Solving Seventh Roots in Polar Form

[morbello]

hi i know its a little later in the day but I am having trouble working out the polar form off the seven roots.

what i have got so far is that they are divided into 60 degrees around the 360 i also need the congurants which when i use the sin(60) sin (120) i have the right numbers but when i work in polar form i seam to get wrong numbers i think its an easy way to work out dregees to squ 3/2 which is the 60 degree.if you know what i mean.

Homework Equations

The Attempt at a Solution

 

[slider142]

360 is divided into only 6 even parts by partitions of 60 degrees (a regular hexagon). To get the 7 7th roots, you need to partition 360 into 7 even parts (a regular heptagon).

 

[morbello]

i have it starting on the 0 degree and then six after,do you know how to get the sin part off the degree as it moveds around the hexagon you like squ3/2 and so on. to get the congurants.

 

[slider142]

i have it starting on the 0 degree and then six after,do you know how to get the sin part off the degree as it moveds around the hexagon you like squ3/2 and so on. to get the congurants.

Six rotations of 60 after the 0th degree is 360 = 0 degrees, leaving you only six roots. If you actually want 6th roots and not 7th roots, then your rotation of 60 is correct. If the number you want to find a sixth root for is n, then simply note the signs in each quadrant. You know that in the first quadrant, the vector will have coordinates of (n^1/6*cos(60), n^1/6*sin(60)). We already know that sin(60) = Sqrt(3)/2 and cos(60) = 1/2 from trigonometry. The rest should follow by changing the appropriate signs.

 

[morbello]

i worked it out to be that but if you change the written squ3/2 around the 360 you get different numbers like i was putting in squ6/3 and 3squ/4 squ6/4 quessing may be i will have to sit and do the maths to find the real squ/ for 120,180,240,320 degrees.im not sure what they want really so talking here and thinking about it.

 

[slider142]

i worked it out to be that but if you change the written squ3/2 around the 360 you get different numbers like i was putting in squ6/3 and 3squ/4 squ6/4 quessing may be i will have to sit and do the maths to find the real squ/ for 120,180,240,320 degrees.im not sure what they want really so talking here and thinking about it.

The other roots will only be sign changes. Draw the picture and you will see the triangles are all the same, just rotated.

 

[morbello]

yes i know i tryed putting in squ3/2 and got it .i just really logged on here to get the fraction for 120 degrees.

well any way I am tired now will do it in the morning hope you are haveing a good time on here night.