If `z=(1)/(2)(sqrt(3)-i)`, then the least possible integral value of `m` such that `(z^(101)+i^(109))^(106)=z^(m+1)` is

Options

A. `11`
B. `7`
C. `8`
D. `9`

Explanation

To solve the problem, we need to find the least possible integral value of \( m \) such that

\[ (z^{101} + i^{109})^{106} = z^{m+1} \]

where \( z = \frac{1}{2}(\sqrt{3} - i) \).

### Step 1: Simplifying \( z \)

First, we can express \( z \) in a more manageable form. We multiply the numerator and denominator by the conjugate of the denominator:

\[ z = \frac{1}{2} \cdot \frac{\sqrt{3} + i}{(\sqrt{3} - i)(\sqrt{3} + i)} = \frac{\sqrt{3} + i}{2(3 + 1)} = \frac{\sqrt{3} + i}{8} \]

Now, we can rewrite \( z \):

\[ z = \frac{\sqrt{3}}{8} + \frac{1}{8}i \]

### Step 2: Finding \( z^{101} \)

Next, we need to find \( z^{101} \). To do this, we can express \( z \) in polar form. The modulus \( r \) of \( z \) is given by:

\[ r = |z| = \sqrt{\left(\frac{\sqrt{3}}{8}\right)^2 + \left(\frac{1}{8}\right)^2} = \sqrt{\frac{3}{64} + \frac{1}{64}} = \sqrt{\frac{4}{64}} = \frac{1}{4} \]

The argument \( \theta \) is given by:

\[ \theta = \tan^{-1}\left(\frac{1/8}{\sqrt{3}/8}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \]

Thus, we can express \( z \) in polar form as:

\[ z = \frac{1}{4} \text{cis} \frac{\pi}{6} \]

Now, we can find \( z^{101} \):

\[ z^{101} = \left(\frac{1}{4}\right)^{101} \text{cis}\left(\frac{101\pi}{6}\right) = \frac{1}{4^{101}} \text{cis}\left(\frac{101\pi}{6}\right) \]

### Step 3: Simplifying \( i^{109} \)

Next, we simplify \( i^{109} \):

\[ i^{109} = i^{4 \cdot 27 + 1} = (i^4)^{27} \cdot i^1 = 1^{27} \cdot i = i \]

### Step 4: Combining \( z^{101} \) and \( i^{109} \)

Now we combine \( z^{101} \) and \( i^{109} \):

\[ z^{101} + i^{109} = \frac{1}{4^{101}} \text{cis}\left(\frac{101\pi}{6}\right) + i \]

### Step 5: Finding \( (z^{101} + i)^{106} \)

Next, we need to compute \( (z^{101} + i)^{106} \). Since \( z^{101} \) is very small (as \( \frac{1}{4^{101}} \) is a very small number), we can approximate:

\[ (z^{101} + i)^{106} \approx i^{106} = (i^4)^{26} \cdot i^2 = 1^{26} \cdot (-1) = -1 \]

### Step 6: Setting up the equation

Now we set up the equation:

\[ -1 = z^{m+1} \]

### Step 7: Finding \( m \)

Since \( z = \frac{1}{4} \text{cis}\left(\frac{\pi}{6}\right) \), we can express \( z^{m+1} \):

\[ z^{m+1} = \left(\frac{1}{4}\right)^{m+1} \text{cis}\left(\frac{(m+1)\pi}{6}\right) \]

For \( z^{m+1} = -1 \), we need:

1. \( \left(\frac{1}{4}\right)^{m+1} = 1 \) implies \( m+1 = 0 \) or \( m = -1 \) (not possible since \( m \) must be integral).
2. The angle \( \frac{(m+1)\pi}{6} \) must equal \( \pi + 2k\pi \) for some integer \( k \).

Thus:

\[ \frac{(m+1)\pi}{6} = \pi + 2k\pi \implies m+1 = 6 + 12k \implies m = 5 + 12k \]

The least integral value of \( m \) occurs when \( k = 0 \):

\[ m = 5 \]

### Final Answer

Thus, the least possible integral value of \( m \) is:

\[ \boxed{5} \]