Complex Numbers Question - None

Question

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

Accepted Answer

Step 1: Given that $\frac{3 + i \sin heta}{4 - i \cos heta}$, where $ heta \in [0, 2\pi]$, is a real number.

Step 2: Rationalize by multiplying by $\frac{4 + i \cos heta}{4 + i \cos heta}$:

$\frac{3 + i \sin heta}{4 - i \cos heta} \cdot \frac{4 + i \cos heta}{4 + i \cos heta} = \frac{(3 + i \sin heta)(4 + i \cos heta)}{16 + \cos^2 heta}$

Step 3: Expand the numerator: $(12 - \sin heta \cos heta) + i(4 \sin heta + 3 \cos heta)$

Step 4: For the expression to be real, the imaginary part must be zero: $4 \sin heta + 3 \cos heta = 0$

Step 5: This gives $ an heta = -\frac{3}{4}$

Step 6: For $\sin heta + i \cos heta$, the argument is $\pi - an^{-1}\left(\frac{3}{4} ight)$

∴ Answer is (a).

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