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1989 Paper 2 Q3
The real numbers \(x\) and \(y\) are related to the real numbers \(u\) and \(v\) by \[ 2(u+\mathrm{i}v)=\mathrm{e}^{x+\mathrm{i}y}-\mathrm{e}^{-x-\mathrm{i}y}. \] Show that the line in the \(x\)-\(y\) plane given by \(x=a\), where \(a\) is a positive constant, corresponds to the ellipse \[ \left(\frac{u}{\sinh a}\right)^{2}+\left(\frac{v}{\cosh a}\right)^{2}=1 \] in the \(u\)-\(v\) plane. Show also that the line given by \(y=b\), where \(b\) is a constant and \(0<\sin b<1,\) corresponds to one branch of a hyperbola in the \(u\)-\(v\) plane. Write down the \(u\) and \(v\) coordinates of one point of intersection of the ellipse and hyperbola branch, and show that the curves intersect at right-angles at this point. Make a sketch of the \(u\)-\(v\) plane showing the ellipse, the hyperbola branch and the line segments corresponding to:
- \(x=0\);
- \(y=\frac{1}{2}\pi,\) \(\quad 0\leqslant x\leqslant a.\)
Solution: \begin{align*} && 2(u+iv) &= e^{a+iy} - e^{-a-iy} \\ && &=(e^a \cos y - e^{-a} \cos y) + (e^a \sin y + e^{-a} \sin y)i \\ &&&= 2 \sinh a \cos y + 2\cosh a \sin y i\\ \Rightarrow && \frac{u}{\sinh a} &= \cos y \\ && \frac{v}{\cosh a} &= \sin y \\ \Rightarrow && 1 &= \left(\frac{u}{\sinh a}\right)^{2}+\left(\frac{v}{\cosh a}\right)^{2} \end{align*} \begin{align*} && 2(u+iv) &= e^{x+ib} - e^{-x-ib} \\ &&&= 2\sinh x \cos b + 2\cosh x \sin b i \\ \Rightarrow && \frac{u}{\cos b} &= \sinh x \\ && \frac{v}{\sin b} &= \cosh x \\ \Rightarrow && 1 &= \left (\frac{v}{\sin b} \right)^2 - \left (\frac{u}{\cos b} \right)^2 \end{align*} Therefore all the points lie of a hyperbola, and since \(\frac{v}{\sin b} > 0 \Rightarrow v > 0\) it's one branch of the hyperbola. (And all points on it are reachable as \(x\) varies from \(-\infty < x < \infty\). \begin{align*} 2(u+iv) &= e^{a+ib} - e^{-a-ib} \\ &= 2 \sinh a \cos b + 2 \cosh a \sin b i \end{align*} so we can take \(u = \sinh a \cos b, v = \cosh a \sin b\). \begin{align*} \frac{\d }{\d u} && 0 &= \frac{2 u}{\sinh^2 a} + \frac{2v}{\cosh^2 a} \frac{\d v}{\d u} \\ \Rightarrow && \frac{\d v}{\d u} &= -\frac{u}{v} \coth^2 a \\ \\ && \frac{\d v}{\d u} \rvert_{(u,v)} &= -\frac{\sinh a \cos b}{\cosh a \sin b} \coth^2 a \\ &&&= -\cot b \coth a \\ \frac{\d }{\d u} && 0 &= \frac{2 v}{\sin^2 b} \frac{\d v}{\d u} - \frac{2u}{\cos^2 b} \\ \Rightarrow && \frac{\d v}{\d u} &= \frac{u}{v} \tan^2 b \\ && \frac{\d v}{\d u} \rvert_{(u,v)} &= \frac{\sinh a \cos b}{\cosh a \sin b} \tan^2 b \\ &&&= \tanh a \tan b \end{align*} Therefore they are negative reciprocals and hence perpendicular.
