If ` z = ( sqrt(3) + i)/( 2)` then `( z^(101) + i ^(103))^(105)` equals

Options

A. z
B. `z^(2)`
C. `z^(3)`
D. none of these

Answer

To solve the problem, we start with the complex number \( z = \frac{\sqrt{3} + i}{2} \). We need to find the value of \( (z^{101} + i^{103})^{105} \).

### Step-by-Step Solution:

1. **Express \( z \) in polar form**:
\[ z = \frac{\sqrt{3}}{2} + \frac{1}{2} i \]
The modulus \( r \) of \( z \) is given by:
\[ r = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = \sqrt{1} = 1 \]
The argument \( \theta \) is:
\[ \theta = \tan^{-1}\left(\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \]
Therefore, we can express \( z \) as:
\[ z = e^{i\frac{\pi}{6}} \]

2. **Calculate \( z^{101} \)**:
\[ z^{101} = \left(e^{i\frac{\pi}{6}}\right)^{101} = e^{i\frac{101\pi}{6}} \]

3. **Simplify \( \frac{101\pi}{6} \)**:
To simplify \( \frac{101\pi}{6} \), we can subtract multiples of \( 2\pi \):
\[ \frac{101\pi}{6} = \frac{101\pi}{6} - 16\pi = \frac{101\pi - 96\pi}{6} = \frac{5\pi}{6} \]
Thus,
\[ z^{101} = e^{i\frac{5\pi}{6}} \]

4. **Calculate \( i^{103} \)**:
Recall that \( i = e^{i\frac{\pi}{2}} \), so:
\[ i^{103} = \left(e^{i\frac{\pi}{2}}\right)^{103} = e^{i\frac{103\pi}{2}} \]
Simplifying \( \frac{103\pi}{2} \):
\[ \frac{103\pi}{2} = \frac{103\pi}{2} - 51\pi = \frac{103\pi - 102\pi}{2} = \frac{\pi}{2} \]
Thus,
\[ i^{103} = e^{i\frac{\pi}{2}} \]

5. **Combine \( z^{101} \) and \( i^{103} \)**:
\[ z^{101} + i^{103} = e^{i\frac{5\pi}{6}} + e^{i\frac{\pi}{2}} \]

6. **Find a common representation**:
To add these, we can convert them to rectangular form:
\[ e^{i\frac{5\pi}{6}} = -\frac{\sqrt{3}}{2} + \frac{1}{2}i \]
\[ e^{i\frac{\pi}{2}} = 0 + 1i \]
Therefore,
\[ z^{101} + i^{103} = \left(-\frac{\sqrt{3}}{2} + \frac{1}{2}i + i\right) = -\frac{\sqrt{3}}{2} + \frac{3}{2}i \]

7. **Calculate \( (z^{101} + i^{103})^{105} \)**:
We need to find the modulus and argument of \( -\frac{\sqrt{3}}{2} + \frac{3}{2}i \):
\[ r = \sqrt{\left(-\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{3}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{9}{4}} = \sqrt{3} \]
The argument \( \theta \) is:
\[ \theta = \tan^{-1}\left(\frac{\frac{3}{2}}{-\frac{\sqrt{3}}{2}}\right) = \tan^{-1}\left(-\frac{3}{\sqrt{3}}\right) = -\frac{\pi}{3} + \pi = \frac{2\pi}{3} \]
Thus, we can express it as:
\[ -\frac{\sqrt{3}}{2} + \frac{3}{2}i = \sqrt{3} e^{i\frac{2\pi}{3}} \]

8. **Raise to the power of 105**:
\[ (z^{101} + i^{103})^{105} = \left(\sqrt{3} e^{i\frac{2\pi}{3}}\right)^{105} = (\sqrt{3})^{105} e^{i\frac{210\pi}{3}} = 3^{52.5} e^{i70\pi} \]
Since \( e^{i70\pi} = e^{i0} = 1 \):
\[ (z^{101} + i^{103})^{105} = 3^{52.5} \]

### Final Answer:
\[ (z^{101} + i^{103})^{105} = 3^{52.5} \]