Complex numbers

Viewing 3 posts - 1 through 3 (of 3 total)
  • Author
    Posts
  • #61058
    Akash Arjun
    Participant

    be a set of complex numbers, where n is a positive integer. How many distinct elements are in the set A ?

    #61089
    venkat jothi
    Participant

    here omega is the cube root of unity

    • This reply was modified 1 month, 4 weeks ago by venkat jothi.
    #61116
    Saumik Karfa
    Moderator

    $\left(\frac{1+\sqrt{3} i}{2}\right)$

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

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

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

    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.
Viewing 3 posts - 1 through 3 (of 3 total)
  • You must be logged in to reply to this topic.