Worksheet02: Dickson 005
Current Work (editable)
Hence the amplitude of the quotient of R(cos \beta + i sin \beta) by r(cos \theta + i sin \theta) is equal to the difference \beta - \theta of their amplitudes, while the modulus of the quotient is equal to the quotient R/r of their moduli. The case \beta = 0 gives the useful formula 1/cos \theta + i sin \theta = cos \theta - i sin \theta. 7. De Moivre's Theorem. If n is any positive whole number, (3) (cos \theta + i sin \theta)^{n} = cos n\theta + i sin n\theta. This relation is evidently true when n = 1, and when n = 2 it follows from formula (2) with \alpha = \theta. To proceed by mathematical induction, suppose that our relation has been established for the values 1, 2, ..., m of n. We can then prove that it holds also for the next value m + 1 of n. For, by hypothesis, we have (cos \theta + i sin \theta)^{m} = cos m\theta + i sin m\theta. Multiply each member by cos \theta + i sin \theta, and for the product on the right substitute its value from (2) with \alpha = m \theta. Thus (cos \theta + i sin \theta)^{m + 1} = (cos \theta + i sin \theta)(cos m\theta + i sin m\theta), = cos (\theta + m \theta) + i sin (\theta + m \theta), which proves (3) when n = m + 1. Hence the induction is complete. Examples are furnished by the results at the end of §5: (cos 120° + i sin 120°)^{2} = cos 240° + i sin 240°, (cos 120° + i sin 120°)^{3} = cos 360° + i sin 360°. 8. Cube Roots. To find the cube roots of a complex number, we first express the number in its trigonometric form. For example, 4 $${2} + 4 $${2} i = 8(cos 45° + i sin 45°). If it has a cube root which is a complex number, the latter is expressible in the trigonometric form (4) r(cos \theta + i sin \theta). The cube of the latter, which is found by means of (3), must be equal to the proposed number, so that r^{3}(cos 3\theta + i sin 3\theta) = 8(cos 45° + i sin 45°).
Answer Key (non-editable)
Hence \begin{Thm}the amplitude of the quotient of $R(\cos \beta + i \sin \beta)$ by $r(\cos \theta + i \sin \theta)$ is equal to the difference $\beta - \theta$ of their amplitudes, while the modulus of the quotient is equal to the quotient $R/r$ of their moduli\end{Thm}. The case $\beta = 0$ gives the useful formula \[ \frac{1}{\cos \theta + i \sin \theta} = \cos \theta - i \sin \theta. \] \Par{7. De Moivre's Theorem.} \begin{Thm} If $n$ is any positive whole number, \[ \Tag{(3)} (\cos \theta + i \sin \theta)^{n} = \cos n\theta + i \sin n\theta. \] \end{Thm} This relation is evidently true when $n = 1$, and when $n = 2$ it follows from formula \Eq{(2)} with $\alpha = \theta$. To proceed by mathematical induction, suppose that our relation has been established for the values $1$, $2$, \dots, $m$ of $n$. We can then prove that it holds also for the next value $m + 1$ of $n$. For, by hypothesis, we have \[ (\cos \theta + i \sin \theta)^{m} = \cos m\theta + i \sin m\theta. \] Multiply each member by $\cos \theta + i \sin \theta$, and for the product on the right substitute its value from \Eq{(2)} with $\alpha = m\theta$. Thus \begin{align*} (\cos \theta + i \sin \theta)^{m + 1} &= (\cos \theta + i \sin \theta)(\cos m\theta + i \sin m\theta), \\ &= \cos (\theta + m \theta) + i \sin (\theta + m \theta), \end{align*} which proves \Eq{(3)} when $n = m + 1$. Hence the induction is complete. \begin{Remark} Examples are furnished by the results at the end of §5: \begin{align*} (\cos 120° + i \sin 120°)^{2} = \cos 240° + i \sin 240°, \\ (\cos 120° + i \sin 120°)^{3} = \cos 360° + i \sin 360°. \end{align*} \end{Remark} \Par{8. Cube Roots.} To find the cube roots of a complex number, we first express the number in its trigonometric form. For example, \[ 4 \sqrt{2} + 4 \sqrt{2} i = 8(\cos 45° + i \sin 45°). \] If it has a cube root which is a complex number, the latter is expressible in the trigonometric form \[ \Tag{(4)} r(\cos \theta + i \sin \theta). \] The cube of the latter, which is found by means of \Eq{(3)}, must be equal to the proposed number, so that \[ r^{3}(\cos 3\theta + i \sin 3\theta) = 8(\cos 45° + i \sin 45°). \]
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