z^10- z^5 - 992=0. find num of roots whose real part is negative.a) 3b) 4c) 5d) 6

The answer is C , it has 5 roots

we can write eqn as (Z⁵)² - Z⁵ - 992 =0

so, we have roots of above eqn. as Z⁵ = 32 and -31

for  Z⁵ = 32   :    Z⁵ = 2⁵*( cos 0 + i sin 0 )

                            Z = 2*[cos(2kpi/5) + i.sin(2kpi/5)]       for k=.......-4,-3,-2,-1,0,1,2,3........

for k=2,3 we have roots as -1.618 +1.175i and -1.618 - 1.175i  whose real parts are -ve

for  Z⁵ = -31   :    Z⁵ = 31*( cos pi + i sin pi )

                            Z = 31^1/5[cos((2k+1)pi/5) + i.sin((2k+1)pi/5)]       for k=.......-4,-3,-2,-1,0,1,2,3........

for k=1,2,3 we have roots as: 31^1/5*[ -0.309 +0.951i] and -31^1/5 and 31^1/5*[ -0.309 -0.951i]  whose real parts are -ve

so we have 5 roots with real part as negative