btw.....I think it would be extremely helpful to newbies like me if the invlaplace help page explained why the answers are given in trig hyperbolic functions instead of the more traditional (at least for EEs, perhaps not for mathematicians?) and cover how the results can be converted to exponentials via the convert instruction.  The intuitive action is to try simplify --- I had no idea "convert" existed.  Simple documentation like that goes a long way and saves a lot of aggravation!!! 

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!