5/(a+bj) + 2/(1+3j) = 1

find a and b

where j = square root of -1.

I'm running out of ideas, and nothing seems to be working.

10 years ago, I would have enjoyed doing that
now I can't be arsed

have you tried googling it?

complex numbers are the devils work

I thought the square root of -1 was i anyway

where j = square root of -1.

Don't think you can have a square root of a negative number unless you are talking complex numbers and then it is i.

have you tried googling it?

Yeah, but googles usually useless for anything more complicated then 2×3.

They're ok chvck, but I still don't really understand the square root of -j.

And j or i is interchangeable, I think they use i in engineering and j in mathematics (or other way round). This question uses j. Change it to i if you really cant cope 😀

sounds like an electronics problem. When I last looked at maths like that, the root of minus 1 was i. The elcetronics guys used j

I thought the square root of -1 was i anyway

The letter "i" has been trade-marked by Apple,

Complex maths had to move on…

treat any occurrence of i^2 as -1,

multiply 1/(a+bj) top and bottom by (a-bj) gives
(a-bj). 1/ (a-bj)(a+bj) =
(a-bj). 1/ (a^2+b^2)

this gets all the j terms on the top line to be rearranged

with great restraint I avoid meerkat remark…

i is equivalent to j. For some reason mathematicians prefer i, physicists and engineers prefer j. Dunno why.

I also haven't been able to find a solution. Though if you assume that a and b probably range from -10 to 10, a very boring hour with a scientific calculator (or Google) will probably set you straight.

multiply by both sides by the brackets containing a and b.

Rearrange,

Solve.

Probably.

Yeah simonfbarnes, already tried that.

The first method I tried was multiplying the second part by the conjugate of the second part, which gives (2-6j)/10 so you then have

[(1/5)-(3/5j)] + 5/(a+bj) = 1

Which should suggest that a = 4/5 and b = -3/5 if you follow it through, but it didn't work out. Unless I made a mistake anyone can spot..

My second method was to multiply the first part by the second parts denominator, and the second part by the first parts denominator (like you used to do with really basic fractions). I eventually managed to pull out some simultaneous equations and got a = -5 and b = 0, which again, did not work.

But clearly, stw is much smarter then bikeradar.

http://www.bikeradar.com/forum/viewtopic.php?p=15620702#15620702

so 5/(a+bj) + 2/(1+3j) = 1
becomes
5(a-bj) 2(1-3j)
——– + ——- = 1
(a^2+b^2) (1+9)

sorry I can't get the formatting right 🙁
but you can rearrange it into a quadratic wossname

I don't think that's right simonfbarnes, doesn't make sense at all – I think you've just squared everything?

How about this – my fourth method (3rd was to ask you guys..)

Multiply everything by [(a+bj) + (1+3j)] so you get 5 + 2 = (a+bj) + (1+3j)

I think that's sort of what you were on to jonb

Which gives a + bj + 1 + 3j = 7

a + bj + 3j = 6

Split up real and imaginary to get

bj = -3j

b = -3

a = 6

???

Complex number notation was the junction in my life where i decided to follow a mechanical engineering route! Christ I hate complex numbers now 🙁

I'm doing electronics A2, physics A2, maths A2, further maths AS, and further maths A2 (second year at college). Did computing AS in my first year, but was boring and too easy, so decided to do the whole of further in my second year.

Electronics and physics are fun, and its nice to work with something a bit more real then a lot of the maths we do (imaginary numbers, for instance). However, all the formulas and boring bits get annoying. I'll do maths in uni.

multiply both sides by (a+bj) and also (1+3j) to remove all the fractions – defo not ((a+bj)+(1+3j)):

5(1+3j) + 2(a+bj) = (1+3j)(a+bj)

multiply out and collect terms:

5 + 15j + 2a +2bj = a +bj +3aj +3b(jxj)

remember that jxj = -1:

5 + 2a +15j + 2bj = a – 3b + 3aj + bj

now you can separate out real and imaginary parts to give simultaneous equations for a and b:

5 + 2a = a – 3b and 15 + 2b = 3a + b

solve for a and b.

any questions?

i make a = 4 and b = -3, but it's a while since i last did this

Yeah I've seen what I did wrong there, you're right. Cheers.

Has no one else noticed the BJ in the middle of that equation?

2BJ in fact.

Bonus!

physicists and engineers prefer j. Dunno why.

Because i is already taken, and likely to confuse if you use it as root(-1)

a and b are in the first set of brackets just after the 5!

Duuuuh you bunch of bloody muppets!

x 😀

I'm lazy, can't stand Maths, (due to my high school maths teacher)

try this though http://www.wolframalpha.com/ copy and paste it into there.. and watch it go..

I have no idea if it's right though

You know what, I was fine with the imaginary numbers part and generating the two equaltions. Can I solve the symiltaneous equation? Can I buggary! That's supposed to be the easy part! Any one care to show their workings for that bit?

Prat! I was tripping myself up with the signs.

Simultaneous equations is surely rather straightforward. Just rearrange one to get an expression for one of the variables, the substitute into the other equation.
e.g. given the equations above:
5 + 2a = a – 3b and 15 + 2b = 3a + b
rearrange first to:
a = -5 – 3b
sub into second:
15 + 2b = 3x(-5 – 3b) + b
I presume you can solve that for b, and then put that back in to get a?

Sometimes the elimination method is nicer for solving simple simultaneous equations though…

We do all realise that sockpuppet's soltuion was right though, right? 😯