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I don't understand why Mathematica has to be such a pain in the ***.

ComplexExpand[(I*ω)^(-3/4), TargetFunctions->{Re, Im}]

The result is

$$\frac{\cos \left(\frac{3}{4} \tan ^{-1}(0,\omega )\right)}{\left(\omega ^2\right)^{3/8}}-\frac{i \sin \left(\frac{3}{4} \tan ^{-1}(0,\omega )\right)}{\left(\omega ^2\right)^{3/8}}$$

Now, all good BUT the arctangent.

I have tried EVERYTHING I could to FORCE Mathematica to understand that $\omega$ is a NATURAL number, and nothing.

It doesn't get it. Assuming, Assumption, Conditions...

Is there a way to get rid of that useless arctangent terms, which is, by the way, nothing but $\frac{\pi}{2}$ considering that indeed $\omega >0$ ?

Thank you!

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  • $\begingroup$ I think you're looking for Simplify with the Assumptions option: Simplify[ComplexExpand[(I*\[Omega])^(-3/4)], Assumptions -> {\[Omega] > 0}] $\endgroup$
    – jjc385
    May 15, 2017 at 15:35
  • $\begingroup$ @jjc385 That is very beautiful but then it gives me an expression in which real part and imaginary part are not separated anymore! $\endgroup$
    – Enrico M.
    May 15, 2017 at 15:36
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    $\begingroup$ Ahh done! I found it out! $\endgroup$
    – Enrico M.
    May 15, 2017 at 15:37

1 Answer 1

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Since I found the answer, I think it's good to give it.

Just use this:

ExpToTrig[Simplify[ComplexExpand[(I*ω)^(-3/4)], Assumptions -> {ω > 0}]]

And the output is: $$\frac{\sin \left(\frac{\pi }{8}\right)}{\omega ^{3/4}}-\frac{i \cos \left(\frac{\pi }{8}\right)}{\omega ^{3/4}}$$

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    $\begingroup$ I'm glad you figured it out! Have you also seen the Assuming function? Note that it only has an effect on functions like Simplify which take an Assumptions option. $\endgroup$
    – jjc385
    May 15, 2017 at 15:41
  • $\begingroup$ There is so much I still have to learn... Thank you! $\endgroup$
    – Enrico M.
    May 15, 2017 at 15:42

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