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  4. If z=(√3+i/2), then the value of z69 is

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Q. If $z=\frac{\sqrt{3}+i}{2}$, then the value of $z^{69}$ is

KEAMKEAM 2019

Solution:

Given, $z=\frac{\sqrt{3}}{2}+\frac{i}{2}$
$i z=\frac{-1}{2}+i \frac{\sqrt{3}}{2}=\omega$
And $z^{69}=z^{4(17)} z=(i z)^{4(17)} z=(\omega)^{68} z \,\,\, \left(\because i^{4 n}=1\right)$
$=\frac{\omega^{69}}{i}=\frac{\left(\omega^{3}\right)^{23}}{i}=-i$