# Solve for a and b if:

• Mar 5th 2012, 09:05 AM
Scoplex
Solve for a and b if:
Solve for a and b if: (jb)/(1-j)=(a-j)2j

Any help will be much appreciated.
• Mar 8th 2012, 01:42 AM
princeps
Re: Solve for a and b if:
Quote:

Originally Posted by Scoplex
Solve for a and b if: (jb)/(1-j)=(a-j)2j

Any help will be much appreciated.

$\displaystyle \frac {b\cdot j}{1-j}=2+2a \cdot j \Rightarrow b \cdot j =(2+2a \cdot j)(1-j) \Rightarrow$

$\displaystyle \Rightarrow b \cdot j = (2a+2) +(2a-2) \cdot j \Rightarrow$

$\displaystyle \Rightarrow 2a-2 = b ~\text {and}~2a+2=0$
• Mar 8th 2012, 05:58 AM
Wilmer
Re: Solve for a and b if:
Quote:

Originally Posted by princeps
$\displaystyle \frac {b\cdot j}{1-j}=2+2a \cdot j$

How did you arrive at 2 + 2aj?
Isn't it 2aj - 2j^2 ?
• Mar 8th 2012, 07:16 AM
princeps
Re: Solve for a and b if:
Quote:

Originally Posted by Wilmer
How did you arrive at 2 + 2aj?
Isn't it 2aj - 2j^2 ?

$\displaystyle j^2=-1$
• Mar 8th 2012, 08:18 AM
Wilmer
Re: Solve for a and b if:
(j*b) / (1 - j) = 2*j*(a - j)

We're to solve for a and b;
b is straightforward: b = [2j(a - j)(1 - j)] / j

a = [2j(j - 1) - b] / [2(j -1)]

That checks out perfectly with the initial equation.
And there's loads of solutions; examples with all positive integers:
(a,b,j) = (1,2,2), (2,4,3), (1,8,3), (3,6,4), (4,8,5)

Can you explain why you jump to j^2 = -1 ? (Speechless)
• Mar 8th 2012, 08:52 AM
princeps
Re: Solve for a and b if:
I think that $\displaystyle j$ is imaginary unit .
• Mar 8th 2012, 09:52 AM
Wilmer
Re: Solve for a and b if:
Quote:

Originally Posted by princeps
I think that $\displaystyle j$ is imaginary unit .

Why? Absolutely no mention of that in original post.
Anyway, ask me if I care (Nerd)