24 randRange( 0, DENOMINATOR - 1 ) BASE_ANGLE_NUMERATOR / DENOMINATOR * 2 * PI 1 randFromArray( [ true, false ] ) polarForm( BASE_RADIUS, BASE_ANGLE, EULER_FORM ) polarForm( BASE_RADIUS, BASE_ANGLE, true ) cos( BASE_ANGLE ) * BASE_RADIUS sin( BASE_ANGLE ) * BASE_RADIUS eulerFormExponent( BASE_ANGLE ) piFraction( BASE_ANGLE, true ) randRange( 2, 20 ) BASE_ANGLE_NUMERATOR * EXPONENT ANGLE_MULTIPLE_NUMERATOR / DENOMINATOR * 2 * PI eulerFormExponent( BASE_ANGLE * EXPONENT ) piFraction( ANGLE_MULTIPLE, true ) 1 ( BASE_ANGLE_NUMERATOR * EXPONENT ) % DENOMINATOR ANSWER_ANGLE_NUMERATOR / DENOMINATOR * 2 * PI piFraction( ANSWER_ANGLE, true ) piFraction( ANSWER_ANGLE, true ) polarForm( ANSWER_RADIUS, ANSWER_ANGLE, true ) polarForm( ANSWER_RADIUS, ANSWER_ANGLE, EULER_FORM ) piFraction( ANGLE_MULTIPLE - ANSWER_ANGLE, true )

Determine the value of the following complex number power. Your answer will be plotted in orange.

(\color{BLUE}{BASE_REP}) ^ {EXPONENT}
graphInit({ range: [ [-5, 5], [-5, 5] ], scale: 30, tickStep: 1, axisArrows: "<->" }); drawComplexChart( 5, DENOMINATOR ); circle( [ BASE_REAL, BASE_IMAG ], 1 / 4, { fill: KhanUtil.BLUE, stroke: "none" }); graph.currComplexPolar = new ComplexPolarForm( DENOMINATOR, 5, EULER_FORM ); redrawComplexPolarForm();
Radius: 1
Angle: 0
1
[ graph.currComplexPolar.getAngleNumerator(), graph.currComplexPolar.getRadius() ]
var angle = guess[0]; var radius = guess[1]; if (angle === 0 && radius === 1) { return ""; } return angle === ANSWER_ANGLE_NUMERATOR && radius === ANSWER_RADIUS;
redrawComplexPolarForm(guess[0], guess[1]);
redrawComplexPolarForm(guess[0], guess[1]);

All powers of 1 are 1.

Let's express our complex number in Euler form first.

\color{BLUE}{BASE_REP} = \color{BLUE}{BASE_EULER_REP}

Since (a ^ b) ^ c = a ^ {b \cdot c}, (\color{BLUE}{BASE_EULER_REP}) ^ {EXPONENT} = e ^ {EXPONENT \cdot (BASE_E_EXPONENT)}

The angle of the result is EXPONENT \cdot BASE_ANGLE_REP, which is ANGLE_MULTIPLE_REP.

ANGLE_MULTIPLE_REP is more than 2 \pi. It is a common practice to keep complex number angles between 0 and 2 \pi, because e^{2 \pi i} = (e^{\pi i}) ^ 2 = (-1) ^ 2 = 1. We will now subtract the nearest multiple of 2 \pi from the angle.

ANGLE_MULTIPLE_REP - NEAREST_TWO_PI_MULTIPLE = ANSWER_ANGLE_REP

Our result is ANSWER_EULER.

Converting this back from Euler form, we get ANSWER.