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<math>{(a+bi)^{-1}}={1\over(a + bi)} = {a\over a^2+b^2}+ \left( {-b\over a^2+b^2}\right)i</math> | <math>{(a+bi)^{-1}}={1\over(a + bi)} = {a\over a^2+b^2}+ \left( {-b\over a^2+b^2}\right)i</math> | ||
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| + | [http://en.wikipedia.org/wiki/Complex_number Wikipedia(Complex Numbers)] | ||
Revision as of 04:29, 2 September 2008
Finding the Inverse of Complex Numbers
$ {(a+bi)^{-1}}={1\over(a + bi)} = {a\over a^2+b^2}+ \left( {-b\over a^2+b^2}\right)i $
