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How do you find the complex cube roots of 1 plus 0i? - Answers

z = 1 + 0i So |rz| = 1 and az = 0 radians. which allows you to write z = rz*cos(az) + i*sin(az) Then, if y = z1/3 then |y| = |z1/3| = |11/3| = 1 and ay is the angle in [0, 360) such that 3*ay = 0 mod(2*pi) that is, ay = 0, 2pi/3 and 4pi/3 And therefore, Root 1 = cos(0) + i*sin(0) Root 2 = cos(2pi/3) + i*sin(2pi/3) and Root 3 = cos(4pi/3) + i*sin(4pi/3).



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How do you find the complex cube roots of 1 plus 0i? - Answers

https://math.answers.com/math-and-arithmetic/How_do_you_find_the_complex_cube_roots_of_1_plus_0i

z = 1 + 0i So |rz| = 1 and az = 0 radians. which allows you to write z = rz*cos(az) + i*sin(az) Then, if y = z1/3 then |y| = |z1/3| = |11/3| = 1 and ay is the angle in [0, 360) such that 3*ay = 0 mod(2*pi) that is, ay = 0, 2pi/3 and 4pi/3 And therefore, Root 1 = cos(0) + i*sin(0) Root 2 = cos(2pi/3) + i*sin(2pi/3) and Root 3 = cos(4pi/3) + i*sin(4pi/3).



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https://math.answers.com/math-and-arithmetic/How_do_you_find_the_complex_cube_roots_of_1_plus_0i

How do you find the complex cube roots of 1 plus 0i? - Answers

z = 1 + 0i So |rz| = 1 and az = 0 radians. which allows you to write z = rz*cos(az) + i*sin(az) Then, if y = z1/3 then |y| = |z1/3| = |11/3| = 1 and ay is the angle in [0, 360) such that 3*ay = 0 mod(2*pi) that is, ay = 0, 2pi/3 and 4pi/3 And therefore, Root 1 = cos(0) + i*sin(0) Root 2 = cos(2pi/3) + i*sin(2pi/3) and Root 3 = cos(4pi/3) + i*sin(4pi/3).

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      z = 1 + 0i So |rz| = 1 and az = 0 radians. which allows you to write z = rz*cos(az) + i*sin(az) Then, if y = z1/3 then |y| = |z1/3| = |11/3| = 1 and ay is the angle in [0, 360) such that 3*ay = 0 mod(2*pi) that is, ay = 0, 2pi/3 and 4pi/3 And therefore, Root 1 = cos(0) + i*sin(0) Root 2 = cos(2pi/3) + i*sin(2pi/3) and Root 3 = cos(4pi/3) + i*sin(4pi/3).

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