Lecture 2
\(z = (x,y)\) belongs to the complex plane with \(x,y\in \R\).
So \(z=x\left(1,0\right)+y\left(0,1\right)\), i. e. \[z = x+i\, y\]
where \(x\) and \(y\) are the real and imaginary parts, respectively.
\(i\) is the complex number represented by \(\left(0,1\right)\). We say \(\R \subset \C\) by identifying the complex number \(x+i\cdot 0\) with the real number \(x\). |
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Set \[\left(x_1,y_1\right)+\left(x_2,y_2\right) = \left(x_1+x_2,y_1+y_2\right)\]
That is
We can use \(\times\) or \(\cdot\) or juxtaposition.
\[{\small \left(x_1, y_1\right) \cdot \left(x_2, y_2\right) = \left(x_1 x_2 - y_1 y_2, y_1 x_2 + x_1 y_2\right)}\]
That is
The definition of \(\times\) formally applies if we use the usual rules for algebra in \(\R\) and set \(i=\sqrt{-1}\).
With the operations \(+\) and \(\times\), \(\C\) is a field.
Exercise: Check that \(\C\) is closed under \(+\) and \(\times\).
(F2)(i) Existence of additive identity: \(0 = 0+i\,0\)
(F2)(ii) Existence of additive inverse: \[-z = -(x+i\,y) = (-x)+i\,(-y)\]
(F5)(i) Existence of multiplicative identity: \(1 = 1+i\,0\)
(F5)(ii) Existence of multiplicative inverse: \(z = x+i\,y \neq 0\), \[ \begin{align*} z^{-1} &= \dfrac{1}{x+i\,y}\cdot \dfrac{x-i\,y}{x-i\,y} \\ &= \dfrac{x}{x^2+y^2} - i \dfrac{y}{x^2 + y^2} \end{align*} \]
Since \(\C\) is a field, there holds: \[z_1 z_2 = 0 \implies \text{ either } z_1=0 \text{ or } z_2=0\]
(called null-factor or cancelation law).
\[ \begin{align*} \left(z_1 z_2\right)^{-1} &= z_1^{-1}z_2^{-1} \\ &= \dfrac{1}{z_1}\cdot \dfrac{1}{z_2}=\dfrac{1}{z_1 z_2}, \text{ with } z_1,z_2\neq 0 \end{align*} \]
\(i^2=-1\), \(\left(-i\right)^2=-1\).
These are the only two solutions of \(z^2=-1\) in \(\C\).