Question 1.26: Evaluate each of the following. (a) [3(cos 40° + isin 40°)][...
Evaluate each of the following. (a) \left[3\left(\cos 40^{\circ}+i \sin 40^{\circ}\right)\right]\left[4\left(\cos 80^{\circ}+i \sin 80^{\circ}\right)\right] , (b) \frac{\left(2 \operatorname{cis} 15^{\circ}\right)^7}{\left(4 \operatorname{cis} 45^{\circ}\right)^3} , (c) \left(\frac{1+\sqrt{3} i}{1-\sqrt{3} i}\right)^{10}
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\begin{aligned} (a) {\left[3\left(\cos 40^{\circ}+i \sin 40^{\circ}\right)\right]\left[4\left(\cos 80^{\circ}+i \sin 80^{\circ}\right)\right] } &=3 \cdot 4\left[\cos \left(40^{\circ}+80^{\circ}\right)+i \sin \left(40^{\circ}+80^{\circ}\right)\right] \\ &=12\left(\cos 120^{\circ}+i \sin 120^{\circ}\right) \\ &=12\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)=-6+6 \sqrt{3} i \end{aligned}
\begin{aligned}(b) \frac{\left(2 \operatorname{cis} 15^{\circ}\right)^7}{\left(4 \operatorname{cis} 45^{\circ}\right)^3} &=\frac{128 \operatorname{cis} 105^{\circ}}{64 \operatorname{cis} 135^{\circ}}=2 \operatorname{cis}\left(105^{\circ}-135^{\circ}\right) \\ &=2\left[\cos \left(-30^{\circ}\right)+i \sin \left(-30^{\circ}\right)\right]=2\left[\cos 30^{\circ}-i \sin 30^{\circ}\right]=\sqrt{3}-i \end{aligned}
\begin{aligned}(b) \frac{\left(2 \operatorname{cis} 15^{\circ}\right)^7}{\left(4 \operatorname{cis} 45^{\circ}\right)^3} &=\frac{128 \operatorname{cis} 105^{\circ}}{64 \operatorname{cis} 135^{\circ}}=2 \operatorname{cis}\left(105^{\circ}-135^{\circ}\right) \\ &=2\left[\cos \left(-30^{\circ}\right)+i \sin \left(-30^{\circ}\right)\right]=2\left[\cos 30^{\circ}-i \sin 30^{\circ}\right]=\sqrt{3}-i \end{aligned}
(c) \left(\frac{1+\sqrt{3} i}{1-\sqrt{3} i}\right)^{10}=\left\{\frac{2 \operatorname{cis}\left(60^{\circ}\right)}{2 \operatorname{cis}\left(-60^{\circ}\right)}\right\}^{10}=\left(\operatorname{cis} 120^{\circ}\right)^{10}=\operatorname{cis} 1200^{\circ}=\operatorname{cis} 120^{\circ}=-\frac{1}{2}+\frac{\sqrt{3}}{2} i
Another Method.
\begin{aligned} \left(\frac{1+\sqrt{3} i}{1-\sqrt{3} i}\right)^{10} &=\left(\frac{2 e^{\pi i / 3}}{2 e^{-\pi i / 3}}\right)^{10}=\left(e^{2 \pi i / 3}\right)^{10}=e^{20 \pi i / 3} \\ &=e^{6 \pi i} e^{2 \pi i / 3}=(1)[\cos (2 \pi / 3)+i \sin (2 \pi / 3)]=-\frac{1}{2}+\frac{\sqrt{3}}{2} i \end{aligned}
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