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I don't know if this question has a solution, but I'll ask it anyway.I often come across complex numbers in the form

$$ \frac{(a_0 + ib_0)}{(a_1 + ib_1)(a_2 + ib_2)(a_3 + ib_3)} $$

but I need these in the normal form of

$$ a + ib $$

Now the way I solve this is that I multiply out the denominator and then rationalise the result but the problem is because the factors $a_i$ and $b_i$ are often a little complex this method is prone to mistakes.So I would like to know if there is an easier way I can do this by hand or if there is a way that I can check that my final answer is correct.

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Try writing your complex numbers in polar form $z=re^{i\theta}$, where multiplication is fast, then converting back to $z=a+bi$ form. This is explained here: en.wikipedia.org/wiki/Complex_number –  Brett Frankel Jan 8 '13 at 19:16

3 Answers 3

up vote 2 down vote accepted

As Brett Frankel points out, you have the following identity:
$$ a_i+i\cdot b_i = r_i e^{i\cdot \theta_i} $$ With $r_i^2 = a_i^2+b_i^2$ and $\theta = \arg(a_i+i\cdot b_i)$ where $\arg(\cdot)$ denotes the argument.

From that it follows, that your expression can be rewritten as \begin{align} a+i\cdot b &= \frac{r_0e^{i\cdot \theta_0}}{r_1e^{i\cdot \theta_1}\cdot r_2e^{i\cdot \theta_2}\cdot r_3e^{i\cdot \theta_3}} \\ &= \frac{r_0}{r_1\cdot r_2\cdot r_3} \cdot e^{i\cdot (\theta_0-\theta_1-\theta_2-\theta_3)} \\ &=: r \cdot e^{i\cdot \theta}\end{align}

And to come back from polar coordinate we have the identity $$ r \cdot e^{i\cdot \theta} = r ( \cos(\theta) +i\cdot \sin(\theta))$$ Which gives you $a$ and $b$.

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You just need to multiply your number and divide it by the conjugate number of the denominator!.

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The question was whether or not there was an easier way than expanding the brackets. –  Daniel Littlewood Jan 8 '13 at 19:16
@ Daniel Littlewood: In this case, you can try several forms of complex numbers: polar, trigonometric. But I believe that there is no other way than this. Can rewrite the denominator again here to see what I can do with it. –  ZE1 Jan 8 '13 at 19:23

The solution is

$\frac{a_0\,\left( a_1\,\left( a_2\,a_1-b_1\,b_1\right) +b_1\,\left( -a_1\,b_1-a_1\,b_1\right) \right) -b\_0\,\left( a_1\,\left( -a_1\,b_1-a_1\,b_1\right) -b_1\,\left( a_1\,a_1-b_1\,b_1\right) \right) }{\left( {b_1}^{2}+{a_1}^{2}\right) \,\left( {b_1}^{2}+{a_1}^{2}\right) \,\left( {b_1}^{2}+{a_1}^{2}\right) }+\frac{i\,\left( a_0\,\left( a_1\,\left( -a_1\,b_1-a_1\,b_1\right) -b_1\,\left( a_1\,a_1-b_1\,b_1\right) \right) +b\_0\,\left( a_1\,\left( a_1\,a_1-b_1\,b_1\right) +b_1\,\left( -a_1\,b_1-a_1\,b_1\right) \right) \right) }{\left( {b_1}^{2}+{a_1}^{2}\right) \,\left( {b_1}^{2}+{a_1}^{2}\right) \,\left( {b_1}^{2}+{a_1}^{2}\right) }$

Just multiply the brackets in the denominator and note that $\frac{1}{i}=-i$.

If you want to check for mistakes, why not use wxMaxima or Wolfram Alpha?

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We disagree on the meaning of "easier way". –  xavierm02 Jan 8 '13 at 20:59

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