de-CH
utf-8
math math-format
Argument bestimmen
komplex-02-03
multiple
64
randRangeExclude( 2, 22,[5,7,11,12,13,15,17,19] ) -3.14159265359 + K * 3.14159265359 / 12 randRange( 2, 9 ) cos( ANGLE ) * RADIUS sin( ANGLE ) * RADIUS

Sei \color{red}z = RADIUS \cdot e^{i \varphi}. Bestimmen Sie \color{purple}\dfrac{\varphi}{\pi} mit -\pi < \varphi \leq \pi.

graphInit({ range: [[-12, 12], [-12, 12]], scale: 20, tickStep: 0.5, labelStep: 4, }); label( [12.2, 0.5], "\\operatorname{Re}", "left" ); label( [0, 12], "\\operatorname{Im}", "right" ); circle( [REAL, IMAG], 3 / 15, { fill: RED, stroke: "none" }); label( [REAL, IMAG], "", "below", { color: RED, labelDistance: 10 } ); path([ [0,0], [REAL, IMAG]], { stroke: PURPLE });
x \color{purple}\dfrac{\varphi}{\pi} = -1+K/12

Der Punkt hat die \color{orange}x-Koordinate RADIUS \cdot \cos \left(\varphi \right) und die \color{blue}y-Koordinate RADIUS \cdot \sin \left(\varphi \right).

path([ [0,0], [REAL, 0]], { stroke: ORANGE, strokeWidth: 3.2 }); path([ [0, IMAG], [0, 0]], { stroke: BLUE, strokeWidth: 3.2 }); path([ [0, IMAG], [REAL, IMAG]], { stroke: BLACK, strokeWidth: 2.2, strokeDasharray: "." });

Suche nun (in Tabellen), welcher Winkel -\pi < \varphi \leq \pi am Einheitskreis den Punkt \left(\color{orange}\frac{\cos\left(\varphi \right)}{RADIUS}, \color{blue}\frac{\sin\left(\varphi \right)}{RADIUS}\right) liefert.

Dies ist hier \varphi = fractionReduce(K-12,12) \cdot \pi und damit \color{purple}\dfrac{\varphi}{\pi} =fractionReduce(K-12,12).