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fp1 complex numbers

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Original post by timebent
not sure if the method i used is called factor theorem or inspection
http://www.examsolutions.net/a-level-maths-papers/Edexcel/Further-Maths/Further-Maths-FP1/2012-January/paper.php#Q5
but for this question if it gives me one complex number and there's 3 roots then i know it's got to be a pair

and i used inspection or factor theorem or whatever it's called and tried 2 and got lucky so where would i get my marks assuming that this is my working

z2=3−iz_2 =3-i because it has a pair

let z3−8z2+22z−20=f(z)z^3 -8z^2 +22z-20=f(z)

f(2)=8-32+44-20=0

therefore z=2 thus z-2 is a factor


It's called the factor theorem. I'm guessing what you could've done if you finished Chapter 1 you'd know this - solve a cubic equation. Find the quadratic equation which has the roots 3 + i and 3 - i.
Then, just do z-a or a variable and work out a by comparing the coefficients.


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Original post by timebent
not sure if the method i used is called factor theorem or inspection
http://www.examsolutions.net/a-level-maths-papers/Edexcel/Further-Maths/Further-Maths-FP1/2012-January/paper.php#Q5
but for this question if it gives me one complex number and there's 3 roots then i know it's got to be a pair

and i used inspection or factor theorem or whatever it's called and tried 2 and got lucky so where would i get my marks assuming that this is my working

z2=3−iz_2 =3-i because it has a pair

let z3−8z2+22z−20=f(z)z^3 -8z^2 +22z-20=f(z)

f(2)=8-32+44-20=0

therefore z=2 thus z-2 is a factor


Try the method I posted. If you manage to find the quadratic equation which has the complex roots 3 +/- i, then I give you the next hint if you struggle to find the answer using my method.


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Original post by timebent
x


Also, according to the mark scheme you'd get full marks using factor theorem.


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Reply 83
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B_9710
18
Original post by timebent
not sure if the method i used is called factor theorem or inspection
http://www.examsolutions.net/a-level-maths-papers/Edexcel/Further-Maths/Further-Maths-FP1/2012-January/paper.php#Q5
but for this question if it gives me one complex number and there's 3 roots then i know it's got to be a pair

and i used inspection or factor theorem or whatever it's called and tried 2 and got lucky so where would i get my marks assuming that this is my working

z2=3−iz_2 =3-i because it has a pair

let z3−8z2+22z−20=f(z)z^3 -8z^2 +22z-20=f(z)

f(2)=8-32+44-20=0

therefore z=2 thus z-2 is a factor


You have to justify it a bit better, z2=3−i z_2 = 3-i because all of the coefficient of the cubic are real. For polynomials with real coefficients if the complex number w w is a root, then so is its complex conjugate w∗ w^* .
The more formal way to find the real root is to recognise that (z−(3−i))(z−(3+i)) (z-(3-i))(z-(3+i)) is a factor of the cubic. If you expand this product out you will get a real quadratic factor.
(edited 7 years ago)
Reply 84
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timebent
OP
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Original post by Chittesh14
It's called the factor theorem. I'm guessing what you could've done if you finished Chapter 1 you'd know this - solve a cubic equation. Find the quadratic equation which has the roots 3 + i and 3 - i.
Then, just do z-a or a variable and work out a by comparing the coefficients.


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Original post by Chittesh14
Try the method I posted. If you manage to find the quadratic equation which has the complex roots 3 +/- i, then I give you the next hint if you struggle to find the answer using my method.


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Original post by Chittesh14
Also, according to the mark scheme you'd get full marks using factor theorem.


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Awesome thanks
Original post by B_9710
You have to justify it a bit better, z2=3−i z_2 = 3-i because all of the coefficient of the cubic are real. For polynomials with real coefficients if the complex number w w is a root, then so is its complex conjugate w∗ w^* .
The more formal way to find the real root is to recognise that (z−(3−i))(z−(3+i)) (z-(3-i))(z-(3+i)) is a factor of the cubic. If you expand this product out you will get a real quadratic factor.


ah i see, so if the coefficient of the cubic are real then if Z1 is a root so is its conjugate?
Original post by timebent
Awesome thanks


ah i see, so if the coefficient of the cubic are real then if Z1 is a root so is its conjugate?


