Dividing/Multiplying Imaginary Numbers

In summary, the conversation discusses the tediousness of multiplying and dividing imaginary numbers in AC circuit calculations. The two methods mentioned are using the complex conjugate for Cartesian form and dividing the magnitudes and subtracting the arguments for polar form. The conversation also mentions the use of technology, specifically the TI-83+ calculator, to simplify these calculations. It is also noted that polar form is the same as phasors. However, there is currently no simpler method for these calculations.
  • #1
cstoos
62
0
Is there a more convenient way to multiply and divide imaginary numbers than converting back and forth from phasors? (I guess I should say "when also having to add and subtract them")

I always find AC circuit calculations to be tedious and problem filled when I do it that way.

For example if I had to simplify something like:



(6+j3)/(6+j8)+(5+j9)/(2+j7)
 
Mathematics news on Phys.org
  • #2
Have you tried rationalizing the denominator? (I suppose "rationalize" is the wrong word here, but it's the same procedure)
 
  • #3
I think you are referring to multiplying by the complex conjugate...correct? A possibility yet still tedious when dealing with say, three impedences in parallel.

There is no way to directly divide? Seems like there should be.

Oh, and the above problem is completely made up, so no, I haven't tried that. I was just using it as a visual referrence as to what I was talking about.
 
  • #4
There are only two ways that I know about for dividing complex numbers. One way involves the complex conjugate for complex numbers in Cartesian form. The other way is to divide the magnitudes and subtract the arguments (angles) for complex numbers in polar form.
 
  • #5
Mark44 said:
There are only two ways that I know about for dividing complex numbers. One way involves the complex conjugate for complex numbers in Cartesian form. The other way is to divide the magnitudes and subtract the arguments (angles) for complex numbers in polar form.
I believe the "polar form" is what cstoos is referring to as "phasors".
 
  • #6
cstoos said:
Is there a more convenient way to multiply and divide imaginary numbers than converting back and forth from phasors? (I guess I should say "when also having to add and subtract them")...
(6+j3)/(6+j8)+(5+j9)/(2+j7)
Well, yes, there is technology!
On the TI-83+ (in a+ bi mode)
...
(6+3i)/(6+8i) + (5+9i)/(2+7i) (MATH-> FRAC)
=
524/265 - (329/530)i.
 
  • #7
You and your fancy calculators...

And yes, HallsofIvy is correct in his statement. Polar form is the same a phasors.

For example 6+3i would be written 6.7/26.6


Well, I guess my question was answered. No simpler way to do it. Now one of you just needs to find a new operation that will make my life easier.
 

1. What are imaginary numbers?

Imaginary numbers are numbers that can be written as a real number multiplied by the imaginary unit, i. They are used to represent quantities that cannot be expressed as real numbers, such as the square root of a negative number.

2. How do you divide imaginary numbers?

To divide imaginary numbers, you first need to rationalize the denominator by multiplying both the numerator and denominator by the conjugate of the denominator. Then, you can simplify the expression and combine any like terms.

3. Can you divide a real number by an imaginary number?

Yes, you can divide a real number by an imaginary number. This results in a complex number, which has both a real and imaginary part. The imaginary part will have the same value as the original imaginary number, and the real part will be equal to the real number divided by the absolute value of the imaginary number.

4. How do you multiply imaginary numbers?

To multiply imaginary numbers, you can use the FOIL method, just as you would with real numbers. Multiply the first terms, the outer terms, the inner terms, and then the last terms. Then, combine any like terms and simplify the expression.

5. Can you multiply two imaginary numbers with different imaginary units?

Yes, you can multiply two imaginary numbers with different imaginary units. This will result in a complex number with both real and imaginary parts. The imaginary parts will be the product of the two imaginary units, and the real part will be the product of the real coefficients.

Similar threads

Replies
3
Views
264
Replies
5
Views
2K
  • General Discussion
Replies
16
Views
1K
Replies
55
Views
3K
Replies
19
Views
2K
Replies
4
Views
409
Replies
2
Views
575
Replies
5
Views
2K
Replies
4
Views
3K
  • General Math
Replies
2
Views
3K
Back
Top