Select Page

# Complex numbers

Home Forums Math Olympiad Complex numbers

Viewing 2 posts - 1 through 2 (of 2 total)
• Author
Posts
• #60847
Akash Arjun
Participant

#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 $(ii)$]

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

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

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

$=-2$

Viewing 2 posts - 1 through 2 (of 2 total)
• You must be logged in to reply to this topic.