Question

The imaginary part of $\frac{\cos 50^{\circ}+i\sin 50^{\circ}}{\cos 50^{\circ}-i\sin 50^{\circ}}$ is equal to

• A. $\cos 10^{\circ}$
• B. $\sin 80^{\circ}$
• C. $\cos 50^{\circ}$
• D. $\sin 40^{\circ}$
• E. $\cos 40^{\circ}$

Hint:

Logic Tip: When dividing complex numbers in the format $\frac{z}{\bar{z}}$, the result is always $e^{i(2\theta)}$. Here, $\theta = 50^{\circ}$, so the result is immediately $e^{i 100^{\circ}}$.

Answer 1

The Correct Option is A

Concept:
Euler's formula states that $e^{i\theta} = \cos \theta + i\sin \theta$. This makes multiplication and division of complex numbers in polar form much simpler by allowing us to use exponent rules.
Step 1: Convert numerator and denominator to Euler's form.
The numerator is $\cos 50^{\circ} + i\sin 50^{\circ}$, which translates to $e^{i 50^{\circ}}$. [cite_start]The denominator is $\cos 50^{\circ} - i\sin 50^{\circ}$, which translates to $e^{-i 50^{\circ}}$[cite: 86].
Step 2: Perform the division using exponent rules.
$$\frac{e^{i 50^{\circ}}}{e^{-i 50^{\circ}}} = e^{i 50^{\circ} - (-i 50^{\circ})}$$ $$= e^{i 100^{\circ}}$$
Step 3: Extract the imaginary part.
Converting back to rectangular form: $$e^{i 100^{\circ}} = \cos 100^{\circ} + i\sin 100^{\circ}$$ The imaginary part is $\sin 100^{\circ}$.
Step 4: Use trigonometric identities to match the options.
We need to express $\sin 100^{\circ}$ in terms of an acute angle to match the given options. $$\sin 100^{\circ} = \sin(90^{\circ} + 10^{\circ})$$ Using the identity $\sin(90^{\circ} + \theta) = \cos \theta$: $$\sin 100^{\circ} = \cos 10^{\circ}$$ [cite_start]This matches Option A[cite: 88, 97].