Position to momentum space in three dimensions

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SUMMARY

This discussion focuses on transforming the hydrogen ground state wave function from position space to momentum space using the integral Ψ(p)=(2πħ)^(-3/2) [SIZE=16px][FONT=Noto Sans](πa^3)^(-1/2)∫∫∫e^(prcosθ/ħ) e^(-r/a) senθ r^2 dΦdθdr. The user successfully derived an expression involving the imaginary unit, specifically (ip/ħ-1/a)^(-2), but struggled with isolating the imaginary part. The professor confirmed the final result as Ψ(p)=(8πħ^4)/[(2πħ)^(-3/2) a^2 (4πa)^(1/2) (p^2+ħ^2/a^2)^2], which the user found challenging to achieve.

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Cleo
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Homework Statement
Mathematical complication solving Q.M. problem.
Relevant Equations
Im [(ip/ħ-1/a)^(-2)]
Hi! I am trying to change the hydrogen ground state wave funcion from position to momentum space, so i solved the integral
Ψ(p)=(2πħ)^(-3/2)
(πa^3)^(-1/2)
∫∫∫e^(prcosθ/ħ) e^(-r/a) senθ r^2 dΦdθdr
and got 4πħ(2πħ)^(-3/2) p^(-1) (πa^3)^(-1/2) I am [(ip/ħ-1/a)^(-2)], which according to the professor's solution is ok, but then i don't know which is the imaginary part that i need to pick. p is the momentum, i is the imaginary unit, a is Bohr radius, r is the radial coordinate, and I am means imaginary part of what is in the brackets.
So basically mi question is: How do i separate the imaginary part from the real part in this expression?:
(ip/ħ-1/a)^(-2)


The professor said that the final result is Ψ(p)=(8πħ^4)/[(2πħ)^(-3/2) a^2 (4πa)^(1/2) (p^2+ħ^2/a^2)^2], but i have no clue of how he did that final step. I guess it shouldn't be very difficoult but i have been long trying to figure out how to do it and for some reason i don't get it. It would be very nice if some of you could help me. Thanks!
 
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To get ##i## out of the denominator, multiply the top and bottom by the complex conjugate:
$$\frac{1}{a+bi} = \frac{a-bi}{(a+bi)(a-bi)} = \frac{a}{a^2+b^2} - i \frac{b}{a^2+b^2}$$
 
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Ahhh thanks a lot!
 

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