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# Factorising complex equations... watch

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
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.
3. Thanks
4. For the 3rd, i meant (z+1) as the factor
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.
6. Since you're dealing with complex numbers, remember you can use the following identity,

z² + a² = (z + ia)(z - ia)
7. (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.
8. (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?
9. I don't think s/he meant the mathematical definition of "complex".
10. (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
==========
11. I think the person means complex as in difficult and not the mathematical sense Fermat...
12. (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....
13. I've only ever seen z notation in the P4 complex numbers chapter, so surely Mr. Fermat was correct in his efforts?
14. Of course they are complex functions, because the variable z is used.
15. 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.
16. to clear things up, i am using z as a complex number i.e in the form of x+yj
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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