How do I solve sol1 in terms of DC and T (and tau)?

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Say I have ,

sol1 := C*vout*vin/(Iout*L*k^2)

and I want to simplify this expression by replacing  with .

Then I need to simplify it further by noting that , and that , so that the answer is only in terms of , , and .

I tried eval(sol1, vin/Iout = omega*Lm) initially to step through the first substitution, but that just returned the same expression.

Thank you.

I can't figure how to simplify

 V:= V1*R2/(s*C*R1*R2+R1_R2)

into

V = V1 *R2/(R1+R2)/(C*R1*R2*s/(R1+R2)+1)

with Maple

Basically, divide numerator and denominator by the s^0 coefficient.

This is the sort of thing that drives me nuts with Maple.

in the example below, how do I factor (L-Lm) so that the answer is in the form L1 :=L(1-k) instead of -L(k-1)?

Lm := (simplify(k*sqrt(L*L)) assuming (0 < L));
                           Lm := k L

L1 := (factor(L - Lm) assuming (0 < L and 0 < Lm and 0 <= k));
                        L1 := -L (k - 1)

in this simple case, it is easy to understand the result.  However, these kind of  representations in more complex solutions may obfuscate the result making it difficult to interpret its meaning, where as the meaning may be more obvious if the simplification or factorization led to a result formatted more similarly to the way a human would do it.  How do I get Maple to simplify things the more "traditional" way?

How do I simplify :

Zin := Rin + omega*L*I + (omega*L)^2/(RL + omega*L*I);
 

so that there are no imaginary terms in the denominator?  

I've tried, 

assum := Rin::real, omega::real, L::real, RL::real, 0 < Rin, 0 < omega, 0 < RL, 0 < L
Zin := Rin + omega*L*I + (omega*L)^2/(RL + omega*L*I);
 
(simplify(Zin) assuming assum);

but this doesn't work.