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Exercise 2.9 · Q24

Q.The value of (1+3i1−3i)10\left(\dfrac{1+\sqrt3i}{1-\sqrt3i}\right)^{10} is

(1) cis⁡2π3\operatorname{cis}\dfrac{2\pi}3
(2) cis⁡4π3\operatorname{cis}\dfrac{4\pi}3
(3) −cis⁡2π3-\operatorname{cis}\dfrac{2\pi}3
(4) −cis⁡4π3-\operatorname{cis}\dfrac{4\pi}3
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Converting numerator and denominator to polar form turns the quotient into a single cis⁡\operatorname{cis} expression, after which de Moivre's theorem and angle reduction finish the problem.

Step 1. Polar form of 1+3i1+\sqrt3i. r=12+(3)2=2r=\sqrt{1^2+(\sqrt3)^2}=2, Quadrant I, reference angle tan⁡−13=π3\tan^{-1}\sqrt3=\dfrac\pi3. So 1+3i=2cis⁡π31+\sqrt3i=2\operatorname{cis}\dfrac\pi3.

Step 2. Polar form of 1−3i1-\sqrt3i. Same modulus r=2r=2, Quadrant IV, so 1−3i=2cis⁡(−π3)1-\sqrt3i=2\operatorname{cis}\left(-\dfrac\pi3\right).

Step 3. Form the ratio.

1+3i1−3i=2cis⁡π32cis⁡(−π3)=cis⁡(π3−(−π3))=cis⁡2π3.\dfrac{1+\sqrt3i}{1-\sqrt3i}=\dfrac{2\operatorname{cis}\frac\pi3}{2\operatorname{cis}\left(-\frac\pi3\right)}=\operatorname{cis}\left(\dfrac\pi3-\left(-\dfrac\pi3\right)\right)=\operatorname{cis}\dfrac{2\pi}3.

Step 4. Raise to the 10th power using de Moivre. …

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