If z = x + iy (x, y in R, x neq -1/2),the num | vedclass

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(B) The given equation is $|z|^n = (z^2 + z)|z|^{n-2} + 1$.Dividing by $|z|^{n-2}$ (assuming $z 
eq 0$),we get $|z|^2 = z^2 + z + \frac{1}{|z|^{n-2}}

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$1$

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(B) The given equation is $|z|^n = (z^2 + z)|z|^{n-2} + 1$.Dividing by $|z|^{n-2}$ (assuming $z 
eq 0$),we get $|z|^2 = z^2 + z + \frac{1}{|z|^{n-2}}$.Since $|z|^2 = z\bar{z}$,we have $z\bar{z} = z^2 + z + \frac{1}{|z|^{n-2}}$.Rearranging,$z^2 - z\bar{z} + z + \frac{1}{|z|^{n-2}} = 0$.For $z^2 + z$ to be real,$z^2 + z = \bar{z}^2 + \bar{z}$,which implies $(z - \bar{z})(z + \bar{z} + 1) = 0$.Given $x 
eq -1/2$,$z + \bar{z} + 1 = 2x + 1 
eq 0$,so $z = \bar{z}$,meaning $z$ is real $(z = x)$.The equation becomes $x^n = x^2|x|^{n-2} + x|x|^{n-2} + 1$.Since $x$ is real,$|x|^n = (x^2 + x)|x|^{n-2} + 1$.If $x > 0$,$x^n = x^n + x^{n-1} + 1 \Rightarrow x^{n-1} = -1$,which is impossible for $x > 0$.If $x  0$. Then $k^n = (k^2 - k)k^{n-2} + 1 = k^n - k^{n-1} + 1$.This implies $k^{n-1} = 1$,so $k = 1$,which means $x = -1$.Thus,there is only $1$ value of $z$ satisfying the equation.

If $z = x + iy$ $(x, y \in R, x \neq -1/2)$,the number of values of $z$ satisfying $|z|^n = z^2|z|^{n-2} + z|z|^{n-2} + 1$ $(n \in N, n > 1)$ is

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