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Core Pure · section 2.1 · Year 12

1. f(z)=z3−3z2+z+5\mathrm{f}(z) = z^3 - 3z^2 + z + 5

Given that 2−i2 - \mathrm{i} is a root of the equation f(z)=0\mathrm{f}(z) = 0, solve f(z)=0\mathrm{f}(z) = 0 completely.

(5)
  1. Real coefficients, so the complex roots are a conjugate pair: 2+i2 + \mathrm{i} is also a root.B1
  2. (z−(2−i))(z−(2+i))=(z−2)2−i2\bigl(z-(2-\mathrm{i})\bigr)\bigl(z-(2+\mathrm{i})\bigr) = (z-2)^2 - \mathrm{i}^2M1
  3. =z2−4z+5= z^2 - 4z + 5A1
  4. z3−3z2+z+5=(z2−4z+5)(z+1)z^3 - 3z^2 + z + 5 = (z^2 - 4z + 5)(z + 1)constant terms: 5×1=55 \times 1 = 5 ✓M1
  5. z=−1,  z=2+i,  z=2−iz = -1,\; z = 2 + \mathrm{i},\; z = 2 - \mathrm{i}A1

Total5

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