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sweetymango
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#1
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#1
Can anyone give any tips/techniques when factorising complext equations
e.g:
z^2+6z+13,
z^2+z+1,
z^3+1

Many Thanks
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JamesF
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#2
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#2
Use the quadratic formula for the first 2, and for the 3rd, take out a factor of (z-1) first then it leaves another quadratic.
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sweetymango
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#3
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#3
Thanks :rolleyes:
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JamesF
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#4
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For the 3rd, i meant (z+1) as the factor
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shift3
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#5
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#5
Here's a general method for solving equations like the 3rd one.

f(x) = x³+ax²+bx+c
Take the factors of c and use them in f(x). If the result is zero, that number is a factor.
eg:
f(x) = x³+1
Factors of one: 1 (and -1)
f(1) = 1+1 = 2, so this isn't a factor.
f(-1) = -1+1 = 0, so this is a factor.

So.. x=-1 => (x+1)=0
Now divide x³+1 by x+1 to get a quadratic equation, and factorize that to get all the required factors.
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Fermat
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#6
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Since you're dealing with complex numbers, remember you can use the following identity,

z² + a² = (z + ia)(z - ia)
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shift3
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#7
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(Original post by Fermat)
Since you're dealing with complex numbers, remember you can use the following identity,

z² + a² = (z + ia)(z - ia)
Complex equations.
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benm
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#8
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(Original post by sweetymango)
Can anyone give any tips/techniques when factorising complext equations
e.g:
z^2+6z+13,
z^2+z+1,
z^3+1

Many Thanks
z = a + bj

z^2+6z+13
= (a+bj)^2 + 6(a + bj) + 13
= a^2 + 2abj - b^2 + 6a + 6bj + 13
= (a^2 - b^2 + 6a + 13) + (2ab + 6b)j

Is that how you mean?
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shift3
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#9
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I don't think s/he meant the mathematical definition of "complex".
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Fermat
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#10
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(Original post by shift3)
Complex equations.
Well, complex equations will involve complex numbers, and should you have the complex equation,

z² + a² = w²

then this can be factorised and rewritten as,

(z+ia)(z-ia) = w²
==========
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Juwel
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#11
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I think the person means complex as in difficult and not the mathematical sense Fermat...
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Fermat
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#12
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(Original post by ZJuwelH)
I think the person means complex as in difficult and not the mathematical sense Fermat...
Aaaargh!
I think I'll go back to sleep.
Silly Zzzzzzzz....
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RobbieC
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#13
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I've only ever seen z notation in the P4 complex numbers chapter, so surely Mr. Fermat was correct in his efforts?
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jpowell
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#14
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Of course they are complex functions, because the variable z is used.
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nollaig
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#15
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Can anyone give any tips/techniques when factorising complext equations
e.g:
z^2+6z+13,
z^2+z+1,
z^3+1
The above are functions of z, not equations. Accepting the standard notation that z = x+iy where x,y are real numbers and i = sqrt(-1) then they can, correctly, be described as complex functions.
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sweetymango
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#16
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to clear things up, i am using z as a complex number i.e in the form of x+yj
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sweetymango
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#17
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#17
(Original post by benm)
z = a + bj

z^2+6z+13
= (a+bj)^2 + 6(a + bj) + 13
= a^2 + 2abj - b^2 + 6a + 6bj + 13
= (a^2 - b^2 + 6a + 13) + (2ab + 6b)j

Is that how you mean?
well i meant writing them in the form as products of its factors, I've found that the method is to use the quadratic formula
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