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# Imaginary numbers help?

1. How do you do question bii? I don't get what it's asking me to do
2. (2-i) is a root, also they're asking for a quadratic factor, so that quadractic multiplied by (z-(2-i)) should give you the original equation. Need anymore help?
Spoiler:
Show
Expand, compare coefficients. Should contain p and q or something along those lines.
3. (Original post by Gatewaymerge)
(2-i) is a root, also they're asking for a quadratic factor, so that quadractic multiplied by (z-(2-i)) should give you the original equation. Need anymore help?
Spoiler:
Show
Expand, compare coefficients. Should contain p and q or something along those lines.
Why do you do (z-(2-i))?
4. (Original post by Faheemcg9)
Why do you do (z-(2-i))?
Because 2-i is a root of the equation. It's like saying (x-1)(x+1)=0 the solutions (roots) are 1 and -1, whereas the factors are (x-1) and (x+1), do you get it? So because 2-i is a root (z-(2-i))=0 which implies z=2-i, which is a root. You already have p and q from previous parts. Now you have your factor, just say (z-(2-i))(az^2+bz+d) = z^3+pz+q and now you compare coefficients and work out a, b and c after expanding.
5. (Original post by Gatewaymerge)
Because 2-i is a root of the equation. It's like saying (x-1)(x+1)=0 the solutions (roots) are 1 and -1, whereas the factors are (x-1) and (x+1), do you get it? So because 2-i is a root (z-(2-i))=0 which implies z=2-i, which is a root. You already have p and q from previous parts. Now you have your factor, just say (z-(2-i))(az^2+bz+d) = z^3+pz+q and now you compare coefficients and work out a, b and c after expanding.
I sort of get it. I'll finish the question tomorrow

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