Calculating Re(z)

1. Feb 17, 2015

whatisreality

In the equation 1/z = 1/R +1/(iwL), why isn't the real part just R?

Unless my rearrangement is wrong, z=R+iwl, so Re(z) should definitely be R!

2. Feb 17, 2015

HallsofIvy

Staff Emeritus
Yes, your rearrangement is wrong. $\frac{1}{a}= \frac{1}{b}+ \frac{1}{c}$ does NOT give "a= b+ c". As a check, suppose b= c= 2. The 1/a= 1/2+ 1/2= 1 so a= 1 But b+ c= 4, not 1.

From $$\frac{1}{z}= \frac{1}{R}+ \frac{1}{iWL}$$, multiply on both sides by all denominators, RiWLz, to get RiWL= iWLz+ Rz= (R+ iWL)z. Solve that for z to find the real and imaginary part.

3. Feb 17, 2015

wabbit

So you're saying 1/(R+iwL)=1/R+1/(iwL) ?

4. Feb 17, 2015

whatisreality

Yes. I did actually think that was weird, not sure why I didn't check it. Let me try that rearrangement again...

5. Feb 17, 2015

whatisreality

z = -R2ω2L2/(iωL+R) is my second attempt!

Edit: maybe not. L squared too. And that makes life more difficult! What do I do about i on the denominator? Multiply by the conjugate?

Last edited: Feb 17, 2015
6. Feb 17, 2015

HallsofIvy

Staff Emeritus
Pretty standard: multiply both numerator and denominator by the complex conjugate of the denominator.
$$\frac{a+ bi}{c+ di}= \frac{a+ bi}{c+ di}\frac{c- di}{c- di}$$$$= \frac{ac+ bd+ (bc- ad)i}{c^2+ d^2}=$$$$\frac{ac+ bd}{c^2+ d^2}+ \frac{bc- ad}{c^2+ d^2} i$$

Here you want to multiply numerator and denominator by $$R- i\omega L$$.

7. Feb 17, 2015

whatisreality

z = RiwL/(R+iwL)
Realise the denominator:
(iR2wL+Rw2L2)/(R2+w2L2)

8. Feb 17, 2015

wabbit

You can do better than that (your first line is OK though).
Try this as a warm up : forget the exercise, just compute $$\frac{1}{a+ib}$$. Then go back to z.