TIFR CSE 2011 | Part A | Question: 13

Question

If $z=\dfrac{\sqrt{3}-i}{2}$ and $\large(z^{95}+ i^{67})^{97}= z^{n}$, then the smallest value of $n$ is

• A. $1$
• B. $10$
• C. $11$
• D. $12$
• E. None of the above

Comments

$(cos (x) + i*sin (x))^n$=$cos(nx)+i*sin(nx)$ could also be used to make things simpler.

Answer 1

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

$\large z^2 =\dfrac 14 \left ( 3 -1 -2\sqrt{3}\,i \right ) = \dfrac 12 \left (1 - \sqrt{3}\,i \right )$

$\large z^4 =\dfrac 14 \left (1 - 3 - 2 \sqrt{3}\,i \right ) =\dfrac 12 \left (-1-\sqrt{3}\,i \right )$

$\large z^8 = \dfrac 14 \left (1 - 3 + 2\sqrt{3}\,i \right ) = \dfrac 12 \left ( -1+\sqrt{3}\,i\right )$

$\large z^{16} = \dfrac 14 \left (1-3-2\sqrt{3}\,i \right ) = \dfrac 12 \left (-1-\sqrt{3}\,i \right ) = \color{red}{z^4}$

$\large z^{32} =z^{16} \times z^{16} = z^4 \times z^4 = z^8$

$\large z^{64} = z^{32} \times z^{32} =z^8 \times z^8 = z^{16} = z^4$

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$\large z^{95} =z^{64} \times z^{16} \times z^{15}$

$\large = z^4 \times z^4 \times z^{15}$

$\large =z^{16} \times z^7$

$\large =z^4 \times z^7$

$\large =z^8 \times z^2 \times z$

$\large =\dfrac 12 \left ( -1+\sqrt{3}\,i\right ) \times \dfrac 12 \left (1 - \sqrt{3}\,i \right ) \times \dfrac 12 \left ( \sqrt{3}-i \right )$

$\large =\dfrac 12 \left (\sqrt{3} + i \right )$

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$\large i^{67} =i^{64} \times i^3$

$\large = 1 \times (-i)$

$\large =-i$

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$\large z^{95} + i^{67} =\dfrac 12 \left (\sqrt{3} + i \right ) - i$

$\large =\dfrac 12 \left (\sqrt{3} - i \right )$

$\large =z$

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$\large \left (z^{95} + i^{67} \right )^{97} = z^{97} = z^{95} \times z^{2}$

$\large =\dfrac 12 \left (\sqrt{3} + i \right ) \times \dfrac 12 \left (1 - \sqrt{3}\,i \right )$

$\large = \dfrac 12 \left ( \sqrt{3}-i \right )$

$\large = z$

Hence, option a is the correct answer.

Comments

Any simpler method please provide?

Answer 2

Polar form of $\frac{(\sqrt{3}-i)}{2}$ is $\cos\left ( \frac{\pi}{6} \right ) - i \sin \left ( \frac{\pi}{6} \right )$

So, $(z^{95} + i^{67})^{97} = ((\cos\left ( \frac{\pi}{6} \right ) - i \sin \left ( \frac{\pi}{6} \right ))^{95} – i)^{97}$ $\;\;\;\;[\because i^{67}=-i]$

$=(\cos\left ( \frac{95\pi}{6} \right ) - i \sin \left ( \frac{95\pi}{6} \right )\; -\;i)^{97}$ [Using De Moivre's Theorem]

$= \left ( \frac{\sqrt{3}}{2} + \frac{i}{2} - i \right )^{97}$

$= \left ( \frac{\sqrt{3}}{2} - \frac{i}{2}  \right )^{97}$

$=(\cos\left ( \frac{\pi}{6} \right ) - i \sin \left ( \frac{\pi}{6} \right ))^{97}$

$=\cos\left ( \frac{97\pi}{6} \right ) - i \sin \left ( \frac{97\pi}{6} \right )$  [Using De Moivre's Theorem]

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

$=z$

Hence, $n=1$

Answer 3

Given $z$=$\frac{\sqrt3-i}{2} \implies z=cos(330^{\circ})+isin(330^{\circ})$.

 $i^{67}={i^{(2)}}^{33} \times i=-1^{33} \times i=-i$

$z^{95}=(cos(330^{\circ})+isin(330^{\circ})^{95}=cos(95 \times 330^{\circ})+isin(95*330^{\circ})$[ By De Moivre’s Theorem]

=$\frac{\sqrt3+i}{2}$

$z^{95}+i^{67}=\frac{\sqrt3+i}{2}-i=\frac{\sqrt3-i}{2}=z$

$z^{97}=cos(97 \times 330^{\circ})+isin(97*330^{\circ})=\frac{\sqrt3-i}{2} \implies n=1$

Comments

how  z$^{2}=\frac{\sqrt{3}-i}{2}$

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Updated answer.

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it would be great if you update the answer with some new idea instead of completely changing your answer with the answer which is already written 5 hours ago on the same page. Previously, you have used cube root of unity. right ? If you use the same concept of cube root of unity here, it will be the shortest solution among all the solutions which are given here.

Answer 4

According to the definition of cube root of unity,

Consider, $\omega = \frac{-1 +i\sqrt{3}}{2} $ and $\omega^2 = \frac{-1 -i\sqrt{3}}{2} $ with $\omega^3 = 1$ and $1+\omega + \omega^2 = 0$

Here, $i \omega^2 = z$

Now, $(z^{95} + i^{67})^{97} = ((i\omega^2)^{95} + i^{67})^{97} = (i^{95}\omega^{2*95} – i)^{97}$ $[\because i^2=-1]$

$= (-i\omega^{190} -i)^{97} = (-i (\omega^3)^{63}\omega \;– i)^{97} = (-i \omega\; – i)^{97} = (-i(\omega + 1))^{97}$

$=(-i(-\omega^2))^{97} = (i\omega^2)^{97} = i^{97}\omega^{194} = i(\omega^3)^{64}\omega^2 = i\omega^2 = z$

Hence, $n=1$