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Hello fellow mathematicians! I came across a question earlier which required completing the square of a quadratic with non-real coefficients. For context, I have started an extension module on complex numbers online.

$$ p(z) = (i+1)z^2+4z+3+4i$$ where $i$ is the imaginary unit and $z \in C$.

From prior knowledge, I would have to divide by the coefficient of the $z^2$ term first however I do not know whether this is appropriate as imaginary and real numbers exist in different dimensions so I have some doubts about performing division of any sort. More generally, do complex numbers follow all the same arithmetic as real numbers do, or are there any important exceptions I should know about?

Shooting Stars
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    "I have started an extension module on complex numbers online."... whut? – Clemens Bartholdy Nov 10 '22 at 09:59
  • I meant that there are websites such as brilliant, skillshare etc; – Shooting Stars Nov 10 '22 at 10:05
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    The complex numbers are a field: https://math.stackexchange.com/q/670236/42969 – Martin R Nov 10 '22 at 10:05
  • In the complex numbers , we have , for example , no order. But it is an algebraically closed field and has very useful properties (for more details google analytic functions). If we go to quaternions (or even higher dimensions) we lose so many properties that it is not worth the effort to define division or anything else there. – Peter Nov 10 '22 at 10:07
  • @MartinR thank you that was helpful – Shooting Stars Nov 10 '22 at 10:17
  • @Peter This is quite interesting because if we do lose so many properties, how are higher fields or dimensions useful to us anymore? – Shooting Stars Nov 10 '22 at 10:20
  • @ShootingStars Some mathematicians surely will claim that such exotics like quternions are very useful, but I do not see any merit to use them. – Peter Nov 10 '22 at 10:25

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Yes, you can divide by any nonzero complex number. For addition, subtraction, multiplication, and division, there are no surprises. Although when you divide by a complex number $a+ib$, getting rid of $i$ in the denominator by multiplying both the numerator and the denominator by the complex conjugate number $a-ib$ might be a good idea.

colt_browning
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