# Easiest way to simplify very big complex number

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.

-
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

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 $b$.

-

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

-
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