# Solve the binomial equation : z^5 - 243 = 0 .

giorgiana1976 | Student

z^5 = 243*1

We'll re-write the equation writing the value 1 as the polar form of a complex number:

1 = 1 + 0*i (rectangular form)

r = sqrt(1^2 + 0^2)

r = 1

tan a = y/x

tan a = 0

a = 0

1 = 1*(cos 0 + i*sin 0) (polar form)

z^5 = 243(cos 0 + i*sin 0)

z = (243)^(1/5)(cos a + i*sin a)^(1/5)

We'll apply Moivre rule:

z = (243)^(1/5)[cos (2k*pi/5) + i*sin (2k*pi/5)]

We'll calculate the 5 roots of the equation, substituting k by 0,1,2,3,4.

k = 0

z0 = 3(cos 0 + i*sin 0)

z0 = 3

k = 1

z1 = 3[cos (2pi/5)+ i*sin(2pi/5)]

k = 2

z2 = 3[cos (4pi/5)+ i*sin(4pi/5)]

k = 3

z3 = 3[cos (6pi/5)+ i*sin(6pi/5)]

k  =4

z4 = 3[cos (8pi/5)+ i*sin(8pi/5)]

We have obtained the vertices of regular pentagon, inscribed in the circle C(0 , 3).

neela | Student

z^5 = 243- 0

z^5 =3^5 *1

z^5 = 3^5 {cos2npi+isin2npi}

We take the 5th root.

z = 3(cos2npi +isin2npi)^(1/5)

z = 3{cos2npi/2 +isinn2pi/5  for n = 0,1,2..4. by De Mivre's theorem.

Z1 = 3, for n =0

z2 = 3(cos2pi/5 +isin2p) for n = 1.

z3 = 3(cos4pi/5 +isin4pi/5) , for n = 2.

z4 = 3(cos6pi/5 +isin6pi/5) for n = 3.

z5 = 3(cos8pi/5 +isin8oi/5), for n = 4.