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if i have a complex variable z=x+iy, is there an inbuilt Mathematica function that extracts the real or imaginary parts of z. Something like Re[z] which would give Re[z]=x or Im[z]=y. Obviously this is a simple example that I wish to apply to a more complex situation.

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I have a variable (which I would like to define as 1/qo but it is tag protected)

qoInv = (-1 + I*f/zi)/(f - d + f*d*I/zi)

I would like to extract the real and imaginary parts of qoInv so I can solve a simultaneous equation by equating Re[qoInv]= 1/Ro and Im[qoInv]=1/zo, how do I go about doing this?

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closed as off-topic by J. M. will be back soon Oct 2 '16 at 12:57

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  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – J. M. will be back soon
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Am I missing something? Re and Im are built-ins that do just that... $\endgroup$ – ciao Aug 22 '16 at 7:13
  • $\begingroup$ That's odd.. I tried to use them and it didn't work. Maybe I did something wrong. I'll try again. $\endgroup$ – NormalsNotFar Aug 22 '16 at 7:13
  • $\begingroup$ I updated my post to show my result.. Is this supposed to be how it works? $\endgroup$ – NormalsNotFar Aug 22 '16 at 7:18
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    $\begingroup$ Per my comment, that's the expected output. You can either specify the domains, like FullSimplify[Re[x + I*y], {x, y} \[Element] Reals], or make a function to do the work and/or pattern transform for you. $\endgroup$ – ciao Aug 22 '16 at 7:24
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    $\begingroup$ Or use ComplexExpand $\endgroup$ – Simon Rochester Aug 22 '16 at 7:29

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