found it....evalf will evaluate  radicals numerically.

@tomleslie 

How do I simplify the expression to resolve the radicals in one "simplify" command?  right now, I can only do it by selecting "approximate" (from the side panel) on the result from the simplify expression.  I tried simplify(convert(temp2,exp) , radical)  but that didn't work.

@tomleslie thank you.  You misunderstood---I was wondering why the invlaplace resulted in sinh and cosh terms, and why i wouln't simplify to the "other" answer if in fact they are the same.  I had to plot them to find out that they were indeed the same.  I was not surprise of the different form of the answer --- just that I couldn't make them look the same!

@acer 

I'd prefer to have the mantissa greater than or equal to 1, so I'd prefer 6e-10 to be represented as 600e-12.

Thank you!

@acer   wow..thank you for your response. I've never worked with modules, so I'll have to take some time to understand what you did.

Thanks again!

@vv thank you !

@dharr thank you. your response helped a lot.

Here's the worksheet.....I'm not sure what happened that last time.

Thank you

simplify_complex.mw

@Joe Riel Hi, thank again for your support.  I thought I had tried that, but I must have had some other issue then  because readding the * to the first line got me over the "algebraic format" error.  I think I have something that works now....Thank you.

@mmcdara very instructive response.  Thank you!!

@Joe Riel hi, yes - I defined impedance, then intended to use the Im(Z) to get the reactance.  I did find a few mistakes which I lster corrected - snd figured out how to make the subs and solve the equations.  All good.

thank you for the help!

@acer Thank you for your help.    What is the reason assume was not working in my case?  Is there any time when I should use one  vs. the other, or should I always use assuming?

Also, off-topic but par to the same worksheet,

how to I do something like this:

solve(subs(f=13.56e6, algosubs(Lp=186/(2*Pi*f),Q=50, eq2),Rsrc). 

just trying to sove eq2 substituting values for f and Lp, and Q.

Ultimately, I'd like to solve for Rsrc, C1, and C2, given f, Lp, and Q, using eq1, eq2, and xfer_re=some amplitude

 I'm struggling with how to do this.

Thank you

@Joe Riel  that is awesome. Thank you.

What documentation do I look at to learn more about what Syrup can do? 

and...these types of commands:

map(f -> op(0,f) = op(f), indets(Ckt, 'function(numeric)'));

subsindets(Ckt, 'function(numeric)', f -> (op(0,f)));

I guess these are Maple instructions, right? --- other than looking into each instruction individually (eg. map, --> , indets, subsindets, etc), is there something that documents something like what you did here to transform Ckts into ckts?

@Joe Riel   thank you.  I don't understand fully how the conversion bit was done, but I will work through it.  

In the last line, 

(**) subs(Syrup:-Solve(ckt), v[4]);
                                                              V
                                             ------------------------------------
                                              3  3  6      2  2  4          2
                                             C  L  s  + 6 C  L  s  + 9 C L s  + 1

I'm having trouble understanding what I'm looking at.  What am I looking at in the denominator? I see the component values in  s....oh! wait, the 3 3 6 2 2 4 and 2 are the exponents to the  C's, L's, and s's ---just looked weird!  never mind.  Thank you!

@tomleslie I appreciate the example you provided above.  I like the ladder syntax  for simple circuits because it makes it easy to find any mistakes.  But, I agree, for more complex circuits, I would probably use the Spice netlist format.

The example you provided covers a lot of the most common things that I typically use Maple for (filter design, frequency response, etc -- obtaining transfer functions and plotting frequency responses out.  Thank you  --it is a nice referene for me to  refer to going forward.

Thanks.