FP1 Complex Numbers Watch

S-T
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#1
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#1
How would I go about solving

a(squared) + b(squared) = -16

Please explain simply!
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Glutamic Acid
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#2
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#2
It can't be solved: there's one equation with two variables. Is there any more information?
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darkraver
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#3
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Not to high jack this thread but I'm doing Higher Maths and complex numbers/imaginary numbers are not part of the syllabus however I'm bored and decided to study this myself as an extra thing.

I'm still in the very early stage of imaginary numbers, just wanted to verify if the following is correct?

√(-212) = √(212 × -1) = √(212)√(-1) = √(4×53)√(-1) = 2√(53)√(-1) = (2√53)i

Thanks in advance!
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Glutamic Acid
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#4
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(Original post by darkraver)
Not to high jack this thread but I'm doing Higher Maths and complex numbers/imaginary numbers are not part of the syllabus however I'm bored and decided to study this myself as an extra thing.

I'm still in the very early stage of imaginary numbers, just wanted to verify if the following is correct?

√(-212) = √(212 × -1) = √(212)√(-1) = √(4×53)√(-1) = 2√(53)√(-1) = (2√53)i

Thanks in advance!
That's correct.
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jayshah31
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#5
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#5
(Original post by darkraver)
Not to high jack this thread but I'm doing Higher Maths and complex numbers/imaginary numbers are not part of the syllabus however I'm bored and decided to study this myself as an extra thing.

I'm still in the very early stage of imaginary numbers, just wanted to verify if the following is correct?

√(-212) = √(212 × -1) = √(212)√(-1) = √(4×53)√(-1) = 2√(53)√(-1) = (2√53)i

Thanks in advance!
Yes.
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darkraver
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#6
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Thank you very much =)
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Guinny
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#7
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#7
Cannot be solved. Unless you are able to assume that ab = 15...

The questions looks familiar.
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S-T
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#8
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#8
(Original post by darkraver)
Not to high jack this thread but I'm doing Higher Maths and complex numbers/imaginary numbers are not part of the syllabus however I'm bored and decided to study this myself as an extra thing.

I'm still in the very early stage of imaginary numbers, just wanted to verify if the following is correct?

√(-212) = √(212 × -1) = √(212)√(-1) = √(4×53)√(-1) = 2√(53)√(-1) = (2√53)i

Thanks in advance!
Yh thats right
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S-T
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#9
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#9
(Original post by Guinny)
Cannot be solved. Unless you are able to assume that ab = 15...

The questions looks familiar.
Yes its that one, ab=15 and z(sqaured) + 16 - 30i = 0.
z is (a+bi)

Any chance of understanding now
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S-T
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#10
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#10
(Original post by Glutamic Acid)
It can't be solved: there's one equation with two variables. Is there any more information?
sorry I didn't realise the first part helped, :rolleyes: Ive posted it up now.
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Guinny
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#11
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#11
Ohhh.. same question I did before :]

ab = 15
a2 - b2 = -16

Can you deduce one root where both the Imaginary and Real are positive?
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S-T
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#12
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#12
(Original post by Guinny)
Ohhh.. same question I did before :]

ab = 15
a2 - b2 = -16

Can you deduce one root where both the Imaginary and Real are positive?
I'm not sure but the mark scheme just says the roots are:

(3+5i) and (-3-5i)
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prudentstudent
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#13
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#13
Sub in a = 16/b

end up with a quartic which you should be able to solve.
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the_neg_master_
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#14
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#14
guys, just a quick question about Newton-Rhapson..
i know how to do the Newton Rhapson but what does 'e' means?
this is the question:


Using 1.6 as starting value, apply the Newton-Rhapson procedure to:
 f(x) = e^{2x} -15x -2
to obtain a second approximation to \alpha

i know that i will differentiate f(x) but how do i differentiate 'e'? and what does 'e' means?
im still doing my AS.. haven't done C2 yet.
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boyvn91
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#15
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#15
e is the Exponential, you can find the information on wikipedia.
When you differentiate e, you will get e

thus, in this case f'(x) = 2e^(2x) -15
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boyvn91
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#16
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#16
e is the Exponential, you can find the information on wikipedia.
When you differentiate e, you will get e

thus, in this case f'(x) = 2e^(2x) -15
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Hedgeman49
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#17
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#17
(Original post by the_neg_master_)
guys, just a quick question about Newton-Rhapson..
i know how to do the Newton Rhapson but what does 'e' means?
this is the question:


Using 1.6 as starting value, apply the Newton-Rhapson procedure to:
 f(x) = e^{2x} -15x -2
to obtain a second approximation to \alpha

i know that i will differentiate f(x) but how do i differentiate 'e'? and what does 'e' means?
im still doing my AS.. haven't done C2 yet.
e is a constant (approximately 2.7). However you will not need to know how to do this as exponentials are not on the new FP1 syllabus. You will learn how to differentiate e^2x in C3.
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the_neg_master_
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#18
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#18
(Original post by Hedgeman49)
e is a constant (approximately 2.7). However you will not need to know how to do this as exponentials are not on the new FP1 syllabus. You will learn how to differentiate e^2x in C3.
cheers bud.. and quick question, what does acknowledge means on your UCAS offers.. it says: Durham(Castle) - acknowledge..

im just curious coz' im applying next year..
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Rodelero
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#19
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#19
E is a constant where in the case of the function e^x, the rate of change is equal to the value of e^x.
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Oh I Really Don't Care
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#20
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#20
Is the exam in the morning?
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