Sunday, 23 October 2016

When working with complex numbers, how can you solve for x when it's inside Re()?



I'm trying to figure out the impedance of a capacitor. My textbook tells me the answer is iωC and plugging that into the equation does work but I wanted to come up with that answer myself. So I wrote out the equation with what I know:




V0ωCsinωt=Re(V0(cosωt+isinωt)x)



This is where I get stuck. I don't know how to isolate x given that it is inside the Re() function. Trying to get somewhere, I tried this:



x=V0(cosωt+isinωt)V0ωCsinωt=cosωtωCsinωtiωC



Seeing iωC makes me feel like I'm on the right track. Now I just need to figure out how to get rid of the first part of that answer. And I'm guessing that if I knew how to isolate x from the first equation, that would do the trick. So how can I isolate x when it is included in the Re() function?


Answer



Re() is a projective map; Re(a+bi) = a. Thus Re(z) = z-iIm(z). So given RHS = Re(z), we have that RHS+bi = z for some real b. Note that Re(z) = a does not yield a single value of z as a solution, but instead gives a vertical line in the complex plane. Each point on that line will give a different value for x.




V0ωCsinωt+bi=V0(cosωt+isinωt)x



In terms of b, x will be:



V0(cosωt+isinωt)V0ωCsinωt+bi



Assuming that ω, V0, and C are real numbers, they can be "absorbed" into b; b is an arbitrary real number, so dividing by a real number just gives another arbitrary real number. So the above can be rewritten as



(cosωt+isinωt)ωC(sinωt+bi)




Factoring an i out of the numerator, we get



i(sinωticosωt)ωC(sinωt+ib)



Again, this describes a solution set, not a particular x. But if you take b=cosωt, then you recover the given expression. Any motivation for that choice will have to come from further facts about the capacitance rather than mathematical properties.


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