Colorful Absolutes

Algebra Level 3

This is a domain-colored complex plot of the polynomial z 5 1 = 0 z^5 - 1 = 0 .

What is the absolute value of the difference between the two complex numbers which map to the two darkest points closest to each other?

Give your answer to 3 decimal places.


The answer is 1.17557.

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1 solution

Pongkiaet Topar
Sep 5, 2017

plotted graph in complex plane , z n z_{n} be a point from z 5 1 = 0 z^{5} - 1 = 0

which z 1 = c i s ( 0 ) z_{1} = cis\left ( 0 \right ) , z 2 = c i s ( 2 π 5 ) z_{2} = cis\left ( \frac{2\pi }{5} \right ) , z 3 = c i s ( 4 π 5 ) z_{3} = cis\left ( \frac{4\pi }{5} \right ) , z 4 = c i s ( 6 π 5 ) z_{4} = cis\left ( \frac{6\pi }{5} \right ) , z 5 = c i s ( 8 π 5 ) z_{5} = cis\left ( \frac{8\pi }{5} \right )

choose z 3 , z 4 z_{3} , z_{4}

the absolute value of the difference between these two complex numbers is z 3 z 4 \left |z_{3} - z_{4} \right |

z 3 z 4 = ( c o s ( 4 π 5 ) c o s ( 6 π 5 ) ) + i ( s i n ( 4 π 5 ) s i n ( 6 π 5 ) ) \left |z_{3} - z_{4} \right | = \left |\left ( cos\left ( \frac{4\pi }{5} \right ) - cos\left ( \frac{6\pi }{5} \right )\right ) + i \left ( sin\left ( \frac{4\pi }{5} \right ) - sin\left ( \frac{6\pi }{5} \right )\right ) \right |

z 3 z 4 = 0 + 2 i s i n 4 π 5 \left |z_{3} - z_{4} \right | = \left |0 + 2i sin\frac{4\pi }{5} \right |

z 3 z 4 = 2 s i n π 5 = 2 s i n ( 3 6 ) = 5 5 2 1.175 5705045 \left |z_{3} - z_{4} \right | = 2sin\frac{\pi }{5} = 2sin\left ( 36^{\circ} \right ) = \sqrt{\frac{5-\sqrt{5}}{2}} \approx {\color{#D61F06} 1.175}5705045

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