Question

Let Z=1/2(√3-i) then smallest positive integer `n' such that (z^95+i^67)^94=Z^n​

Answer 1

Answer:

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Answer 2

Answer:

$n=10$ is the required least positive integer.

Step-by-step explanation:

From the hypothesis, we have

$z=\frac{\sqrt{3}}{2}-\frac{i}{2}=i\left(-\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)=i \omega$

where $\omega=-\frac{1}{2}-\frac{\mathrm{i} \sqrt{3}}{2}$, which is a cube root of unity.

Now, $z^{95}=(i \omega)^{95}=-i \omega^{2}$

and $i^{67}=i^{3}=-i$

Therefore, $z^{95}+i^{67}=-i\left(1+\omega^{2}\right)=(-i)(-\omega)=i \omega$

$\Rightarrow\left(\mathrm{z}^{95}+\mathrm{i}^{67}\right)^{94}=(\mathrm{i} \omega)^{94}=\mathrm{i}^{2} \omega=-\omega$

Now, $-\omega=z^{n}=(i \omega)^{n}$

$\begin{aligned}&\Rightarrow \mathrm{i}^{\mathrm{n}} \cdot \mathrm{w}^{\mathrm{n}-1}=-1 \\&\Rightarrow \mathrm{n}=2,6,10,14, \ldots \ldots\end{aligned}$

and $n-1=3,6,9, \ldots \ldots$

Hence,$n=10$ is the required least positive integer.