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Exercise 2.9 · Q16

Q.The principal argument of 3−1+i\dfrac3{-1+i} is

(1) −5π6-\dfrac{5\pi}6
(2) −2π3-\dfrac{2\pi}3
(3) −3π4-\dfrac{3\pi}4
(4) −π2-\dfrac\pi2
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We first convert 3−1+i\dfrac3{-1+i} into rectangular form x+iyx+iy by rationalizing with the conjugate, identify its quadrant, then use the quadrant rule for the principal argument Arg⁡∈(−π,π]\operatorname{Arg}\in(-\pi,\pi].

Step 1. Rationalize by the conjugate −1−i-1-i.

3−1+i=3(−1−i)(−1+i)(−1−i)=−3−3i(−1)2−i2=−3−3i1+1=−3−3i2.\dfrac3{-1+i}=\dfrac{3(-1-i)}{(-1+i)(-1-i)}=\dfrac{-3-3i}{(-1)^2-i^2}=\dfrac{-3-3i}{1+1}=\dfrac{-3-3i}2.

Step 2. Write in rectangular form.

3−1+i=−32−32i,x=−32, y=−32.\dfrac3{-1+i}=-\dfrac32-\dfrac32i,\qquad x=-\dfrac32,\ y=-\dfrac32.

Step 3. Locate the quadrant. Both x<0x<0 and y<0y<0, so the point lies in Quadrant III.

Step 4. Compute the reference angle. …

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