ex: 7+9i = (8+i)(1+i)
2007-07-21 01:30:12 · 7 answers · asked by Motafael y 1 in Science & Mathematics ➔ Mathematics
are you asking, if you have 7 + 9i can you factor it to (8+i)(1+i)?
in general the answer is yes
consider
(x+i)(y+i) = (xy - 1) + (x + y)i
so you need to find x , y such that
xy - 1 = 7 ..........(1)
x + y = 9 ...........(2)
you could try and guess like you normally do to factor a quadratic or if you can't then
y = 9 - x from (2) then substitute in (1) gives
x(9 - x) = 8
x^2 - 9x +8 =0 then solve using the quadratic formula
gives
x = 8 or 1
so 7 + 9i = (8+i)(1+i)
the end
.
2007-07-21 02:31:37 · answer #1 · answered by The Wolf 6 · 1⤊ 0⤋
Remember, the i is *not* a variable. It is an imaginary constant, defined as â(-1) = i.
7+9i = (8+i)(1+i)
Expand right side using FOIL
7 + 9i = 8·1 + 8i + i1 + i²
7 + 9i = 8 + 9i + (-1)
7 + 9i = 8 + (-1) + 9i
7 + 9i = 7 + 9i
Both sides are equal, so this proves that the original statement was an identity.
2007-07-21 09:12:45 · answer #2 · answered by Tony The Dad 3 · 0⤊ 1⤋
7+9i
=7+9i+i^2-i^2
=i^2+9i+7+1
=i^2+9i+8
=(i+8)(i+1)
my regards to Mudhish.(4 minutes)
2007-07-21 17:51:27 · answer #3 · answered by osama I 2 · 0⤊ 0⤋
7+9i = (8+i)(1+i)
7+9i = 8 (1+i) + i (1+i)
7+9i = 8 +8i + 1i +i^2
7+9i = 8 +9i +i^2
9i on both the sides of the = get cancelled
7= 8 +i^2
7 - 8 = i^2
- 1 = i ^2
therefore root of i ^2 = +1 and -1
2007-07-21 08:43:50 · answer #4 · answered by ritukiran16 3 · 0⤊ 2⤋
(8 + i) (1 + i)
= 8 + 8 i + 1 i + i²
= 8 + 9 i - 1
= 7 + 9 i
2007-07-21 19:08:07 · answer #5 · answered by Como 7 · 0⤊ 0⤋
analysis is not a verb so your sentence does not convey a specific meaning to me.
The formula you wrote is consistent with the notation I am used to seeing where "i" stands for the squareroot of negative one.
I happen to have an interest in prime integers and so perhaps you are asking if any complex number can be represented as the product of two complex numbers. I think the answer is yes but I don't have a proof.
5 is a prime integer It is also a complex number(5+0i)
5=(2+i)*(2-i)=4+2i-2i-i*i=4+1=5
2007-07-21 09:11:37 · answer #6 · answered by anonimous 6 · 0⤊ 0⤋
7+9i=(8+i)(1+i)
7+9i=8+8i+1i+i^2
7+9i=8+9i+i^2
7+9i-9i=8+i^2
7-8=i^2
-1=i^2
â-1=i
therefore i=1 or i=-1
2007-07-21 08:55:38 · answer #7 · answered by bluemaja_pooh 2 · 0⤊ 2⤋