Since \(\tan \theta=b/a\), Lovelace assumed that \(\sin\theta=b\) and \(\cos\theta=a\) in her attempt to solve Problem 3. However, \(\sin\theta\) and \(\cos\theta\) are actually equal to \(b/r\) and \(a/r\), respectively; so the expression Lovelace derived should have been

\[\begin{array}{lcl} (a+bk)^{m+nk}&=& r^{m+nk} (\cos\theta+k \sin\theta)^{m+nk}\\

&=&r^{m+nk} (e^{k\theta})^{m+nk}\\

&=&r^{m+nk}\cdot e^{k(m\theta)} \cdot e^{-n\theta}\\

&=&r^m\cdot e^{-n\theta} \cdot r^{nk} \cdot e^{k(m\theta)}\\

&=&e^{m \log r-n\theta}\cdot e^{k(n \log r+m\theta)}\\

&=&e^A (\cos B + k \sin B ).\end{array}\]

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