0
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ I think you're looking for Simplify with the Assumptions option: Simplify[ComplexExpand[(I*\[Omega])^(-3/4)], Assumptions -> {\[Omega] > 0}] $\endgroup$ – jjc385 May 15 '17 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! $\endgroup$ – Henry May 15 '17 at 15:36
  • 1
    $\begingroup$ Ahh done! I found it out! $\endgroup$ – Henry May 15 '17 at 15:37
3
$\begingroup$

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

$\endgroup$
  • 3
    $\begingroup$ 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 '17 at 15:41
  • $\begingroup$ There is so much I still have to learn... Thank you! $\endgroup$ – Henry May 15 '17 at 15:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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