Solveeit Logo
Login

Question

Question: If $\frac{3+i\sin\theta}{4-i\cos\theta}$, $\theta \in [0, 2\pi]$, is a real number, then an argument...

If 3+isin⁡θ4−icos⁡θ\frac{3+i\sin\theta}{4-i\cos\theta}, θ∈[0,2π]\theta \in [0, 2\pi], is a real number, then an argument of sin⁡θ+icos⁡θ\sin\theta + i\cos\theta is

A

−tan⁡−1(34)-\tan^{-1}(\frac{3}{4})

B

tan⁡−1(43)\tan^{-1}(\frac{4}{3})

C

π−tan⁡−1(43)\pi - \tan^{-1}(\frac{4}{3})

D

π−tan⁡−1(34)\pi - \tan^{-1}(\frac{3}{4})

Answer

π−tan⁡−1(43)\pi - \tan^{-1}(\frac{4}{3})

Explanation

Solution

Let the given complex number be Z=3+isin⁡θ4−icos⁡θZ = \frac{3+i\sin\theta}{4-i\cos\theta}. For ZZ to be a real number, its imaginary part must be zero. We find the imaginary part by rationalizing the denominator: Z=3+isin⁡θ4−icos⁡θ×4+icos⁡θ4+icos⁡θZ = \frac{3+i\sin\theta}{4-i\cos\theta} \times \frac{4+i\cos\theta}{4+i\cos\theta} The numerator is: (3+isin⁡θ)(4+icos⁡θ)=12+3icos⁡θ+4isin⁡θ+i2sin⁡θcos⁡θ(3+i\sin\theta)(4+i\cos\theta) = 12 + 3i\cos\theta + 4i\sin\theta + i^2\sin\theta\cos\theta =12+i(3cos⁡θ+4sin⁡θ)−sin⁡θcos⁡θ= 12 + i(3\cos\theta + 4\sin\theta) - \sin\theta\cos\theta =(12−sin⁡θcos⁡θ)+i(3cos⁡θ+4sin⁡θ)= (12 - \sin\theta\cos\theta) + i(3\cos\theta + 4\sin\theta) The denominator is: (4−icos⁡θ)(4+icos⁡θ)=42−(icos⁡θ)2=16−i2cos⁡2θ=16+cos⁡2θ(4-i\cos\theta)(4+i\cos\theta) = 4^2 - (i\cos\theta)^2 = 16 - i^2\cos^2\theta = 16 + \cos^2\theta So, Z=(12−sin⁡θcos⁡θ)+i(3cos⁡θ+4sin⁡θ)16+cos⁡2θZ = \frac{(12 - \sin\theta\cos\theta) + i(3\cos\theta + 4\sin\theta)}{16 + \cos^2\theta}. The imaginary part of ZZ is Im(Z)=3cos⁡θ+4sin⁡θ16+cos⁡2θ\text{Im}(Z) = \frac{3\cos\theta + 4\sin\theta}{16 + \cos^2\theta}. Since ZZ is real, Im(Z)=0\text{Im}(Z) = 0. As 16+cos⁡2θ>016 + \cos^2\theta > 0, we must have: 3cos⁡θ+4sin⁡θ=03\cos\theta + 4\sin\theta = 0 4sin⁡θ=−3cos⁡θ4\sin\theta = -3\cos\theta Dividing by cos⁡θ\cos\theta (note that cos⁡θ≠0\cos\theta \ne 0, otherwise sin⁡θ=0\sin\theta=0, which is impossible), we get: tan⁡θ=−34\tan\theta = -\frac{3}{4} We are looking for an argument of the complex number w=sin⁡θ+icos⁡θw = \sin\theta + i\cos\theta. Let ϕ\phi be an argument of ww. Then tan⁡ϕ=Im(w)Re(w)=cos⁡θsin⁡θ=cot⁡θ\tan\phi = \frac{\text{Im}(w)}{\text{Re}(w)} = \frac{\cos\theta}{\sin\theta} = \cot\theta. Since tan⁡θ=−3/4\tan\theta = -3/4, we have cot⁡θ=1tan⁡θ=1−3/4=−43\cot\theta = \frac{1}{\tan\theta} = \frac{1}{-3/4} = -\frac{4}{3}. So, tan⁡ϕ=−4/3\tan\phi = -4/3.

The condition tan⁡θ=−3/4\tan\theta = -3/4 implies that θ\theta lies in the second or fourth quadrant, since θ∈[0,2π]\theta \in [0, 2\pi].

Case 1: θ\theta is in the second quadrant. In this case, sin⁡θ>0\sin\theta > 0 and cos⁡θ<0\cos\theta < 0. From tan⁡θ=−3/4\tan\theta = -3/4, we can construct a right triangle with opposite side 3 and adjacent side 4. The hypotenuse is 32+42=5\sqrt{3^2+4^2}=5. So, sin⁡θ=3/5\sin\theta = 3/5 and cos⁡θ=−4/5\cos\theta = -4/5. Then w=sin⁡θ+icos⁡θ=35+i(−45)=35−i45w = \sin\theta + i\cos\theta = \frac{3}{5} + i\left(-\frac{4}{5}\right) = \frac{3}{5} - i\frac{4}{5}. This complex number is in the fourth quadrant. Its principal argument is: ϕ=arctan⁡(−4/53/5)=arctan⁡(−43)=−arctan⁡(43)\phi = \arctan\left(\frac{-4/5}{3/5}\right) = \arctan\left(-\frac{4}{3}\right) = -\arctan\left(\frac{4}{3}\right)

Case 2: θ\theta is in the fourth quadrant. In this case, sin⁡θ<0\sin\theta < 0 and cos⁡θ>0\cos\theta > 0. So, sin⁡θ=−3/5\sin\theta = -3/5 and cos⁡θ=4/5\cos\theta = 4/5. Then w=sin⁡θ+icos⁡θ=−35+i(45)=−35+i45w = \sin\theta + i\cos\theta = -\frac{3}{5} + i\left(\frac{4}{5}\right) = -\frac{3}{5} + i\frac{4}{5}. This complex number is in the second quadrant. Its principal argument is: ϕ=π+arctan⁡(4/5−3/5)=π+arctan⁡(−43)=π−arctan⁡(43)\phi = \pi + \arctan\left(\frac{4/5}{-3/5}\right) = \pi + \arctan\left(-\frac{4}{3}\right) = \pi - \arctan\left(\frac{4}{3}\right)

The possible arguments are −arctan⁡(4/3)-\arctan(4/3) and π−arctan⁡(4/3)\pi - \arctan(4/3). Option (3) matches one of the possible arguments.