| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 2 (Further Pure Core 2) |
| Year | 2018 |
| Session | December |
| Marks | 5 |
| Topic | Complex numbers 2 |
| Type | Complex number arithmetic and simplification |
| Difficulty | Standard +0.8 This is a Further Maths question requiring knowledge of Euler's formula and the relationship between hyperbolic and trigonometric functions. Part (a) is a standard derivation (3 marks), while part (b) requires applying the identity to solve a non-standard equation with complex solutions, involving inverse hyperbolic functions and logarithms. The conceptual leap and technical manipulation place it moderately above average difficulty. |
| Spec | 4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\cosh(iz) = \frac{e^z + e^{-iz}}{2}\) | M1 | Use of correct exponential form for cosh |
| \(= \frac{\cos z + i\sin z + \cos z - i\sin z}{2}\) | M1 | Correct use of Euler's formula at least once |
| \(= \frac{2\cos z}{2} = \cos z\) AG | A1 | Both M marks must be awarded. Must have cosh(iz) = or LHS = for proof must be complete |
| (b) \(\cos z = 2 \Rightarrow \cosh(iz) = 2 \Rightarrow z = (\cosh^{-1}2)i\) | M1 | |
| \(= -i\ln(2 + \sqrt{3})\) | A1 | \(\pm\) inside or outside the ln (ie allow eg ln(2 + √3) or ln(2 - √3) and condone eg ±iln(2 + √3) www) |
**(a)** $\cosh(iz) = \frac{e^z + e^{-iz}}{2}$ | M1 | Use of correct exponential form for cosh
$= \frac{\cos z + i\sin z + \cos z - i\sin z}{2}$ | M1 | Correct use of Euler's formula at least once
$= \frac{2\cos z}{2} = \cos z$ AG | A1 | Both M marks must be awarded. Must have cosh(iz) = or LHS = for proof must be complete
**(b)** $\cos z = 2 \Rightarrow \cosh(iz) = 2 \Rightarrow z = (\cosh^{-1}2)i$ | M1 |
$= -i\ln(2 + \sqrt{3})$ | A1 | $\pm$ inside or outside the ln (ie allow eg ln(2 + √3) or ln(2 - √3) and condone eg ±iln(2 + √3) www)
---\begin{enumerate}[label=(\alph*)]
\item By using Euler's formula show that $\cosh(\text{iz}) = \cos z$. [3]
\item Hence, find, in logarithmic form, a root of the equation $\cos z = 2$.
[You may assume that $\cos z = 2$ has complex roots.] [2]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 2 2018 Q9 [5]}}