Let z = a + ib, b neq 0 be a complex number s | vedclass

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(B) Given $z^{2} = \overline{z} \cdot 2^{1-|z|}$. Taking the modulus on both sides,we get $|z|^{2} = |\overline{z}| \cdot 2^{1-|z|}$.Since $|\overline

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$6$

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(B) Given $z^{2} = \overline{z} \cdot 2^{1-|z|}$. Taking the modulus on both sides,we get $|z|^{2} = |\overline{z}| \cdot 2^{1-|z|}$.Since $|\overline{z}| = |z|$,we have $|z|^{2} = |z| \cdot 2^{1-|z|}$.Since $b 
eq 0$,$z 
eq 0$,so $|z| = 2^{1-|z|}$.By inspection,$|z| = 1$ satisfies the equation $(1 = 2^{1-1} = 2^{0} = 1)$.Substituting $|z| = 1$ into the original equation,$z^{2} = \overline{z}$.Multiplying by $z$,we get $z^{3} = z\overline{z} = |z|^{2} = 1$.Thus,$z$ is a cube root of unity,i.e.,$z = \omega$ or $z = \omega^{2}$,where $\omega = e^{i2\pi/3}$.We want to find the least $n \in N$ such that $z^{n} = (z + 1)^{n}$.Since $1 + \omega = -\omega^{2}$,the equation becomes $\omega^{n} = (-\omega^{2})^{n} = (-1)^{n} \omega^{2n}$.This implies $1 = (-1)^{n} \omega^{n}$,or $\omega^{n} = (-1)^{n}$.If $n = 3$,$\omega^{3} = 1$ and $(-1)^{3} = -1$ (not equal).If $n = 6$,$\omega^{6} = 1$ and $(-1)^{6} = 1$. Thus,$n = 6$ is the least natural number.

Let $z = a + ib, b \neq 0$ be a complex number satisfying $z^{2} = \overline{z} \cdot 2^{1-|z|}$. Then the least value of $n \in N$ such that $z^{n} = (z + 1)^{n}$ is equal to:

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