# Factorising complex equations...

Watch
This discussion is closed.
#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 0
15 years ago
#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.
0
#3
Thanks 0
15 years ago
#4
For the 3rd, i meant (z+1) as the factor
0
15 years ago
#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.
0
15 years ago
#6
Since you're dealing with complex numbers, remember you can use the following identity,

z² + a² = (z + ia)(z - ia)
0
15 years ago
#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. 0
15 years ago
#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?
0
15 years ago
#9
I don't think s/he meant the mathematical definition of "complex".
0
15 years ago
#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
==========
0
15 years ago
#11
I think the person means complex as in difficult and not the mathematical sense Fermat...
0
15 years ago
#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....
0
15 years ago
#13
I've only ever seen z notation in the P4 complex numbers chapter , so surely Mr. Fermat was correct in his efforts?
0
15 years ago
#14
Of course they are complex functions, because the variable z is used.
0
15 years ago
#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.
0
#16
to clear things up, i am using z as a complex number i.e in the form of x+yj 0
#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 0
X
new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### How are you feeling ahead of results day?

Very Confident (21)
8.68%
Confident (30)
12.4%
Indifferent (41)
16.94%
Unsure (65)
26.86%
Worried (85)
35.12%