Complex number query: "If Z = R + j(omega)L + 1/j(omega)C, express Z in (a+jb) form when..."

[vortexring]

In the above image, why (at the "6.25/j" section) in the next expression is the nominator and denominator multiplied by "-j". I can understand that you need a "j^2" in order to make it equivalent to "j^2=-1" but I don't understand the rule where you just multiply top and bottom by "-j" to get what you need, it doesn't seem to follow the usual algebraic rules, why aren't the other values in the expression also multiplied by "-j", why is it isolated to the fraction section of the expression?

Note: This is from an engineering maths book hence the use of "j" instead of "i"

 

[stapel]

View attachment 36615

In the above image, why (at the "6.25/j" section) in the next expression is the nominator and denominator multiplied by "-j". I can understand that you need a "j^2" in order to make it equivalent to "j^2=-1" but I don't understand the rule where you just multiply top and bottom by "-j" to get what you need, it doesn't seem to follow the usual algebraic rules, why aren't the other values in the expression also multiplied by "-j", why is it isolated to the fraction section of the expression?

Note: This is from an engineering maths book hence the use of "j" instead of "i"

They're rationalizing the denominator, so that it doesn't contain the imaginary. It's more a pretty-fying of the term than a mathematical necessity. (It's like back in pre-algebra, when the instructor cared about improper fractions and insisted on turning improper fractions into mixed numbers, but now that you're more advanced, improper fractions are fine, even preferred.)

 

[vortexring]

They're rationalizing the denominator, so that it doesn't contain the imaginary. It's more a pretty-fying of the term than a mathematical necessity. (It's like back in pre-algebra, when the instructor cared about improper fractions and insisted on turning improper fractions into mixed numbers, but now that you're more advanced, improper fractions are fine, even preferred.)

Thank you, it sounds so simple now you point that out, the complex number aspect makes you forget the basics!

 

[topsquark]

View attachment 36615

In the above image, why (at the "6.25/j" section) in the next expression is the nominator and denominator multiplied by "-j". I can understand that you need a "j^2" in order to make it equivalent to "j^2=-1" but I don't understand the rule where you just multiply top and bottom by "-j" to get what you need, it doesn't seem to follow the usual algebraic rules, why aren't the other values in the expression also multiplied by "-j", why is it isolated to the fraction section of the expression?

Note: This is from an engineering maths book hence the use of "j" instead of "i"

The expression for Z is equivalent impedance of a resistor, inductor, and capacitor in series. j would be the current in the circuit, not the imaginary unit. Why would this be a complex number problem? (In the Engineering world j is often used to represent a current, rather than the i used in Physics, but I've never before seen j used to represent [imath]\sqrt{-1}[/imath].)

Also (and I realize you are probably simply quoting this from a source), the last term should be written as [imath]1/(j \omega C)[/imath].

-Dan

 

[Dr.Peterson]

The expression for Z is equivalent impedance of a resistor, inductor, and capacitor in series. j would be the current in the circuit, not the imaginary unit. Why would this be a complex number problem? (In the Engineering world j is often used to represent a current, rather than the i used in Physics, but I've never before seen j used to represent [imath]\sqrt{-1}[/imath].)

I'm not a physicist or electrical engineer, but I've always understood the notation in both to be the reverse of what you say, as Wikipedia says:

In contexts in which use of the letter i is ambiguous or problematic, the letter j is sometimes used instead. For example, in electrical engineering and control systems engineering, the imaginary unit is normally denoted by j instead of i, because i is commonly used to denote electric current.​

And I think taking j as the imaginary unit in this problem is correct. Isn't it? Impedance of inductor and capacitors is imaginary.

Also (and I realize you are probably simply quoting this from a source), the last term should be written as [imath]1/(j \omega C)[/imath].

My understanding, increasingly, is that it is quite common in physics and elsewhere to take the multiplication, when written this way, as having higher priority than the division. I still recommend using the parentheses, but am less inclined to say "should" than I used to be.

 

[topsquark]

I'm not a physicist or electrical engineer, but I've always understood the notation in both to be the reverse of what you say, as Wikipedia says:

In contexts in which use of the letter i is ambiguous or problematic, the letter j is sometimes used instead. For example, in electrical engineering and control systems engineering, the imaginary unit is normally denoted by j instead of i, because i is commonly used to denote electric current.​

Whoops! I just double checked that, and you are right. I got that reversed, somehow. Thanks for the catch!

And I think taking j as the imaginary unit in this problem is correct. Isn't it? Impedance of inductor and capacitors is imaginary.

Yes, they are pure imaginary.

Okay, my excuse is that I'm on a new medication. But still, I'm heading for the corner as we speak. (@Steven G: I need my dunce cap back!)

-Dan