frac{1}{n}i. Show that z_n converges to 0.
We need to find some N_\varepsilon such that n > N_\varepsilon implies |z_n| < \varepsilon.
We know |z_n| = \frac{1}{n}\sqrt{2}. Thus
\begin{align*}
\frac{1}{n}\sqrt{2} &< \varepsilon \\
\frac{1}{\varepsilon}\sqrt{2} &< n
\end{align*}
So we choose N_\varepsilon = \lceil \sqrt{2}/\varepsilon \rceil.
Infinite Series
Given a sequence \{z_n\}^\infty_{n=1} we can take it’s sum and ask if it converges to a complex number.
Same as real variable calculus.
Let \{z_n\} be a sequence. If \sum_{n=1}^\infty converges, then \lim_{n\to\infty}z_n = 0.
A geometric series \sum_{n=0}^\infty z_n converges or diverges depending on |z| as follows
If \sum_{n=0}^\infty has the property \lim_{n\to\infty} |z_{n+1}|/|z_n| = L then
If \lim_{n\to\infty} < 1 then the series is guaranteed to converge absolutely.
We can do some pretty cool stuff and find the radius of convergence
Fractals
Fractals encode information about iterative functions.
For example, given a function f(z) = z^2, what happens if we apply this iterativly to some initial point z_0.
Clearly when |z_0| > 1, the sequence never converges. When |z_0| < 1, the sequence converges. What about when |z| = 1?
The convergent points on the circle have a different cardinality (\aleph) then the points that don’t (2^\aleph).
K = \{z\in\mathbb{C} \mid f^\infty(z) \text{ is bounded}\}