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 March 5th, 2013, 02:35 AM #1 Member   Joined: Mar 2013 Posts: 56 Thanks: 0 Complex number help! Hallo I have to find all complex numbers v and u satisfying 1/v +1/u = 1/vu AND 2vu = u^2 -1 = 0 Please help! I don't know what to do.
 March 5th, 2013, 02:42 AM #2 Math Team   Joined: Apr 2010 Posts: 2,770 Thanks: 356 Re: Complex number help! 1/v +1/u = (u + v)/vu = 1/vu right?
 March 5th, 2013, 03:48 AM #3 Member   Joined: Mar 2013 Posts: 56 Thanks: 0 Re: Complex number help! using the conjugate yes. but where to pull complex numbers from!
 March 5th, 2013, 04:33 AM #4 Math Team   Joined: Apr 2010 Posts: 2,770 Thanks: 356 Re: Complex number help! conjugates? (u + v)/vu = 1/vu gives u + v = 1 right (for uv <> 0). In the second, eliminate v
March 5th, 2013, 08:17 AM   #5
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Re: Complex number help!

Quote:
 Originally Posted by razzatazz using the conjugate yes. but where to pull complex numbers from!
$\frac{1}{u}+ \frac{1}{v}= \frac{u+ v}{uv}$ for any numbers u and v. The "conjugate" has nothing to do with it.
So $\frac{1}{u}+ \frac{1}{v}= \frac{1}{uv}$ becomes $\frac{u+ v}{uv}= \frac{1}{uv}$ so u+ v= 1.

Now what you wrote for your other equation, "2vu = u^2 -1 = 0" is impossible. If 2vu= 0, then either u= 0 so 1/u is impossible, or v= 0 so 1/v= 0. I think what you intended was $2vu+ u^2- 1= 0$ (you missed the "shift" key!). From u+ v= 1, above, v= 1- u so the equation becomes
$2(1- u)u+ u^2- 1$$= 2- 2u^2+ u^2- 1$$= 1- u^2= 0$. Solve that equation for u, then find v.

This involves "complex numbers" only in the sense that the set of complex numbers includes the real numbers and so integers!

 March 5th, 2013, 03:23 PM #6 Member   Joined: Mar 2013 Posts: 56 Thanks: 0 Re: Complex number help! Sorry, I hit the wrong button meant to be 2uv-u^2-1=0 I solved for u = 1 and then with that, v = 1 as well. Is that right? Thank you so so so much for you help!
 March 6th, 2013, 06:26 AM #7 Global Moderator   Joined: Dec 2006 Posts: 17,204 Thanks: 1291 You need u + v = 1. Have you checked your working?
March 12th, 2013, 11:50 AM   #8
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Re: Complex number help!

Quote:
 Originally Posted by razzatazz Sorry, I hit the wrong button meant to be 2uv-u^2-1=0 I solved for u = 1 and then with that, v = 1 as well. Is that right? Thank you so so so much for you help!
If, when you say "I solved for u= 1", you mean $1- u^2= 0$, u= 1 is one solutions but, since that is a quadratic equation, there is another solution.

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