Imaginary fraction questions. lost.

AI Thread Summary
The discussion centers on confusion regarding complex fraction problems encountered while studying for a circuits midterm. The original poster struggles to understand how to simplify expressions involving the imaginary unit "j" and the application of distributive properties. Clarifications are provided about proper notation, emphasizing the importance of using parentheses to avoid misinterpretation of terms like "j2" versus "2j." Additionally, advice is given on rationalizing denominators and ensuring consistent use of "j" or "i" to prevent confusion. Overall, the conversation aims to clarify mathematical processes relevant to circuit analysis.
ahhgidaa
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im studying for my circuits midterm and the proff has handouts with questions and answers but not detailed answers. i can't figure out how he went from a fraction to an answer.

(-j2)(2+j2)/-j2+2+j2 the answer on the paper is 2-j2


i do not know what I am allowed to do with the 2 next to the j with the distributive properties.


another example i was stuck on was i= -j2/1+j the answer here was rad2 at an angle of -135

and the above equation came from -j2+(2-j4) I_2 + (I_1 + I_2)j6=0 with I_1 =1

and insight or help would be appreciated. thakns
 
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ahhgidaa said:
im studying for my circuits midterm and the proff has handouts with questions and answers but not detailed answers. i can't figure out how he went from a fraction to an answer.

(-j2)(2+j2)/-j2+2+j2 the answer on the paper is 2-j2
First off, you need to learn how to write mathematical expressions so that they mean what you intend. I am willing to bet that on the handout it looks like this:

\frac{-2j(2 + 2j)}{-2j + 2 + 2j}

If you write fractions like this without using LaTeX to format them, put parentheses around the entire numerator and the entire denominator.

If you write 2j, people will be likely to mistake this for j2, which is something different.

For this problem, the first thing to do is to simplify the denominator.
ahhgidaa said:
i do not know what I am allowed to do with the 2 next to the j with the distributive properties.


another example i was stuck on was i= -j2/1+j
I'm guessing you mean i = -2j/(1 + j).

Look for an example where they rationalize the denominator by multiplying by the conjugate over itself.
ahhgidaa said:
the answer here was rad2 at an angle of -135

and the above equation came from -j2+(2-j4) I_2 + (I_1 + I_2)j6=0 with I_1 =1

and insight or help would be appreciated. thakns
 
i ended up with -2j-j^2 which i^2 is -1 so i finally got the answer 2-2j and i copied it exactly how it was but ure way of putting it help me with the math. idk why he writes it like that
 
1) Don't mix "j" and "i".

2) -2j- i^2= 1- 2j, not 2- 2j.

3) I think Mark44 meant to say that some would confuse "j2" with "j^2", not "2j".
 
HallsofIvy said:
1) Don't mix "j" and "i".
The OP is in an electronics class - they write j for the imaginary unit, probably because i is used for electrical current. Still, the advice is good. Use one or the other consistently, but don't use both.
Mark44 said:
2) -2j- i^2= 1- 2j, not 2- 2j.

3) I think Mark44 meant to say that some would confuse "j2" with "j^2", not "2j".
Mark44 said:
Right, that's exactly what I meant.
 
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