Complex Numbers
Complex numbers are expressed in the form \( z = a + bi \), where \( a \) and \( b \) are real numbers and \( i \) is the imaginary unit satisfying \[ i^2 = -1. \]
Prove that no order can be defined in the complex field that turns it into an ordered field.
In any ordered field, squares are always nonnegative.
Since \[ i^2 = -1, \] we would obtain \[ 0 \leq -1. \] Adding \( 1 \) to both sides gives \[ 1 \leq 0. \] But also \( 0 \leq 1 \), hence \[ 0 \leq 1 \leq 0, \] implying \[ 0 = 1, \] a contradiction.
Another attempt to define an order on \( \mathbb{C} \) is:
For \[ z = a + bi,\qquad w = c + di, \] define \[ z < w \] whenever either \[ a < c, \] or \[ a = c \quad \text{and} \quad b < d. \]
Proof:
Since \[ i^2 = -1, \] this would imply \[ 0 \leq -1. \] Then \[ 1 = 0 + 1 \leq -1 + 1 = 0. \] Hence \[ 1 \leq 0 \leq 1, \] which again implies \[ 0 = 1, \] a contradiction.
Therefore no ordering exists on the complex numbers that makes \( \mathbb{C} \) an ordered field.
To determine the values of \( r \) and \( \theta \) in polar coordinates:
|
| Formulas and Tables — Schaum's Outline |
Let \[ z = 3 + 4i, \] which may also be written in polar form as \[ z = re^{i\theta}. \]
The Cartesian coordinates satisfy \[ x = r\cos\theta, \qquad y = r\sin\theta. \]
Recall: \[ \tan\theta = \frac{\sin\theta}{\cos\theta}. \]
Since \[ x = 3, \qquad y = 4, \] we obtain \[ 3 = r\cos\theta, \qquad 4 = r\sin\theta. \]
Therefore \[ \tan\theta = \frac{4}{3} = \frac{r\sin\theta}{r\cos\theta}. \]
Hence \[ \theta = \arctan\left(\frac{4}{3}\right), \] and \[ r = \sqrt{3^2 + 4^2} = 5. \]
- Solved.
Solve \[ z = 2e^{i\pi/4}. \]
Recall Euler’s formula: \[ e^{i\theta} = \cos\theta + i\sin\theta. \]
Thus \[ 2e^{i\pi/4} = 2\cos\left(\frac{\pi}{4}\right) + 2i\sin\left(\frac{\pi}{4}\right). \]
Since \[ \cos\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, \] we conclude \[ z = \sqrt{2} + i\sqrt{2}. \]
