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

Ramesh V
70 Points
12 years ago

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 = 311/5[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: 311/5*[ -0.309 +0.951i] and -311/5 and 311/5*[ -0.309 -0.951i]  whose real parts are -ve

so we have 5 roots with real part as negative