1. ## i operator help

I think I should make sure I have this right so I don't screw up the rest of my homework. Here's what I've got,

$(j3)(-j7)(-j)(j)=(j^2)(-j^2)(21)=(-1)(1)(21)=-21$

So, am I right or where did I go wrong. I think it's whether or not I signed that right. Thanks for the help, Jarod.

2. ## Re: i operator help

Originally Posted by Jarod_C
I think I should make sure I have this right so I don't screw up the rest of my homework. Here's what I've got,

$(j3)(-j7)(-j)(j)=(j^2)(-j^2)(21)=(-1)(1)(21)=-21$

So, am I right or where did I go wrong. I think it's whether or not I signed that right. Thanks for the help, Jarod.
Are you using $j$ as most people use $i~?$

Does $(j3)$ mean $j^3\text{ or }3j~?$

3. ## Re: i operator help

Sorry I should have clarified that. This is for the use of electronics. Since "i" is used to represent current we use j instead. Also $j3$ means $j\cdot 3$ or $3j$ Sorry for the confusion.

4. ## Re: i operator help

Originally Posted by Jarod_C
Sorry I should have clarified that. This is for the use of electronics. Since "i" is used to represent current we use j instead. Also $j3$ means $j\cdot 3$ or $3j$ Sorry for the confusion.
In that case $(j3)(-j7)(-j)(j)=(3)(7)(j^3)=-21j$

$\\j^0=1\\j^1=j\\j^2=-1\\j^3=-j$

$j^N=j^{N\text{Mod}~4}$

5. ## Re: i operator help

Thanks, as usual the help here is always fast and courteous.

6. ## Re: i operator help

Quick question, why does this result in $j^3$ instead of $j^4$? That's how I figured my answer of -21 because $j^4=(j^2)(j^2)=(-1)(-1)=1$

One more thing, do I need to make a to thread for a new problem? I thought I read that somewhere once.