We have a complete set of real numbers. Do we need any more?
For the real numbers, we showed there were functions on the rational numbers which did not have rational solutions. We can similarly show that there are functions on real numbers which do not have real solutions.
Consider:
\(f(x)=\sqrt x\)
This has no real solution for \(x<0\).
We define:
\(i:=\sqrt {-1}\)
\(i\) and \(-i\) can be used interchangeably.
\((-i)^2=(-1)^2i^2=i^2=-1\)
Complex numbers can be shown more generally as:
\(a+bi\)
We define the complex conjugate of
\(x=a+bi\)
As
\(\bar x=a-bi\)
Note that
\(x\bar x=(a+bi)(a-bi)=a^2-b^2\)
We can take exponents of imaginary numbers
\(c^{i\theta}=a+bi\)
We know the opposite is true.
\(c^{-i\theta}=a-bi\)
So
\(c^{i\theta}c^{-i\theta}=(a+bi)(a-bi)\)
\(1=a^2-b^2\)
The case where \(c=e\) is of particular note. We explore this later.
Define as an ordered pair of reals
We have a complete set of real numbers. Do we need any more?
For the real numbers, we showed there were functions on the rational numbers which did not have rational solutions. We can similarly show that there are functions on real numbers which do not have real solutions.
Consider:
\(f(x)=\sqrt x\)
This has no real solution for \(x<0\).
We define:
\(i:=\sqrt {-1}\)
\(i\) and \(-i\) can be used interchangeably.
\((-i)^2=(-1)^2i^2=i^2=-1\)
Complex numbers can be shown more generally as:
\(a+bi\)
We define the complex conjugate of
\(x=a+bi\)
As
\(\bar x=a-bi\)
Note that
\(x\bar x=(a+bi)(a-bi)=a^2-b^2\)
We can take exponents of imaginary numbers
\(c^{i\theta}=a+bi\)
We know the opposite is true.
\(c^{-i\theta}=a-bi\)
So
\(c^{i\theta}c^{-i\theta}=(a+bi)(a-bi)\)
\(1=a^2-b^2\)
The case where \(c=e\) is of particular note. We explore this later.
For each of these we have:
\(x=a+bi\)
\(y=c+di\)
Addition is defined as:
\(x+y=a+bi+c+di\)
\(x+y=(a+c)+(b+d)i\)
Subtraction is defined as:
\(x-y=a+bi-c-di\)
\(x-y=(a-c)+(b-d)i\)
Multiplication is defined as:
\(xy=(a+bi)(c+di)\)
\(xy=ac-bd+adi+bci\)
\(xy=(ac-bd)+(ad+bc)i\)
Division is defined as:
\(\dfrac{x}{y}=\dfrac{a+bi}{c+di}\)
\(\dfrac{x}{y}=\dfrac{(a+bi)(c-di)}{(c+di)(c-di)}\)
\(\dfrac{x}{y}=\dfrac{(ac+bd)+(bc-ad)i}{c^2+d^2}\)
We have \(z=a+bi\).
The complex conjugate is:
\(\bar z=a-bi\)
\(|z|=\sqrt {z\bar z}\)
\(|z|=\sqrt {(a+bi)(a-bi)}\)
\(|z|=\sqrt {a^2+b^2}\)
A disk is the area contained by a circle.
An open disk at \((a,b)\) of radius \(r\) is:
\(\{(x,y)\in \mathbb R^2:(x-a)^2+(y-b)^2 < r^2\}\)
For a closed disk it is:
\(\{(x,y)\in \mathbb R^2:(x-a)^2+(y-b)^2 \le r^2\}\)
We defined an open disk at \((a,b)\) of radius \(r\) as:
\(\{(x,y)\in \mathbb R^2:(x-a)^2+(y-b)^2 < r^2\}\)
For a closed disk it is:
\(\{(x,y)\in \mathbb R^2:(x-a)^2+(y-b)^2 \le r^2\}\)
An annulus is a disk, which excludes a smaller disk inside the disk
If the interior disk is just a point, it is a punctured disk.