What is the real part of z when given 1/z = 1/R + 1/(iωL)?

[whatisreality]

Homework Statement

Find the real part of z, given that 1/z = 1/R + 1/(iωL)

Homework Equations

The Attempt at a Solution

I rearranged to find z, and got that z = R + iωL. So that would make the Re(z)=R, wouldn't it?? It's wrong, but I don't have a clue why!

 

[Merlin3189]

When you rearranged, you got the wrong result. Can you show working for that please?
(You can't just turn all the fractions upside down.)

 

[whatisreality]

Ohhhhh soooo stupid... forgot to multiply the LHS when finding a common denominator.

 

[whatisreality]

Actually, hang on... here's the rearrangement.

1/z = 1/R + 1/(iwL) Multiply by R:

R/z = 1 + R/(iwL) Multiply by iwL:

RiwL/z = iwL + R

So z = (RiwL) / (iwL + R)?? There's clearly something wrong but I can't see what... Do I need to realize the denominator?

 

[milesyoung]

So z = (RiwL) / (iwL + R)?? ...

That's true, but you're asked to find Re(z). You can use this useful identity that comes up a lot in electrical engineering:
$$
z = \frac{z \overline{z}}{\overline{z}} = \frac{|z|^2}{\overline{z}}
$$
Along with:
$$
\mathrm{Re}\!\left(\frac{1}{\overline{z}}\right) = \mathrm{Re}\!\left(\frac{1}{z}\right)
$$

 

[Merlin3189]

Yep. Fine so far. You're quite right - what is wrong is that you still can't separate the Re and I am parts.

Miles has put something above (which I don't understand) and if that helps you, great.

For me, the next step is just plain maths, nothing to do with electronics. If you want to separate this expression into Re & Im, you need to have a real denominator and the standard maths 'trick' is to multiply a complex denominator by its complex conjugate.

Every complex number has a conjugate and when you multiply it by the conjugate you get a real number - the i's disappear.
Any fraction can be multiplied by 1 (or x/x since anything divided by itself is 1), so you multiply your fraction by conjugate/conjugate
Now the conjugate is dead simple: conjugate of A + iB is A - iB or conjugate of C - iD is C + iD (negate the I am part)
So you need to multiply your z expression by (R - iωL ) / ( R - iωL)

And just to complete the maths, remember the difference of two squares,
x² - y² = (x-y)(x+y)
So (A + iB)(A - iB) = A² - (i)²B²
= A² - (-1)B²
= A² + B²