RE: Easy way to simplify expr to a+bi?

• To: mathgroup at smc.vnet.net
• Subject: [mg26282] RE: [mg26245] Easy way to simplify expr to a+bi?
• From: "David Park" <djmp at earthlink.net>
• Date: Sun, 10 Dec 2000 00:19:54 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com

```Christopher,

Here is one method.

Z = 1/(1/R1 + 1/(R2 + I*w*L) + I*w*C);

Simplify[ComplexExpand[Re[Z], TargetFunctions ->
{Re, Im}]]
(R1*(R1*R2 + R2^2 + L^2*w^2))/(2*R1*R2 + R2^2 +
L^2*w^2 + R1^2*(1 - 2*C*L*w^2 +
C^2*w^2*(R2^2 + L^2*w^2)))

Simplify[ComplexExpand[Im[Z], TargetFunctions ->
{Re, Im}]]
(R1^2*w*(L - C*R2^2 - C*L^2*w^2))/(2*R1*R2 + R2^2 +
L^2*w^2 + R1^2*(1 - 2*C*L*w^2 +
C^2*w^2*(R2^2 + L^2*w^2)))

I have to admit that manipulating complex functions in Mathematica is not
totally intuitive!

David Park

> From: crcarle at sandia.gov [mailto:crcarle at sandia.gov]
To: mathgroup at smc.vnet.net
> Hi:
>
> I often encounter complex expressions such as:
>
> Z = 1/( 1/R1 + 1/( R2 + I w L ) + I w C )
>
> which I want to simplify to the form of
>
> Z = a + b I
>
> so I can answer various questions about the behavior of either a or b .
>
> I can do this by hand of course, tediously adding the fractions,
> inverting, multiplying by the complex conjugate on top and bottom, then
> simplifying a long list of terms until I get there.
>
> It would be nice if I could do the equivalent of Re[Z] and Im[Z] and
> have Mathematica do the work for me, but that is not the result of those
> functions.
>
> Any ideas?
>
> Thanks.
> --
> _______________________
> Christopher R. Carlen
> Sr. Laser/Optical Tech.
> Sandia National Labs
>
>

```

• Prev by Date: Re: ListContourPlot3D
• Next by Date: Re: problem with Expand
• Previous by thread: Re: Easy way to simplify expr to a+bi?
• Next by thread: Newton's method