Question

Let z = (√3/2) - (i/2)
Then the smallest positive integer n such that (z^95 +i^-67)^94 = z^nis
O 12
O 10
O 9
O 8​

Answer 1

The correct option is (b) 10.

Step-by-step explanation:

From the hypothesis, we have

z=$\sqrt{3$/2-i/2=i $(-\frac{1}{2}- \frac{\sqrt{3} }{2})=iw$

Where $w= -\frac{1}{2} - i\frac{\sqrt{3} }{2}$  which is a cube root of unity.

Now, $z^{95}$ = $(iw)^{95} =-iw^{2}$

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

Therefore, $z^{95} + i^{67} =-i (1+ w^{2} )=(-i)(-w)-iw$

⇒$(z^{95} +i^{67} )^{94} =(iw)^{94} =i^{2}w=-w$

Now, $-w=z^{n} =(iw)^{n}$

⇒$i^{n} .w^{n-1} =-1$

⇒n=2,6,10,14

and n-1=3,6,9

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

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