Question

Real part of $\frac{1+\sin\frac{2\pi}{27}-i\cos\frac{2\pi}{27}}{1+\sin\frac{2\pi}{27}+i\cos\frac{2\pi}{27}}$ is equal to:

• A. $\cos\frac{2\pi}{27}$
• B. $\sin\frac{2\pi}{27}$
• C. $1+\sin\frac{2\pi}{27}$
• D. $1+\cos\frac{2\pi}{27}$
• E. $\sin\frac{2\pi}{27}+\cos\frac{2\pi}{27}$

Hint:

For complex numbers $w$, the real part is denoted as $Re(w)$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We need to find the real part of a complex fraction. A common structure to look for is \( \frac{\bar{z}}{z} \). Let's see if the given expression fits this form.
Step 2: Key Formula or Approach:
Let \( \theta = \frac{2\pi}{27} \). The expression can be written as \( \frac{1+\sin\theta - i\cos\theta}{1+\sin\theta + i\cos\theta} \).
Let the complex number in the denominator be \( z = (1+\sin\theta) + i(\cos\theta) \).
The conjugate of z is \( \bar{z} = (1+\sin\theta) - i(\cos\theta) \), which is exactly the numerator.
So, we need to find the real part of \( \frac{\bar{z}}{z} \).
To simplify this, we can multiply the numerator and the denominator by \( \bar{z} \):
\[ \frac{\bar{z}}{z} = \frac{\bar{z} \cdot \bar{z}}{z \cdot \bar{z}} = \frac{(\bar{z})^2}{|z|^2} \] The real part of this expression will be \( \frac{\text{Re}((\bar{z})^2)}{|z|^2} \).
Step 3: Detailed Explanation:
Let's calculate the components:
1. Denominator: \( |z|^2 \)
\( |z|^2 = (1+\sin\theta)^2 + (\cos\theta)^2 \)
\( = 1 + 2\sin\theta + \sin^2\theta + \cos^2\theta \)
Since \( \sin^2\theta + \cos^2\theta = 1 \),
\( |z|^2 = 1 + 2\sin\theta + 1 = 2 + 2\sin\theta = 2(1 + \sin\theta) \).
2. Numerator: \( (\bar{z})^2 \)
\( \bar{z} = (1+\sin\theta) - i\cos\theta \)
\( (\bar{z})^2 = ((1+\sin\theta) - i\cos\theta)^2 \)
\( = (1+\sin\theta)^2 - 2i(1+\sin\theta)\cos\theta + (i\cos\theta)^2 \)
\( = (1+2\sin\theta+\sin^2\theta) - 2i(\cos\theta + \sin\theta\cos\theta) - \cos^2\theta \)
3. Real Part of Numerator: \( \text{Re}((\bar{z})^2) \)
The real part is \( (1+2\sin\theta+\sin^2\theta) - \cos^2\theta \).
Using \( \cos^2\theta = 1 - \sin^2\theta \),
\( \text{Re}((\bar{z})^2) = 1+2\sin\theta+\sin^2\theta - (1 - \sin^2\theta) \)
\( = 1+2\sin\theta+\sin^2\theta - 1 + \sin^2\theta \)
\( = 2\sin\theta + 2\sin^2\theta = 2\sin\theta(1 + \sin\theta) \).
4. Real Part of the Whole Expression:
\[ \text{Re}\left(\frac{\bar{z}}{z}\right) = \frac{\text{Re}((\bar{z})^2)}{|z|^2} = \frac{2\sin\theta(1 + \sin\theta)}{2(1 + \sin\theta)} \] Assuming \( 1+\sin\theta \neq 0 \) (which is true since \( \theta = \frac{2\pi}{27} \)), we can cancel the term \( (1+\sin\theta) \).
\[ \text{Re}\left(\frac{\bar{z}}{z}\right) = \sin\theta \] Step 4: Final Answer:
Substituting back \( \theta = \frac{2\pi}{27} \), the real part of the expression is \( \sin\frac{2\pi}{27} \).