Factorisation over the Complex Field (1 Viewer)

[trailblazer]

5. Show that the roots of (z-1)^4 + (z+1)^4 = 0 are -+icot(pi/8) and +-icot(3pi/8)

6. Show that the roots of(z-1)^6 + (z+1)^6 = 0 are
+-icot(pi/12), +-icot(5pi/12), +-i

Help would be appreciated. Thanks in advance.
Get In before Tim!

 

[azureus88]

i don't want to type out the full solution cause its pretty long n tedious so i'll just guide you along.

5. (z-1)^4 + (z+1)^4 = 0
[(z-1)/(z+1)]^4= -1 (dividing both sides by (z+1)^4 are rearranging)
(z-1)/(z+1) = (-1)^0.25 = cis[(pi+2kpi)/4]
z{cis[(pi+2kpi)/4] - 1} = - {cis[(pi+2kpi)/4] + 1}

z = - {cis[(pi+2kpi)/4] + 1}/{cis[(pi+2kpi)/4] - 1}

change cis into cos and isin and use double angle to get get rid of the -1 and 1s, then take out cos[(pi+2kpi)/8] on numerator and sin[(pi+2kpi)/8] on demoninator and mutliply top and bottom by i. you will get icot[(pi+2kpi)/4]

6. same stuff. you should be fine.

 

[trailblazer]

i don't want to type out the full solution cause its pretty long n tedious so i'll just guide you along.

5. (z-1)^4 + (z+1)^4 = 0
[(z-1)/(z+1)]^4= -1 (dividing both sides by (z+1)^4 are rearranging)
(z-1)/(z+1) = (-1)^0.25 = cis[(pi+2kpi)/4]
z{cis[(pi+2kpi)/4] - 1} = - {cis[(pi+2kpi)/4] + 1}

z = - {cis[(pi+2kpi)/4] + 1}/{cis[(pi+2kpi)/4] - 1}

change cis into cos and isin and use double angle to get get rid of the -1 and 1s, then take out cos[(pi+2kpi)/8] on numerator and sin[(pi+2kpi)/8] on demoninator and mutliply top and bottom by i. you will get icot[(pi+2kpi)/4]

6. same stuff. you should be fine.

Thanks heaps