reciprocals of complex numbers

[Anonymous No.5730857]

If Z1 and Z2 are complex numbers

and

1/Z = 1/Z1 +1/Z2

Can I say

Z1=Z2/Z1Z2

Z2=Z1/Z1Z2

1/Z = (Z1+Z2)/Z1Z2

so

Z=Z1Z2/(Z1+Z2)

?

[Anonymous No.5730861]

not if Z=0

[Anonymous No.5730878]

>>5730857
1/Z = (Z1+Z2)/Z1Z2
no.

Z1=Z2/Z1Z2
Z2=Z1/Z1Z2
Redistribute your perameters and make that inacurate to simplify to the x -> y

>>5730861
>especially if Z equals zero
>

[Anonymous No.5730879]

>>5730861

We're assuming Z isn't zero because 1/Z was given in the question

[Anonymous No.5730881]

>>5730879
How does this exclude Z=0? In this case 1/Z would be infinity.

[Anonymous No.5730883]

>>5730881

Infinity isn't a real number so obviously I have to assume Z isn't zero or else 1/Z would be undefined.

[Anonymous No.5730886]

>>5730883
OP is talking about complex numbers, not real numbers. Are you illiterate?

[Anonymous No.5730890]

Could somebody please tell me how to work out Z

[Anonymous No.5730901]

>>5730890
It's literally a pile of grabage.
This is science.
Bias is Pleb tier.
>Z is the Curtain

[Anonymous No.5731250]

Your question isn't even readable.

[Anonymous No.5731849]

>>5730886

All real numbers are complex numbers you stupid piece of shit.

And $ZZ_1Z_2\neq 0$ is a reasonable and obvious assumption in the above.

[Anonymous No.5731899]

Yes!

Scroll down to the 2nd frame here:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/imped.html

I was too lazy to explain it myself. Basically if you're working with circuits this is a MUCH faster way to add up parallel impedances than the usual Z = 1 / (1/z1 + 1/z2 ...) etc.

[Anonymous No.5731912]

>>5731849
K im not sure if my comps scripting isnt working or what but can you say that in english?

[Anonymous No.5731927]

>>5731912

Z * Z_1 * Z_2 != 0

In other words none of the variables are zero

[Anonymous No.5732093]

If you have complex numbers in your denominators, perhaps you should clear them by multiplying top and bottom by the complex conjugates of those numbers.

[Anonymous No.5732110]

>>5730857
What you did:
<div class="math"> \frac{1}{z} = \frac{1}{z_1} + \frac{1}{z_2} = \frac{z_1 + z_2}{z_1 z_2} \Rightarrow z = \frac{z_1 z_2}{z_1 + z_2}. </div>
This is right if <span class="math">z, z_1, z_2, z_1+z_2 \neq 0[/spoiler]. However, all of these follow directly from <span class="math"> \frac{1}{z} = \frac{1}{z_1} + \frac{1}{z_2}[/spoiler]. So if <span class="math">\frac{1}{z} = \frac{1}{z_1} + \frac{1}{z_2} [/spoiler] holds true you can calculate <span class="math">z[/spoiler] the way you did without problems.

[Anonymous No.5732121]

Actually, the product over sum formula you end with is used all the time in electronics for parallel impedance circuits.