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Complex numbers - HSN forum

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Complex numbers


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#1 Ali89

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Posted 04 November 2007 - 05:56 PM

Hi

was wondering if anyone could help out with a niggling problem.

Complex numbers as a whole is fine- but i can't figure out (or find a lot of info)- on finding out how to calculate the reciprocal of a complex number!

i.e.

1 / (a + bi)

Thanks in advance folks.

#2 The Wedge Effect

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Posted 06 November 2007 - 09:31 PM

Just use the conjugate complex number rule, so what you end up with is:


(a-bi) / [ (a+bi) * (a-bi) ]

#3 George

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Posted 07 November 2007 - 08:42 AM

Just to give a little bit of explanation, we start with \frac1{a+bi}. If we multiply the a+bi by its complex conjugate, a-bi, then we'll just get a real number.

So if we multiply by 

\begin{align*}\frac{a-bi}{a-bi}\end{align*}

, we'll get a real number on the bottom. Also, we've just multiplied by 1 so our answer is the same as the original expression.

If you remember "rationalising the denominator" from Standard Grade, it's a bit like that.

#4 Ali89

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Posted 10 November 2007 - 01:24 PM

QUOTE(George @ Nov 7 2007, 08:42 AM) View Post
Just to give a little bit of explanation, we start with \frac1{a+bi}. If we multiply the a+bi by its complex conjugate, a-bi, then we'll just get a real number.

So if we multiply by 

\begin{align*}\frac{a-bi}{a-bi}\end{align*}

, we'll get a real number on the bottom. Also, we've just multiplied by 1 so our answer is the same as the original expression.

If you remember "rationalising the denominator" from Standard Grade, it's a bit like that.


cheers guys. Sorry, clearly had a blonde moment as this is pretty easy stuff. Also thought as well that seeing as 1 is a complex number anyway (1 + 0i) - then you could do (1+0i)/(complex number) and that would do the same thing. Slight bit more faffin' around though doing the latter.

Thanks again.





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