If Z neq 0 is a complex number such that Z^2 | vedclass

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Question

(C) Given the equation $Z^2 + Z|Z| + |Z|^2 = 0$. Since $Z 
eq 0$,we can divide by $|Z|^2$: $\left(\frac{Z}{|Z|}ight)^2 + \left(\frac{Z}{|Z|}ight)

Vedclass · Accepted Answer

$\{\omega, \omega^2\}$

Vedclass · Answer

(C) Given the equation $Z^2 + Z|Z| + |Z|^2 = 0$. Since $Z 
eq 0$,we can divide by $|Z|^2$: $\left(\frac{Z}{|Z|}ight)^2 + \left(\frac{Z}{|Z|}ight) + 1 = 0$. Let $u = \frac{Z}{|Z|}$. Note that $|u| = \frac{|Z|}{|Z|} = 1$. The equation becomes $u^2 + u + 1 = 0$. The roots of this quadratic equation are $u = \omega$ and $u = \omega^2$,where $\omega = e^{i2\pi/3}$ and $\omega^2 = e^{i4\pi/3}$. Both $\omega$ and $\omega^2$ satisfy $|u| = 1$. Thus,$\frac{Z}{|Z|} = \omega$ or $\frac{Z}{|Z|} = \omega^2$. This implies $Z = |Z|\omega$ or $Z = |Z|\omega^2$ for any $|Z| > 0$. Since the set of all such $Z$ is not restricted to a finite set of values but rather rays in the complex plane,and none of the provided options represent this set,the question as posed with these options is mathematically inconsistent. However,if we look for the values of $u = Z/|Z|$,the set is $\{\omega, \omega^2\}$.

If $Z \neq 0$ is a complex number such that $Z^2 + Z|Z| + |Z|^2 = 0$,then $Z$ is in the set (Here $\omega$ is a complex cube root of unity).

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