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November 08, 2025, 04:12:17 am
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leflyi

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Complex Number Question
« on: March 06, 2013, 09:40:15 pm »
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Right before I go to bed, not quite sure how to get around this w/o a calculator.




Find a and b
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Jeggz

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Re: Complex Number Question
« Reply #1 on: March 06, 2013, 09:41:36 pm »
+1
Quote from: leflyi on March 06, 2013, 09:40:15 pm
Right before I go to bed, not quite sure how to get around this w/o a calculator.




Find a and b

Have you tried expanding the left hand side, and then just equating the coefficients?  :)
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leflyi

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Re: Complex Number Question
« Reply #2 on: March 06, 2013, 09:44:57 pm »
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In which i get




Still unsure..

Edit : 3+4i to 3-4i changed
« Last Edit: March 06, 2013, 10:52:48 pm by leflyi »
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Conic

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Re: Complex Number Question
« Reply #3 on: March 06, 2013, 09:50:10 pm »
+1
Quote from: leflyi on March 06, 2013, 09:44:57 pm
In which i get




Still unsure..



1:   

2:   

Then go on from there with simultaneous equations.

Edit - typo
« Last Edit: March 06, 2013, 09:53:35 pm by Conic »
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Jeggz

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Re: Complex Number Question
« Reply #4 on: March 06, 2013, 09:56:38 pm »
+1
And then you would come up with the equations
and
If you solve the second one you get b=a/2 and hence you can sub this into the first equation. When you do so, you get. Times everything by , and you get
Let , and you get a nice quadratic equation of  which factorises into (x-4)(x+1). When we replace the a^2, we get . Solve this to get a and then substitute back in a to get b  :)

Edit: beaten by conic!
« Last Edit: March 06, 2013, 09:59:37 pm by myuraa »
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leflyi

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Re: Complex Number Question
« Reply #5 on: March 06, 2013, 10:07:08 pm »
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So how to know if it's pos 2, neg 2, pos 1, neg 1?

Edit: errors in writing
« Last Edit: March 06, 2013, 10:54:02 pm by leflyi »
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Daenerys Targaryen

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Re: Complex Number Question
« Reply #6 on: March 06, 2013, 10:14:47 pm »
+1
When you sub is pos a, it comes out as pos b
neg a gives neg b
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leflyi

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Re: Complex Number Question
« Reply #7 on: March 06, 2013, 10:23:42 pm »
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That is all fine and dandy until the answer I find is a = 2 , b = -1 and a =-2, b = 1
Is that separate or together ?
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Dayman

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Re: Complex Number Question
« Reply #8 on: March 06, 2013, 10:42:59 pm »
+1
How I do it is I make b the subject b=-2/a then substitute in a^2-b^2=4 find a then put it back into b=-2/a
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Ancora_Imparo

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Re: Complex Number Question
« Reply #9 on: March 06, 2013, 10:43:44 pm »
+2
Quote from: leflyi on March 06, 2013, 09:40:15 pm
Right before I go to bed, not quite sure how to get around this w/o a calculator.




Find a and b

Quote from: leflyi on March 06, 2013, 09:44:57 pm
In which i get




Still unsure..

I believe you have accidentally switched the sign next to the between the above quoted posts in this thread.
If you keep the negative sign, you will get the desired result of , which is one solution, and , the other solution.
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Re: Complex Number Question
« Reply #10 on: March 06, 2013, 10:44:28 pm »
+1
Quote from: leflyi on March 06, 2013, 10:23:42 pm
That is all fine and dandy until the answer I find is a = 2 , b = -1 and a =-2, b = 1
Is that separate or together ?

separate solutions

Edit: Beaten yet again!  :P
« Last Edit: March 06, 2013, 10:46:51 pm by myuraa »
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leflyi

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Re: Complex Number Question
« Reply #11 on: March 06, 2013, 10:44:58 pm »
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Yeah! Hahah, I was about to go to bed and it hit me, 2ab = -4, therefore the signs switched..

Thanks all!

Edit: Yeah separate solutions, i understood that bit, I only had just realised that 2ab = -4, not +4.
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Dayman

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Re: Complex Number Question
« Reply #12 on: March 06, 2013, 10:45:31 pm »
-1
Lol
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