put z^5=t n solve

t^2 - t - 992 = 0

u'll get t=32 n t=-31

 

so z⁵=32(cos2n+isin2n) (n is any integer)

so z=2(cos2n/5+isin2n/5)

 

again z⁵=-31(cos2n+isin2n) ( n - integer)

so z=-(31)^1/5(cos2n/5+isin2n/5)

 

dis r the 10 roots of the equation.

for the first 5 roots, real part will be -ve when cos(2npi/5) is -ve

it is 4 n=2 and n=3

For the next 5 roots, real part will be -ve when cos(2npi/5) is +ve.

it is for n=1, n=4, n=5   (we can't put n>5 as dere r only 10 roots)

 

so there r 5 roots in which the real part is -ve.