# Homework Help: Find Re(z)

1. Feb 17, 2015

### whatisreality

1. The problem statement, all variables and given/known data
Find the real part of z, given that 1/z = 1/R + 1/(iωL)

2. Relevant equations

3. 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!

2. Feb 17, 2015

### 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.)

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

### whatisreality

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

4. Feb 17, 2015

### 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 realise the denominator?

5. Feb 17, 2015

### milesyoung

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:
$$\mathrm{Re}\!\left(\frac{1}{\overline{z}}\right) = \mathrm{Re}\!\left(\frac{1}{z}\right)$$

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

### Merlin3189

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