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.
Algebra
The two main operations on complex numbers are as follows.
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 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).
Real and Imaginary functions
The main two unary functions of a complex number separate it into it’s real and inmaginary parts.
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
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
\text{Re}(z) = \frac{z + \overline{z}}{2}
\text{Im}(z) = \frac{z - \overline{z}}{2}
\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
Modulus
The modulus of a complex number is defined
|z| = \sqrt{x^2+y^2}
For all z_1,z_2 \in \mathbb{C},
|z_1 + z_2| \leq |z_1| + |z_2|
Let z \in \mathbb{C}. Then
For all z_1,z_2 \in \mathbb{C},
|z_1 - z_2| \geq ||z_1| - |z_2||
Let z \in \mathbb{C}. Then
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 exponets
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.
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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,Ect.) Openness/Pathconnectedness Find images under mappings Find limits and epsilon delta proofs