What is the purpose of imaginary numbers and how do they work?

[Gokul43201]

In Ohio High Schools, they do not teach you what a voltage (potential difference) is during your regular physics. But they do cover Ohm's Law...and yet, for some (unfathomable to me) reason, they no not use the words "voltage" or "potential difference."

I don't get it

 

[JasonRox]

It's not that bad over here.

I pity you guys.

 

[Gonzolo]

So NanakiIII, my post 37 is just about everything there is to know about multiplicating complex numbers. If you understand that, you are ready to fully understand division, because dividing a complex number is actually multiplying it by 1, that is, a fraction with identical numerator and denominator.

You choose the fraction=1 such that its denominator is identical to the dividing denominator, except for the sign in front of the i. That is what I do in the first step below. Then, it's all multiplication exactly as in post 37.

\frac{a + ib}{c + id} = \frac{a + ib}{c + id} X 1 = \frac{a + ib}{c + id} X \frac{?}{?}

We choose c - id, because the - sign in front of its i is the opposite of the denominator's + in c + id. We call this the conjugate. (c - id is the conjugate of c + id and vice-versa).

When we choose this and multiply, the i's in the denominator disappear :

\frac{a + ib}{c + id} = \frac{a + ib}{c + id} X \frac{(c - id)}{(c - id)} = \frac{ac - iad + ibc - (i^2)bd}{c^2 - icd + icd - (i^2)d^2}

= \frac{ac - (-1)bd + i(bc - ad)}{c^2 -(-1)d^2} = \frac{ac + bd + i(bc - ad)}{c^2 + d^2} = \frac{ac + bd}{c^2 + d^2} + i\frac{bc - ad}{c^2 + d^2}

...which was you mentionned you learned in couple form (x,y).

 

[NanakiXIII]

Okay, it took me a while, but I traced those formulas and see how you got there. But what I don't really see is why you use c-id. Why do you need the opposite of the denominator?

 

[TenaliRaman]

1/(sqrt(2)-1) = (sqrt(2)+1)
i multiplied numerator and denominator by sqrt(2)+1
but why did i multiplied by sqrt(2)+1??
bcos (sqrt(2)+1)(sqrt(2)-1) simplifies to an integer.
This is called rationalising the denominator.

A similar technique has been used above.
Bcos (c-id) (c+id) gives u a real number , we multiply numerator and denominator by the "conjugate" of the denominator.

-- AI

 

[NanakiXIII]

I still don't really understand. But the main thing is clearer now. Thanks.

 

[MiGUi]

The reason your calculator reports an error is the same that a student of ten years says that we can not do 5 - 7 because he doen't know anything about negative numbers.

 

[NanakiXIII]

...what? I think we were past that, but thanks anyway.

 

[Gonzolo]

Okay, it took me a while, but I traced those formulas and see how you got there. But what I don't really see is why you use c-id. Why do you need the opposite of the denominator?

The reason is that when you use it, all the i's in the denominators dissappear, so that what you get in the end is still in the form a + ib.

We always write a complex number with no i in the denominator, so as to immediately recognize the real part (a) and the imaginary part (b).

 

[NanakiXIII]

Thanks, Gonzolo.

 

[JasonRox]

Pick up a short complex number book. It will teach you a lot, and then you can impress all the girls with your knowledge. ;)