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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 $i$? :)2012-05-12
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    @Zarrax that still leaves us with one $i$ term, doesn't it?2012-05-12
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    You can't. It stays complex equation.2012-05-12
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    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.2012-05-12
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    @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.2012-05-13

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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.

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    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. :)2012-05-12
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    @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!2012-05-13
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    I did not realize that. Thank you for clarifying.2012-05-13
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    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 stuck2012-05-16
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    @axell yes, you can. Just like I showed in the above answer.2012-05-16
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    How can I expand $$(-2ig)^{1/2} $$2012-05-19
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    @axell You do of course realize that $(-2ig)^{1/2}$ is the same as $\sqrt{-2ig}$, right? Then since, as I mentioned in the above answer, $\sqrt{-2 i}=\pm(1-i)$, we get that $\sqrt{-2ig}=\pm(1-i)\sqrt{g}$.2012-05-19
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    Yes I noticed that but I don't want the plus minus answers. I don't know how but my lecturer suggest me to factor out the equation to get only 1 solution.2012-05-19
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    @axell What your lecturer suggests you to do is probably to write it in this form: $i \frac{dx}{dt} = i(x\pm \sqrt{g}) \pm \sqrt{g}$ (see above). Here the $i$ term is not in any bracket, as required. If you don't want "the plus minus answer" just pick one and go with it, for example assume $\pm(1-i)=(1-i)$. This will work as well as $(i-1)$. The "plus minus" comes from the fact that if you take the square root of something the answer can either be positive or negative, and still give the same result. Consider for instance $\sqrt{4}$ - the answer is $\pm 2$ since $2^2=4$ but $(-2)^2$ is also 4.2012-05-19
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    @Milosz thanks a lot. Your explanations really help me.2012-05-19
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    @axell no problem, don't forget to upvote and accept if I answered your question! :)2012-05-19
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    @axell in case you don't understand how to accept an answer, you do it by clicking the green "check" mark on the left of the question, next to the vote arrows.2012-05-20