BdMO National Junior 2007/9
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- Posts:1007
- Joined:Sat Dec 09, 2017 1:32 pm
$\sqrt {-1}$ is called the imaginary number '$i$'.Using this can you find the value of $\dfrac {1+i}{1-i}$?
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- Posts:1007
- Joined:Sat Dec 09, 2017 1:32 pm
Re: BdMO National Junior 2007/9
$i=\sqrt{-1}\Rightarrow i^2=-1$
Given expression,
\[\dfrac {1+i}{1-i}\]
\[=\dfrac {(1+i)(1+i)}{(1-i)(1-i)}\]
\[=\dfrac {(1+i)^2}{1-i^2}\]
\[=\dfrac {1+2i+i^2}{1+1}\]
\[=\dfrac {2i}{2}\]
\[=i\]
Given expression,
\[\dfrac {1+i}{1-i}\]
\[=\dfrac {(1+i)(1+i)}{(1-i)(1-i)}\]
\[=\dfrac {(1+i)^2}{1-i^2}\]
\[=\dfrac {1+2i+i^2}{1+1}\]
\[=\dfrac {2i}{2}\]
\[=i\]