Find the fourth roots of 256sqrt 3 iSolutionLet z 43 i14 Pu

Solution

Let z= 4(3 +i)^1/4

Put 3= rcos and 1= rsin

r=2 and =/6

Therefore z= 4.2^1/4 [ Cos+isin]^1/4

So, z= 4*2^1/4 [ cos (2n+)/4 + i Sin (2n+)/4 ] Applying De-Moivre’s theorem

Where n=0,1,2,3

Put n=0

z1= 4*2^1/4 [ cos 7.5 + i Sin 7.5]

Put n=1,

z2= 4*2^1/4 [ -Sin7.5 + i Cos7.5]

Put n=2,

z3= 4*2^1/4[ -Cos7.5 -iSin7.5]

Put n=3,

z4= 4*2^1/4 [ Sin7.5-iSin7.5]