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Exercise 2.9 · Q20

Q.The principal argument of the complex number (1+i3)24i(1−i3)\dfrac{(1+i\sqrt3)^2}{4i(1-i\sqrt3)} is

(1) 2π3\dfrac{2\pi}3
(2) π6\dfrac\pi6
(3) 5π6\dfrac{5\pi}6
(4) π2\dfrac\pi2
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We convert 1+i31+i\sqrt3, 1−i31-i\sqrt3, and 4i4i each to polar form, combine using the argument-of-product/quotient rules, and reduce to the principal range.

Step 1. Polar form of 1+i31+i\sqrt3. r=12+(3)2=2r=\sqrt{1^2+(\sqrt3)^2}=2, and it lies in Quadrant I with reference angle tan⁡−1(3)=π3\tan^{-1}(\sqrt3)=\dfrac\pi3. So 1+i3=2cis⁡π31+i\sqrt3=2\operatorname{cis}\dfrac\pi3.

Step 2. Square it. (1+i3)2=4cis⁡2π3(1+i\sqrt3)^2=4\operatorname{cis}\dfrac{2\pi}3 (modulus squares to 44, argument doubles to 2π3\dfrac{2\pi}3).

Step 3. Polar form of 1−i31-i\sqrt3 and of 4i4i. 1−i3=2cis⁡(−π3)1-i\sqrt3=2\operatorname{cis}\left(-\dfrac\pi3\right) (Quadrant IV); 4i=4cis⁡π24i=4\operatorname{cis}\dfrac\pi2.

Step 4. Combine the denominator 4i(1−i3)4i(1-i\sqrt3). Multiplying: modulus =4×2=8=4\times2=8; argument =π2+(−π3)=π6=\dfrac\pi2+\left(-\dfrac\pi3\right)=\dfrac\pi6. So 4i(1−i3)=8cis⁡π64i(1-i\sqrt3)=8\operatorname{cis}\dfrac\pi6.

Step 5. Divide numerator by denominator. …

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