math help

**slade** · 06-12-2005, 05:38 PM

Originally posted by wad04

ok so i'm trying to finish up this math course i took last year and its due in a few weeks. My tutor said i should be able to figure out the rest and she gave up on me. if you guys know the answers to any of these problems please help me out. and if you know the answer give me a detailed description of how you got it, i'm not just looking for an answer. thanks

Problem: 4-2i/7+3i

possible answers: 11-13i/29 OR 13-17i/29 OR 11-14i/29 OR 17-13i/29

I have a few more but i'll see if this will work.

you can also hit me up on aim if you like. somescreename04

not doing it now, but you multiply it by (7+3i)/(7+3i). that will remove the imaginary numbers from the denominator.

**tropical_fishy** · 06-13-2005, 12:48 AM

Originally posted by slade

not doing it now, but you multiply it by (7+3i)/(7+3i). that will remove the imaginary numbers from the denominator.

no, you multiply by (7-3i)/(7-3i)

It makes a perfect square.

(4-2i/7+3i)(7-3i)/(7-3i)=
(4-2i)(7-3i)/(49-9) (i^2=-1)
(28-12i-14i+6i^2)/(49-9)

and so on. It's late, my brain doesn't want to figure that out, but there yougo.

**gimp** · 06-13-2005, 09:46 AM

Here is another way to do it using phasors. I prefer working with phasors because it's more intuitive as to whats really going on. Say your complex number is a + jb (I use j instead of i but it's the same thing). Plot it in the complex plane. The complex plane is where the y axis is the complex term (b in our case), and the x axis is the real term (a in our case). You draw a line from the orgin to the point (a,b) and this is your phasor. The magnitude is just sqrt(a^2 + b^2). We'll call the angle phi. You get that with tan^-1(b/a). Let's call the magnitude Z. So you can express a+jb as Z at an angle phi. That is a phasor. Or you can write it in polar form, which is Ze^(jphi).

So, here is how you would do your problem using phasors.
4-2j = sqrt(4^2 + (-2)^2) at an angle of tan^-1(-2/4)
= 4.47 at an angle -26.56 degrees

7+3j = sqrt(7^2 + 3^2) at an angle of tan^-1(3/7)
= 7.62 at an angle of 23.2 degrees

Now you just do the math

(4.47 at -26.56 degrees) / (7.62 at 23.2 degrees)

The magnitude is just 4.47/7.62 = .587

The angle is a little different. You have to add the angles together. But, when you have an angle that is in the denominator, it's negative.

So your angle would be -26.56 + -23.2 = -49.76 degrees

Your final answer is .587 at -49.76 degrees in phasor form.
.587e^(j(-49.76deg)) in polar form.

You can convert it back to rectangular form really easily too. Just use a little trig.
Z at an angle phi = Zcos(phi) + jZsin(phi)

In the example, it would be:
.587cos(-49.76) + j(.587)sin(-49.76) = .379 - j(.448)

---------------------------
Your problably thinking "Gee gimp, why the heck would I do it that way?!?!" Well, it's a very practical in electrical stuff.

Phasors also apply to sinusiods. Acos(wt+phi) is represented as Ae^(jphi) or A at phi degrees.

This is really useful when your input is a sinusiod (120 V at 60hz for example) and your circuit has impedences in the form of a+jb. It's also very useful when working with power.

Also, it replaces all that difficult integration and derivation with simple algebra. Differentation in the time domain is just multiplication by jw in the phasor domain.

That's probably not what your teacher wants you to do, but I think it's interesting and I'm bored. I usually draw the phasor diagram to help me visualize what is actually going on in the system. You could just go buy a ti-89. That does everything for you.

**slade** · 06-13-2005, 05:19 PM

Originally posted by tropical_fishy

no, you multiply by (7-3i)/(7-3i)

It makes a perfect square.

(4-2i/7+3i)(7-3i)/(7-3i)=
(4-2i)(7-3i)/(49-9) (i^2=-1)
(28-12i-14i+6i^2)/(49-9)

and so on. It's late, my brain doesn't want to figure that out, but there yougo.

ehh, thats what i meant. make it the reciprocal or w/e so the expanded binomial wont have any imaginary numbers in it.

math help

math help

Comment

Comment

Comment

Comment