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Further Pure 1 help

https://jacktilson.net/edu/maths/markschemes/summer-2005.pdf
If you go to page 26 , to question 7b! How do they get the final answer of u^2+v^2=1/2
Ive done everything on top but i cant seem to get the final answer tbh?
Reply 1
Avatar for Notnek
Notnek
21
Original post by English-help
https://jacktilson.net/edu/maths/markschemes/summer-2005.pdf
If you go to page 26 , to question 7b! How do they get the final answer of u^2+v^2=1/2
Ive done everything on top but i cant seem to get the final answer tbh?

Since they have the same denominator, you can add the fractions to get

u2+v2(u2+v2)2=2\displaystyle \frac{u^2+v^2}{\left(u^2+v^2 \right)^2} = 2

Now try cancelling the fraction.
Original post by notnek
Since they have the same denominator, you can add the fractions to get

u2+v2(u2+v2)2=2\displaystyle \frac{u^2+v^2}{\left(u^2+v^2 \right)^2} = 2

Now try cancelling the fraction.


Ohhh yeahh , so simple , my brain :facepalm:
Original post by TeeEm
FP1, WJEC, June 2005, page 4.jpg


Thank you , much appreciated :smile:
Reply 4
Avatar for h3rmit
h3rmit
15
I don't do FM, so sorry if this is a stupid question, but why you do include the v's on the bottom when taking the real part and include the u's when taking the imaginary part?
Reply 5
Avatar for Notnek
Notnek
21
Original post by h3rmit
I don't do FM, so sorry if this is a stupid question, but why you do include the v's on the bottom when taking the real part and include the u's when taking the imaginary part?

Do you mean top? You may need to clarify what you mean.
Reply 6
Avatar for h3rmit
h3rmit
15
Original post by notnek
Do you mean top? You may need to clarify what you mean.


On the second line, the v's are included in the bottom part (u^2 + v^2). But since they're the imaginary component , why are they included in the real part (and viceversa for the imaginary part and u's)
Reply 7
Avatar for Notnek
Notnek
21
Original post by h3rmit
On the second line, the v's are included in the bottom part (u^2 + v^2). But since they're the imaginary component , why are they included in the real part (and viceversa for the imaginary part and u's)

You don't do FM so I'm not sure what you know / don't know.

Basically in the working they have reached this complex number:

u−ivu2+v2\displaystyle \frac{u-iv}{u^2+v^2}

where uu and vv are real numbers. By splitting the fraction you can rewrite this:

uu2+v2−ivu2+v2=uu2+v2−(vu2+v2)i\displaystyle \frac{u}{u^2+v^2} - \frac{iv}{u^2+v^2} = \frac{u}{u^2+v^2}- \left(\frac{v}{u^2+v^2}\right)i

So the real part of this complex number is uu2+v2\displaystyle \frac{u}{u^2+v^2} and the imaginary part is −vu2+v2\displaystyle \frac{-v}{u^2+v^2}.
Reply 8
Avatar for h3rmit
h3rmit
15
Original post by notnek
You don't do FM so I'm not sure what you know / don't know.

Basically in the working they have reached this complex number:

u−ivu2+v2\displaystyle \frac{u-iv}{u^2+v^2}

where uu and vv are real numbers. By splitting the fraction you can rewrite this:

uu2+v2−ivu2+v2=uu2+v2−(vu2+v2)i\displaystyle \frac{u}{u^2+v^2} - \frac{iv}{u^2+v^2} = \frac{u}{u^2+v^2}- \left(\frac{v}{u^2+v^2}\right)i

So the real part of this complex number is uu2+v2\displaystyle \frac{u}{u^2+v^2} and the imaginary part is −vu2+v2\displaystyle \frac{-v}{u^2+v^2}.


Ah, I was tforgetting v was a real number for some reason, never mind. Cheers :smile:

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