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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