SOLUTION: hi please help me solve this question 7. (a) Solve the equation z^4-2z^3+5z^2-6z+6=0 given it has a root of the form z=ia where a is a real number. (b)

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: hi please help me solve this question 7. (a) Solve the equation z^4-2z^3+5z^2-6z+6=0 given it has a root of the form z=ia where a is a real number. (b)      Log On


   

Question 706140: hi please help me solve this question
7. (a) Solve the equation z^4-2z^3+5z^2-6z+6=0 given it has a root of the form z=ia where a is a real number.

(b) Evaluate the sum i+i^2+i^3+i^4+....+i^174
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
(a) z%5E4-2z%5E3%2B5z%5E2-6z%2B6=0+
i%5E2=-1 , i%5E3=-i and i%5E4=1
If z=ia then
z%5E2=%28ia%29%5E2=i%5E2a%5E2=%28-1%29%28a%5E2%29=-a%5E2 ,
z%5E3=%28ia%29%5E3=i%5E3a%5E3=-ia%5E3 and
z%5E4=%28ia%29%5E4=i%5E4a%5E4=1%2Aa%5E4=a%5E4
If z=ia is a root, then substituting into z%5E4-2z%5E3%2B5z%5E2-6z%2B6=0 we get
a%5E4%2Bi%282a%5E3%29-5a%5E2%2Bi%28-6a%29%2B6=0 --> %28a%5E4-5a%5E2%2B6%29%2Bi%282a%5E3-6a%29=0 --> %28a%5E4-5a%5E2%2B6%29%2Bi%282a%29%28a%5E2-3%29=0 --> %28%28a%5E2-3%29%28a%5E2-2%29%29%2Bi%282a%29%28a%5E2-3%29=0
That means that %28a%5E2-3%29%28a%5E2-2%29=0 and %282a%29%28a%5E2-3%29=0
at the same time (for the same real a value).
highlight%28a=sqrt%283%29%29 and highlight%28a=-sqrt%283%29%29 are the only solutions.

EXTRA:
The imaginary roots are z=-i%2Asqrt%283%29 and z=i%2Asqrt%283%29.
so a factor in the factoring of z%5E4-2z%5E3%2B5z%5E2-6z%2B6 should be

Dividing we find that
z%5E4-2z%5E3%2B5z%5E2-6z%2B6=%28z%5E2%2B3%29%28z%5E2-2z%2B2%29
So the other two complex solutions to z%5E4-2z%5E3%2B5z%5E2-6z%2B6=0
are the solutions to z%5E2-2z%2B2=0
which happen to be z=1+%2B-+i

(b)
ONE WAY TO LOOK AT IT:
i%5E0=1 , i%5E1=i , i%5E2=-1 , i%5E3=-i , i%5E4=1 , i%5E5=i , i%5E6=-1 and so on.
In general, for any integer n
i%5E%284n%29=1 , i%5E%284n%2B1%29=i , i%5E%284n%2B2%29=-1 , i%5E%284n%2B3%29=-i
and i%5E%284n%29%2Bi%5E%284n%2B1%29%2Bi%5E%284n%2B2%29%2Bi%5E%284n%2B3%29=1%2Bi-1-i=0
So adding up the terms of the sum in groups of 4 we get
i%2Bi%5E2%2Bi%5E3%2Bi%5E4=1%2Bi-1-i=0 , i%5E5%2Bi%5E6%2Bi%5E7%2Bi%5E8=1%2Bi-1-i=0 and so on.
Until when?
Dividing 174 by 4 we get a remainder of 2:
174=172%2B2=4%2A24%2B2 so the last group of 4 would end with i%2A172=i%5E%284%2A24%29=1
After that we have i%5E173=i%5E%284%2A24%2B1%29=i and i%2A174=i%5E%284%2A24%2B2%29=-1 , so
i%2Bi%5E2%2Bi%5E3%2Bi%5E4+....+i%5E174+=+%28i%2Bi%5E2%2Bi%5E3%2Bi%5E4%29+....+%28i%5E169%2Bi%5E170%2Bi%5E171%2Bi%5E172%29%2Bi%5E173%2Bi%5E174+=+0%2B0%2B0+...+0%2Bi%5E173%2Bi%5E174=highlight%28i-1%29
or if you prefer highlight%28-1%2Bi%29

ANOTHER WAY:
i+i^2+i^3+i^4+....+i^174 is the sum of a geometric sequence,(or geometric progression, depending on where you are studying math).
From what we may know about geometric sequences and series,
i+i^2+i^3+i^4+....+i^174 = %28i%5E175-i%29%2F%28i-1%29
Since i%5E175=i%5E%284%2A24%2B3%29=-i ,