| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Verify roots satisfy polynomial equations |
| Difficulty | Standard +0.3 This is a straightforward multi-part question requiring standard techniques: solving a quadratic with complex roots using the formula, converting to modulus-argument form, and verifying the roots satisfy another equation using De Moivre's theorem. Part (iii) involves calculation but no novel insight—it's a direct verification exercise. Slightly above average difficulty due to the multi-step nature and need for careful arithmetic with complex numbers. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument4.02b Express complex numbers: cartesian and modulus-argument forms4.02i Quadratic equations: with complex roots4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Use the quadratic formula, completing the square, or the substitution \(z = x + iy\) to find a root and use \(i^2 = -1\) | M1 | |
| Obtain final answers \(-\sqrt{3} \pm i\), or equivalent | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| State that the modulus of both roots is 2 | B1√ | |
| State that the argument of \(-\sqrt{3} + i\) is \(150°\) or \(\frac{5}{6}\pi\) (2.62) radians | B1√ | |
| State that the argument of \(-\sqrt{3} - i\) is \(-150°\) (or \(210°\)) or \(-\frac{5}{6}\pi\) (-2.62) radians or \(\frac{7}{6}\pi\) (3.67) radians | B1√ | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Carry out an attempt to find the sixth power of a root | M1 | |
| Verify that one of the roots satisfies \(z^6 = -64\) | A1 | |
| Verify that the other root satisfies the equation | A1 | [3] |
**(i)**
Use the quadratic formula, completing the square, or the substitution $z = x + iy$ to find a root and use $i^2 = -1$ | M1 |
Obtain final answers $-\sqrt{3} \pm i$, or equivalent | A1 | [2]
**(ii)**
State that the modulus of both roots is 2 | B1√ |
State that the argument of $-\sqrt{3} + i$ is $150°$ or $\frac{5}{6}\pi$ (2.62) radians | B1√ |
State that the argument of $-\sqrt{3} - i$ is $-150°$ (or $210°$) or $-\frac{5}{6}\pi$ (-2.62) radians or $\frac{7}{6}\pi$ (3.67) radians | B1√ | [3]
**(iii)**
Carry out an attempt to find the sixth power of a root | M1 |
Verify that one of the roots satisfies $z^6 = -64$ | A1 |
Verify that the other root satisfies the equation | A1 | [3]7 (i) Find the roots of the equation
$$z ^ { 2 } + ( 2 \sqrt { } 3 ) z + 4 = 0$$
giving your answers in the form $x + \mathrm { i } y$, where $x$ and $y$ are real.\\
(ii) State the modulus and argument of each root.\\
(iii) Showing all your working, verify that each root also satisfies the equation
$$z ^ { 6 } = - 64$$
\hfill \mbox{\textit{CAIE P3 2011 Q7 [8]}}