If (3+i sin θ)/(4-i cos θ), θ ∈ [0,2π], is a real number, then the argument of sin θ + i cos θ is :

Q: Wait, why do we multiply by the conjugate of the denominator? Can't we just equate the numerator's imaginary part to zero?

That is a common trap! If you have a fraction $\frac{N}{D}$, the imaginary part of the whole expression is not just the imaginary part of the numerator. You must rationalize the denominator so that the denominator becomes a purely real number. Only then can you safely say that the imaginary part of the entire expression is zero.

Q: I got $\tan \theta = -\frac{3}{4}$. How do I know which quadrant $\sin \theta + i \cos \theta$ falls into?

Great observation! Since $\tan \theta = -\frac{3}{4}$, $\theta$ could be in the second or fourth quadrant. However, the target complex number is $w = \sin \theta + i \cos \theta$. You must test both possibilities for $(\sin \theta, \cos \theta)$ and check which one aligns with the provided options.

Q: Why is the argument $\pi - \tan^{-1}(\frac{4}{3})$ and not just $\tan^{-1}(-\frac{4}{3})$?

The principal argument of a complex number always lies in the interval $(-\pi, \pi]$. When a complex number is in the second quadrant, the angle is measured from the positive real axis, which is $\pi - \alpha$, where $\alpha$ is the acute angle. Using $\tan^{-1}$ directly on a negative value often gives you a fourth-quadrant angle, which is incorrect for a second-quadrant point.

Q: Is it possible for the denominator to be zero?

Excellent mathematical rigor! We should always check if $4 - i \cos \theta = 0$. Since $4$ is real and $i \cos \theta$ is imaginary, this would require both $4=0$ and $\cos \theta = 0$, which is impossible. Thus, the denominator is never zero, and our division is always valid.

Q: Does the value of $\theta$ matter for the final argument?

The value of $\theta$ is constrained by the condition that the original expression is real. This constraint fixes the ratio of $\sin \theta$ to $\cos \theta$, which in turn fixes the position of $w$ in the Argand plane. So, while $\theta$ can take two specific values, they lead to a consistent argument based on the options provided.

Question

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

Options:

(A) $\pi - 	an^{-1} (\frac{4}{3})$

(B) $-	an^{-1} (\frac{3}{4})$

(C) $\pi - 	an^{-1} (\frac{4}{3})$

(D) $	an^{-1} (\frac{4}{3})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the Argand plane, looking at a complex number that seems to be hiding its true nature. You are given $z = \frac{3+i \sin 	heta}{4-i \cos 	heta}$, and you are told that this number is purely real.

We are looking for the hidden geometry of $w = \sin 	heta + i \cos 	heta$. Let us embark on this journey to uncover the value of its argument.

### The Art of Rationalization

When we see a complex number in the denominator, our first instinct should be to clear the path. We want to see the real and imaginary parts clearly separated.

To do this, we multiply the numerator and the denominator by the conjugate of the denominator, which is $4 + i \cos 	heta$:

$$z = \frac{(3+i \sin 	heta)(4+i \cos 	heta)}{(4-i \cos 	heta)(4+i \cos 	heta)}$$

As we expand the denominator, something beautiful happens. The imaginary terms cancel out, leaving us with $16 + \cos^2 	heta$. This is the beauty of the conjugate—it transforms a complex obstacle into a solid, real foundation.

### Unmasking the Imaginary Part

Now, let us turn our attention to the numerator. Expanding $(3+i \sin 	heta)(4+i \cos 	heta)$ gives us $12 + 3i \cos 	heta + 4i \sin 	heta + i^2 \sin 	heta \cos 	heta$.

Remembering that $i^2 = -1$, we group the real and imaginary parts:

$$z = \frac{12 - \sin 	heta \cos 	heta}{16 + \cos^2 	heta} + i \frac{4 \sin 	heta + 3 \cos 	heta}{16 + \cos^2 	heta}$$

For $z$ to be a real number, the imaginary part must vanish into thin air. This forces the condition $4 \sin 	heta + 3 \cos 	heta = 0$.

Suddenly, the complexity collapses into a simple trigonometric relationship:

$$	an 	heta = -\frac{3}{4}$$

### The Final Destination

We are asked to find the argument of $w = \sin 	heta + i \cos 	heta$. With $	an 	heta = -\frac{3}{4}$, we know that $\sin 	heta$ and $\cos 	heta$ have opposite signs.

This places our complex number $w$ either in the second or the fourth quadrant. If we consider the case where $\sin 	heta = \frac{3}{5}$ and $\cos 	heta = -\frac{4}{5}$, the point lies at $(x, y) = (\frac{3}{5}, -\frac{4}{5})$ in the fourth quadrant.

Alternatively, if $\sin 	heta = -\frac{3}{5}$ and $\cos 	heta = \frac{4}{5}$, the point lies at $(x, y) = (-\frac{3}{5}, \frac{4}{5})$ in the second quadrant. The argument of a complex number in the second quadrant is given by $\pi - 	an^{-1} |y/x|$.

Substituting our values for the second quadrant case, we get:

$$\arg(w) = \pi - 	an^{-1} \left| \frac{4/5}{-3/5} ight| = \pi - 	an^{-1} \left( \frac{4}{3} ight)$$

This result is elegant and precise. You have successfully navigated the complex plane, stripped away the layers of the fraction, and arrived at the truth. The final answer is **$\pi - 	an^{-1}(\frac{4}{3})$** (or **$-	an^{-1}(\frac{4}{3})$** depending on the quadrant choice).

Animated Solution for Mathematics - Complex Numbers: If $\frac{3 + i s i n θ}{4 - i c o s θ}, θ \in [0, 2 π]$ , is a real number, then the argument of $sin θ + i cos θ$ is :

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Visualized Solution

Given Complex Number $z$

Rationalizing the Denominator

Multiplying by the Conjugate

Simplifying the Denominator

Expanding the Numerator

Grouping Real and Imaginary Parts

Condition for a Real Number

Solving for $tan θ$

Defining the Target Number $w$

Analyzing the Quadrant

Mapping $w$ to the Argand Plane

Calculating the Argument

Final Answer Selection

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

The Art of Rationalization

Unmasking the Imaginary Part

The Final Destination

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Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 4−icosθ3+isinθ​,θ∈[0,2π], is a real number, then the argument of sinθ+icosθ is :