Yeah, that's right.

A cubic equation can have:

Either; all three roots as real
Or, one root is real and the other two roots form a complex conjugate pair.





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Original post by timebent
Awesome thanks


ah i see, so if the coefficient of the cubic are real then if Z1 is a root so is its conjugate?


Also, I'm not sure if you tried my method. But, here is how I'd have done it:

ImageUploadedByStudent Room1469220402.561607.jpg

First thing (if you're confused) - I worked out the quadratic equation which has those complex roots.


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Reply 87
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B_9710
18
Original post by timebent
Awesome thanks


ah i see, so if the coefficient of the cubic are real then if Z1 is a root so is its conjugate?

Yes.
Original post by B_9710
Yes.


Help please.

Part (b)

ImageUploadedByStudent Room1469223206.557464.jpg

I've come to so many equations and if I check the answer and sub he values, they all work. But, I can't get to the actual answer...


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Reply 89
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timebent
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Original post by Chittesh14
Also, I'm not sure if you tried my method. But, here is how I'd have done it:

ImageUploadedByStudent Room1469220402.561607.jpg

First thing (if you're confused) - I worked out the quadratic equation which has those complex roots.


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hmm i'll have a look tomorrow i think i know what you've done
Original post by B_9710
Yes.


awesome thanks :biggrin:
Reply 90
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B_9710
18
Original post by Chittesh14
Help please.

Part (b)

ImageUploadedByStudent Room1469223206.557464.jpg

I've come to so many equations and if I check the answer and sub he values, they all work. But, I can't get to the actual answer...


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Part a?
Original post by B_9710
Part a?


Part (b)
Original post by timebent
hmm i'll have a look tomorrow i think i know what you've done


awesome thanks :biggrin:

Okay :P.
Reply 93
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B_9710
18
Original post by Chittesh14
Part (b)


It's just simultaneous equations.
You have z=p−3q+(3p+q)i z= p-3q +(3p+q)i and p+2q=0 p+2q=0 .
It says that the modulus of z z is 102 10\sqrt 2 . So it's just Pythagoras. If you have a complex number a+bi a+bi then it's modulus is a2+b2 \sqrt{a^2+b^2} .
Original post by B_9710
It's just simultaneous equations.
You have z=p−3q+(3p+q)i z= p-3q +(3p+q)i and p+2q=0 p+2q=0 .
It says that the modulus of z z is 102 10\sqrt 2 . So it's just Pythagoras. If you have a complex number a+bi a+bi then it's modulus is a2+b2 \sqrt{a^2+b^2} .


I know, I did that 100 times. Idk why I'm getting it wrong. Maybe I got the values of z wrong (other way round).
Reply 95
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Devify
19
Original post by Chittesh14
I know, I did that 100 times. Idk why I'm getting it wrong. Maybe I got the values of z wrong (other way round).

How about you take a picture of your workings and maybe we can see where you keep making the mistakes?
Original post by Devify
How about you take a picture of your workings and maybe we can see where you keep making the mistakes?


Yeah tomorrow, I'm too busy playing games.. lol.
I'll probably see my mistake tomorrow morning anyway somehow.
If not, I'll be mad and post it here, hoping for satisfaction.
Reply 97
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timebent
OP
2
Original post by Chittesh14
Help please.

Part (b)

ImageUploadedByStudent Room1469223206.557464.jpg

I've come to so many equations and if I check the answer and sub he values, they all work. But, I can't get to the actual answer...


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how does one do part A???
Original post by timebent
how does one do part A???


I'll explain that once I get out of bed lol


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Reply 99
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timebent
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Original post by Chittesh14
I'll explain that once I get out of bed lol


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how late u getting up xD

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