If `z_(0)=(1-i)/2`, then the value of the product `(1+z_(0))(1+z_(0)^(2))(1+z_(0)^(2^(2))(1+z_(0)^(2^(3)))…..(1+z_(0)^(2^(n)))` must be

Options

A. `(1-i)(1+1/(2/(2^(n-1))))`, if `n gt 1`
B. `(1-i)(1-1/2^(2^(n))), if n gt 1`
C. `(1-i)(1-1/(2^(n-1))), "if" n gt 1`
D. `(1-i)(1+1/2^(2^(n))), "if" n gt 1`

Answer

To solve the problem, we need to evaluate the product:

\[ P = (1 + z_0)(1 + z_0^2)(1 + z_0^{2^2})(1 + z_0^{2^3}) \ldots (1 + z_0^{2^n}) \]

where \( z_0 = \frac{1 - i}{2} \).

### Step 1: Rewrite the product using \( z_0 \)

Let’s denote the product as \( P \):

\[ P = (1 + z_0)(1 + z_0^2)(1 + z_0^{2^2})(1 + z_0^{2^3}) \ldots (1 + z_0^{2^n}) \]

### Step 2: Multiply both sides by \( 1 - z_0 \)

Now, we multiply both sides of the equation by \( (1 - z_0) \):

\[ (1 - z_0)P = (1 - z_0)(1 + z_0)(1 + z_0^2)(1 + z_0^{2^2}) \ldots (1 + z_0^{2^n}) \]

Using the identity \( (1 - z)(1 + z) = 1 - z^2 \), we can simplify the left-hand side:

\[ (1 - z_0)P = (1 - z_0^2)(1 + z_0^2)(1 + z_0^{2^2}) \ldots (1 + z_0^{2^n}) \]

### Step 3: Simplify the product

Continuing this process, we can express the product as:

\[ (1 - z_0)P = (1 - z_0^{2^{n+1}}) \]

### Step 4: Solve for \( P \)

Now, we can isolate \( P \):

\[ P = \frac{1 - z_0^{2^{n+1}}}{1 - z_0} \]

### Step 5: Calculate \( z_0^{2^{n+1}} \)

Next, we need to calculate \( z_0^{2^{n+1}} \):

1. First, calculate \( z_0^2 \):

\[ z_0^2 = \left(\frac{1 - i}{2}\right)^2 = \frac{(1 - i)^2}{4} = \frac{1 - 2i + i^2}{4} = \frac{1 - 2i - 1}{4} = \frac{-2i}{4} = -\frac{i}{2} \]

2. Now, calculate \( z_0^{2^{n+1}} \):

Using the polar form, we can express \( z_0 \):

\[ z_0 = \frac{1}{2} - \frac{i}{2} = \frac{1}{\sqrt{2}} e^{-i\pi/4} \]

Thus,

\[ z_0^{2^{n+1}} = \left(\frac{1}{\sqrt{2}} e^{-i\pi/4}\right)^{2^{n+1}} = \frac{1}{2^{(n+1)/2}} e^{-i\pi 2^{n+1}/4} \]

### Step 6: Substitute back into \( P \)

Substituting \( z_0^{2^{n+1}} \) into the equation for \( P \):

\[ P = \frac{1 - \frac{1}{2^{(n+1)/2}} e^{-i\pi 2^{n+1}/4}}{1 - \frac{1 - i}{2}} \]

### Step 7: Simplify the denominator

The denominator simplifies to:

\[ 1 - \frac{1 - i}{2} = \frac{2 - (1 - i)}{2} = \frac{1 + i}{2} \]

### Step 8: Final expression for \( P \)

Thus, we have:

\[ P = \frac{2(1 - \frac{1}{2^{(n+1)/2}} e^{-i\pi 2^{n+1}/4})}{1 + i} \]

### Conclusion

The final value of the product \( P \) is:

\[ P = \frac{2(1 - \frac{1}{2^{(n+1)/2}} e^{-i\pi 2^{n+1}/4})}{1 + i} \]