to get -1 + 0 i
z^5 = 1 cos 180 + i sin 180
That should get you started
Solve the equation z^5=-1 giving the solutions in polar form
I've got z^5 =1 but just not sure as its a negative
2 answers
let z^5 = rcosθ + rsinθ i
= -1 + 0
we have r = 1
θ = 180°
so z^5 = 1(cos180° + isin180°)
by de Moivre's theorme
z = 1^(1/5)(cos 180/5 + i sin180/5)
= 1(cos 36° + isin36°) or cis36°
repeating at 360/5 or 72°
z = cis 36°
z = cis108°
z = cis180° -----> there is our -1
z= cis252°
z = cis324°
= -1 + 0
we have r = 1
θ = 180°
so z^5 = 1(cos180° + isin180°)
by de Moivre's theorme
z = 1^(1/5)(cos 180/5 + i sin180/5)
= 1(cos 36° + isin36°) or cis36°
repeating at 360/5 or 72°
z = cis 36°
z = cis108°
z = cis180° -----> there is our -1
z= cis252°
z = cis324°
2 answers