If z is a complex number,then the number of c | vedclass

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(B) Given equations are $z^{1985}+z^{100}+1=0$ and $z^3+2z^2+2z+1=0$. First,factorize $z^3+2z^2+2z+1=0$: $(z^3+1) + (2z^2+2z) = 0$ $(z+1)(z^2-z+1) + 2

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$2$

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(B) Given equations are $z^{1985}+z^{100}+1=0$ and $z^3+2z^2+2z+1=0$. First,factorize $z^3+2z^2+2z+1=0$: $(z^3+1) + (2z^2+2z) = 0$ $(z+1)(z^2-z+1) + 2z(z+1) = 0$ $(z+1)(z^2-z+1+2z) = 0$ $(z+1)(z^2+z+1) = 0$ The roots are $z = -1$,$z = \omega$,and $z = \omega^2$,where $\omega$ is the complex cube root of unity. Now,check these roots in the first equation $f(z) = z^{1985}+z^{100}+1=0$: $1$. For $z = -1$: $(-1)^{1985} + (-1)^{100} + 1 = -1 + 1 + 1 = 1 
eq 0$. $2$. For $z = \omega$: $\omega^{1985} + \omega^{100} + 1 = \omega^{1985 \pmod 3} + \omega^{100 \pmod 3} + 1 = \omega^2 + \omega + 1 = 0$. $3$. For $z = \omega^2$: $(\omega^2)^{1985} + (\omega^2)^{100} + 1 = \omega^{3970} + \omega^{200} + 1 = \omega^1 + \omega^2 + 1 = 0$. Thus,the common roots are $z = \omega$ and $z = \omega^2$. The number of common roots is $2$.

If $z$ is a complex number,then the number of common roots of the equations $z^{1985}+z^{100}+1=0$ and $z^3+2z^2+2z+1=0$ is equal to:

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