SPS SPS FM 2021 March — Question 8 4 marks

Exam BoardSPS
ModuleSPS FM (SPS FM)
Year2021
SessionMarch
Marks4
TopicComplex Numbers Arithmetic
TypeReal and imaginary part expressions
DifficultyStandard +0.3 This is a straightforward Further Maths complex numbers question requiring standard algebraic manipulation. Part (i) involves multiplying by the conjugate to separate real and imaginary parts—a routine technique. Part (ii) asks to verify that the modulus squared equals 1, which follows immediately from the result in (i) or by direct calculation. Both parts are mechanical applications of basic complex number operations with no novel insight required, making this slightly easier than average even for FM.
Spec4.02a Complex numbers: real/imaginary parts, modulus, argument4.02e Arithmetic of complex numbers: add, subtract, multiply, divide

The function f is defined, for any complex number \(z\), by $$\text{f}(z) = \frac{iz - 1}{iz + 1}.$$ Suppose throughout that \(x\) is a real number.
  1. Show that $$\text{Re f}(x) = \frac{x^2 - 1}{x^2 + 1} \quad \text{and} \quad \text{Im f}(x) = \frac{2x}{x^2 + 1}.$$ [2]
  2. Show that \(\text{f}(x)\text{f}(x)^* = 1\), where \(\text{f}(x)^*\) is the complex conjugate of \(\text{f}(x)\). [2]

The function f is defined, for any complex number $z$, by
$$\text{f}(z) = \frac{iz - 1}{iz + 1}.$$

Suppose throughout that $x$ is a real number.

\begin{enumerate}[label=(\roman*)]
\item Show that
$$\text{Re f}(x) = \frac{x^2 - 1}{x^2 + 1} \quad \text{and} \quad \text{Im f}(x) = \frac{2x}{x^2 + 1}.$$ [2]
\item Show that $\text{f}(x)\text{f}(x)^* = 1$, where $\text{f}(x)^*$ is the complex conjugate of $\text{f}(x)$. [2]
\end{enumerate}

\hfill \mbox{\textit{SPS SPS FM 2021 Q8 [4]}}