Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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: – Brett Frankel Jan 8 '13 at 19:16
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 $b$.

share|cite|improve this answer

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.