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Tagged: complex numbers

- This topic has 2 replies, 3 voices, and was last updated 1 month, 4 weeks ago by Saumik Karfa.

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- March 29, 2020 at 11:15 pm #61058Akash ArjunParticipant
be a set of complex numbers, where n is a positive integer. How many distinct elements are in the set A ?

March 30, 2020 at 10:53 am #61089venkat jothiParticipanthere omega is the cube root of unity

- This reply was modified 1 month, 4 weeks ago by venkat jothi.

March 30, 2020 at 2:19 pm #61116Saumik KarfaModerator$\left(\frac{1+\sqrt{3} i}{2}\right)$

$=\left[\frac 12+\frac{\sqrt 3}{2}\iota\right]$

$=[\cos \frac{\pi}{3}+\iota\sin \frac{\pi}{3}]$

$=[e^{\frac{\iota\pi}{3}}]$

Therefore $\left(\frac{1+\sqrt{3} i}{2}\right)^{n}=[e^{\frac{\iota\pi}{3}}]^n$

$=[e^{\frac{n\iota \pi}{3}}]$

Elements of $A$ are

Now for $n=1$

$a_1=e^{\frac{\iota \pi}{3}}$

for $n=2$

$a_2=e^{\frac{2\iota \pi}{3}}$

for $n=3$

$a_3=e^{\frac{3\iota \pi}{3}}=e^{\iota\pi}=-1$

for $n=4$

$a_4=e^{\frac{4\iota \pi}{3}}=e^{\iota\pi}\cdot e^{\frac{\iota \pi}{3}}= -e^{\frac{\iota \pi}{3}}$

for $n=5$

$a_5=-e^{\frac{2\iota \pi}{3}}$

for $n=6$

$a_6=e^{\frac{6\iota \pi}{3}}=(-1)^2=1$

Now for $n=7,8,\ldots$ the elements will repeat.

Therefore only $6$ distinct element in $A$.

- This reply was modified 1 month, 4 weeks ago by Saumik Karfa.
- This reply was modified 1 month, 4 weeks ago by Saumik Karfa.
- This reply was modified 1 month, 4 weeks ago by Saumik Karfa.

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