Complex numbers

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  • #60847
    Akash Arjun
    Participant

    Please help me to solve this problem

    #61011
    Saumik Karfa
    Moderator

    We have , $(1+\iota)^2=1+2\iota + \iota^2=1+2\iota-1=2\iota\ldots\ldots (i)$

    And,

    $\frac{1+\iota}{1-\iota}$

    $=\frac{(1+\iota)^2}{(1-\iota^2)}$ [Multiplying the numerator and denominator by $(1+\iota)$, the conjugate of $(1-\iota)$]

    $=\frac{2\iota}{1-(-1)}$ [by $(i)$ and $\iota^2=-1$]

    $=\frac{2\iota}{2}=\iota \ldots\ldots (ii)$

    Now,

    $\frac{(1+\iota)^{2011}}{(1-\iota)^{2009}}$

    $=\bigg(\frac{1+\iota}{1-\iota}\bigg)^{2009} \times (1+\iota)^2$

    $= (\iota)^{2009} . 2\iota$ [by $(i)$ and $(ii)$]

    $=2\times \iota^{2010}$

    $=2\times (\iota^2)^{1005}$

    $=2\times (-1)^{1005}$

    $=-2$

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