# square root of complex differential equation

Let's say I have complex equation

$$i \frac{dx}{dt} = i x+ (-2ig)^{1/2}$$

$i$ is a complex number and $g$ is just some constant

How do I eliminate the $i$?

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divide both sides by $i$? :) –  Zarrax May 12 '12 at 15:37
@Zarrax that still leaves us with one $i$ term, doesn't it? –  Milosz Wielondek May 12 '12 at 16:18
You can't. It stays complex equation. –  user2468 May 12 '12 at 16:29
You won't eliminate $i$ entirely from the equation, but you'll have $dx/dt - x =$(complex) constant which you can solve using the usual methods, getting a complex-valued solution. There are no real solutions. –  Zarrax May 12 '12 at 17:05
@axell don't forget to accept an answer if it answers your question! This way you mark your question for others to see as answered. –  Milosz Wielondek May 13 '12 at 14:18

Notice that $\sqrt{-2 i}=\pm(1-i)$. Hence the equation becomes $$i \frac{dx}{dt} = i(x\pm \sqrt{g}) \pm \sqrt{g}$$
and it becomes clear that there's no way of completely eliminating $i$ from the equation.
I don't think $\sqrt{-2i}=\pm(1-i)$ is relevant. I think it can be noticed right away: $$i\frac{dx}{dt}=ix+(-2ig)^{\frac{1}{2}}=ix+i^{\frac{1}{2}}(-2g)^{\frac{1}{2}}$$ Observing that $i$ is not to the same power throughout both sides of the equation automatically convinced me it's not going anywhere. But we may see things differently. :) –  000 May 12 '12 at 19:02
@Limitless I was extensively elaborate on purpose, to make it more clearer for the OP. Seeing that each term is has (different powers of) $i$ in it, might mislead an untrained eye to believe that the $i$ can be factored out and eliminated. Having that said, your argument is of course 100% valid, only perhaps not as obvious! –  Milosz Wielondek May 13 '12 at 0:13
Sorry for the late reply guys. Thank you for all the responses. So, if I can't solve the equation to get complete real equation, can I at least bring out the $$i$$ from the square root? I mean the complex number without the fractional power outside the bracket? I was thinking of using Binomial expansion but the fractional power made me stuck –  axell May 16 '12 at 3:51