I have an equation which results in an answer of the form

$\frac{i a}{\sqrt{c-d}}$

is there any way to get Mathematica to present it in it's real form? like


I know that I can simply take Re[f[x]] to get the real part, but I don't just want the real part. If the answer is complex, I definitely want to know that, but when it is not complex and can be written as a fully real number, I would like it to give me that form.

My actual is example is far more complicated than this and the inside of the radical will always be positive if it is written as d - c, but I don't think i can put that assumption in because it is so complicated.

The actual example is :

(I*2^(5/2 + l)*Subscript[a, 1]^(5/2 + l)*
  (Subscript[a, 0] + Subscript[a, 1])^(4 + 2*l))/
 Sqrt[Gamma[4 + 2*l]*(16^(2 + l)*(3 + 2*l)*Subscript[a, 0]^(3 + 2*l)*
   Subscript[a, 1]^(5 + 2*l) - 2*(2 + l)*
   (Subscript[a, 0] + Subscript[a, 1])^(8 + 4*l))]

where I already have the assumptions :

{Subscript[a, 0] > 0, Subscript[a, 1] > 0, l >= 0, Element[l, Integers]}

and obtained the result (written above) when using FullSimplify on my system of equations.

  • $\begingroup$ The assumption here is that d>c? $\endgroup$
    – Eli Lansey
    Mar 22, 2012 at 20:37
  • $\begingroup$ Why not just do Simplify[I a/Sqrt[c - d], {d > c}]? Is your actual problem more complicated than this example? $\endgroup$
    – rm -rf
    Mar 22, 2012 at 20:37
  • $\begingroup$ Problem is there is no real form. There's a real part and there may be a form that "has the least I in it", which sounds like a very subjective criterion to me. $\endgroup$
    – David
    Mar 22, 2012 at 23:32

2 Answers 2


That results only holds if you assume that d > c. I.e.

Simplify[(I a)/Sqrt[c - d], Assumptions -> d > c]

The number is only real if $d > c$ so you really need to tell this to Mathematica:

Refine[(I a)/Sqrt[c - d], d > c]

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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