petitpauline
New member
- Joined
- Sep 15, 2010
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Hi I need help with this exercice
z^4 - 2z^3 + 3z^2 - 2z +2 =0
answers are i and -i
thank a lot
z^4 - 2z^3 + 3z^2 - 2z +2 =0
answers are i and -i
thank a lot
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Set of Complex Numbers
- Thread starter petitpauline
- Start date
Hello, petitpauline!
\(\displaystyle \text{Let }z = i\)
\(\displaystyle \text{We have: }\;i^4 - 2(i^3) + 3(i^2) - 2(i) + 2 \;=\;1 + 2i - 3 - 2i + 2 \;=\;0\)
\(\displaystyle \text{We know that }i\text{ is a root . . . and so is }-i.\)
. . \(\displaystyle \text{Hence, }(z - i)(z + i) \:=\:z^2+1\text{ is a factor of the polynomial.}\)
\(\displaystyle \text{We find that: }\:z^4 - 2z^3 + 3z^2 - 2z + 2 \;=\;(z^2+1)(z^2 - 2z + 2)\)
. . \(\displaystyle \text{and }\:z^2-2z+2 \:=\:0\,\text{ has two more roots: }\:z \:=\:1 \pm i\)
\(\displaystyle \text{The }four\text{ roots are: }\:\pm i,\;1 \pm 1\)
\(\displaystyle z^4 - 2z^3 + 3z^2 - 2z +2 \:=\:0\)
Answers are \(\displaystyle i\) and \(\displaystyle -i\)
But those aren't the only answers!
\(\displaystyle \text{Let }z = i\)
\(\displaystyle \text{We have: }\;i^4 - 2(i^3) + 3(i^2) - 2(i) + 2 \;=\;1 + 2i - 3 - 2i + 2 \;=\;0\)
\(\displaystyle \text{We know that }i\text{ is a root . . . and so is }-i.\)
. . \(\displaystyle \text{Hence, }(z - i)(z + i) \:=\:z^2+1\text{ is a factor of the polynomial.}\)
\(\displaystyle \text{We find that: }\:z^4 - 2z^3 + 3z^2 - 2z + 2 \;=\;(z^2+1)(z^2 - 2z + 2)\)
. . \(\displaystyle \text{and }\:z^2-2z+2 \:=\:0\,\text{ has two more roots: }\:z \:=\:1 \pm i\)
\(\displaystyle \text{The }four\text{ roots are: }\:\pm i,\;1 \pm 1\)
petitpauline said:Hi I need help with this exercice
z^4 - 2z^3 + 3z^2 - 2z +2 =0
answers are i and -i
petitpauline,
what is the intent of the problem? Are you stating that after you solve it,
you will find that answers are i and -i? That is how I took it. You did not
state that two of the known roots are i and -i as part of the problem.
You left the problem unclear/ambiguous as to what was given.