Pre-U Pre-U 9794/1 2018 June — Question 3 4 marks

Exam BoardPre-U
ModulePre-U 9794/1 (Pre-U Mathematics Paper 1)
Year2018
SessionJune
Marks4
TopicComplex Numbers Arithmetic
TypeRoots of unity and special equations
DifficultyEasy -1.2 This is a straightforward application of factorising a difference of cubes and solving a quadratic equation. Students need only factor (z-1)(z²+z+1)=0 and apply the quadratic formula to z²+z+1=0, requiring minimal problem-solving beyond routine algebraic manipulation.
Spec4.02r nth roots: of complex numbers

3 Given that \(z = 1\) is the real root of the equation \(z ^ { 3 } - 1 = 0\), find the two complex roots.

Question 3: Complex roots of \(z^3 - 1 = 0\)
\(z = 1\) is a root implies \((z-1)\) is a factor B1 (Must be stated or used)
\(\frac{z^3 - 1}{z - 1} = z^2 + z + 1\) M1 (Must reach a 3 term quadratic. Accept alternative methods e.g. coefficient matching)
\(z = \frac{-1 \pm \sqrt{-3}}{2}\) M1 (Allow one error of substitution in correct quadratic formula if stated. If formula not stated and error present M0)
\(z = \frac{-1 \pm \text{i}\sqrt{3}}{2}\) A1
Total: 4 marks
**Question 3: Complex roots of $z^3 - 1 = 0$**

$z = 1$ is a root implies $(z-1)$ is a factor **B1** (Must be stated or used)

$\frac{z^3 - 1}{z - 1} = z^2 + z + 1$ **M1** (Must reach a 3 term quadratic. Accept alternative methods e.g. coefficient matching)

$z = \frac{-1 \pm \sqrt{-3}}{2}$ **M1** (Allow one error of substitution in correct quadratic formula if stated. If formula not stated and error present M0)

$z = \frac{-1 \pm \text{i}\sqrt{3}}{2}$ **A1**

**Total: 4 marks**
3 Given that $z = 1$ is the real root of the equation $z ^ { 3 } - 1 = 0$, find the two complex roots.

\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2018 Q3 [4]}}