Question 706140
(a) {{{z^4-2z^3+5z^2-6z+6=0 }}}
{{{i^2=-1}}} , {{{i^3=-i}}} and {{{i^4=1}}}
If {{{z=ia}}} then
{{{z^2=(ia)^2=i^2a^2=(-1)(a^2)=-a^2}}} ,
{{{z^3=(ia)^3=i^3a^3=-ia^3}}} and
{{{z^4=(ia)^4=i^4a^4=1*a^4=a^4}}}
If {{{z=ia}}} is a root, then substituting into {{{z^4-2z^3+5z^2-6z+6=0}}} we get
{{{a^4+i(2a^3)-5a^2+i(-6a)+6=0}}} --> {{{(a^4-5a^2+6)+i(2a^3-6a)=0}}} --> {{{(a^4-5a^2+6)+i(2a)(a^2-3)=0}}} --> {{{((a^2-3)(a^2-2))+i(2a)(a^2-3)=0}}}
That means that {{{(a^2-3)(a^2-2)=0}}} and {{{(2a)(a^2-3)=0}}}
at the same time (for the same real {{{a}}} value).
{{{highlight(a=sqrt(3))}}} and {{{highlight(a=-sqrt(3))}}} are the only solutions.
 
EXTRA:
The imaginary roots are {{{z=-i*sqrt(3)}}} and {{{z=i*sqrt(3)}}}.
so a factor in the factoring of  {{{z^4-2z^3+5z^2-6z+6}}} should be
{{{(z+i*sqrt(3))(z-i*sqrt(3))=z^2-(i*sqrt(3))^2=z^2-i^2(sqrt(3))^2=z^2-(-1)3=z^2+3}}}
Dividing we find that
{{{z^4-2z^3+5z^2-6z+6=(z^2+3)(z^2-2z+2)}}}
So the other two complex solutions to {{{z^4-2z^3+5z^2-6z+6=0}}}
are the solutions to {{{z^2-2z+2=0}}}
which happen to be {{{z=1 +- i}}}
 
(b)
ONE WAY TO LOOK AT IT:
{{{i^0=1}}} , {{{i^1=i}}} , {{{i^2=-1}}} , {{{i^3=-i}}} , {{{i^4=1}}} , {{{i^5=i}}} , {{{i^6=-1}}} and so on.
In general, for any integer {{{n}}}
{{{i^(4n)=1}}} , {{{i^(4n+1)=i}}} , {{{i^(4n+2)=-1}}} , {{{i^(4n+3)=-i}}}
and {{{i^(4n)+i^(4n+1)+i^(4n+2)+i^(4n+3)=1+i-1-i=0}}}
So adding up the terms of the sum in groups of 4 we get
{{{i+i^2+i^3+i^4=1+i-1-i=0}}} , {{{i^5+i^6+i^7+i^8=1+i-1-i=0}}} and so on.
Until when?
Dividing 174 by 4 we get a remainder of 2:
{{{174=172+2=4*24+2}}} so the last group of 4 would end with {{{i*172=i^(4*24)=1}}}
After that we have {{{i^173=i^(4*24+1)=i}}} and {{{i*174=i^(4*24+2)=-1}}} , so
{{{i+i^2+i^3+i^4}}}+....+{{{i^174 = (i+i^2+i^3+i^4)}}}+....+{{{(i^169+i^170+i^171+i^172)+i^173+i^174 = 0+0+0}}}+...+{{{0+i^173+i^174=highlight(i-1)}}}
or if you prefer {{{highlight(-1+i)}}}
 
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 = {{{(i^175-i)/(i-1)}}}
Since {{{i^175=i^(4*24+3)=-i}}} ,
{{{(i^175-i)/(i-1)=(-2i)/(i-1)=(-2i(i+1))/(i-1)(i+1)=(-2)(i^2+i)/(i^2-1^2)=-2(i^2+i)/(-1-1)=-(2)(i^2+i)/(-2)=i^2+i=-1+i}}}