Given z is a complex number with modulus 1. Then the equation `((1 + i a)/(1 - i a))^(4) = z ` has

Options

A. all roots real and distinct
B. two roots real and two imaginary
C. three roots real and one imaginary
D. one root real and three imaginary

Answer

To solve the equation \(\left(\frac{1 + ia}{1 - ia}\right)^{4} = z\) given that \(z\) is a complex number with modulus 1, we will follow these steps:

### Step 1: Analyze the given equation
We start with the equation:
\[ \left(\frac{1 + ia}{1 - ia}\right)^{4} = z \]
Since \(z\) has a modulus of 1, we can deduce that the modulus of the left-hand side must also be 1.

### Step 2: Calculate the modulus of the left-hand side
We need to find the modulus of \(\frac{1 + ia}{1 - ia}\):
\[ \text{Modulus} = \frac{|1 + ia|}{|1 - ia|} \]
Calculating the modulus of the numerator and denominator:
- For \(1 + ia\):
\[ |1 + ia| = \sqrt{1^2 + a^2} = \sqrt{1 + a^2} \]
- For \(1 - ia\):
\[ |1 - ia| = \sqrt{1^2 + a^2} = \sqrt{1 + a^2} \]
Thus, we have:
\[ \frac{|1 + ia|}{|1 - ia|} = \frac{\sqrt{1 + a^2}}{\sqrt{1 + a^2}} = 1 \]

### Step 3: Raise the modulus to the power of 4
Now, raising the modulus to the power of 4:
\[ \left(\frac{|1 + ia|}{|1 - ia|}\right)^{4} = 1^{4} = 1 \]
This confirms that the left-hand side has a modulus of 1.

### Step 4: Set the modulus equal to the modulus of \(z\)
Since we know that the modulus of \(z\) is also 1, we have:
\[ 1 = |z| = 1 \]
This condition is satisfied.

### Step 5: Analyze the argument of the complex number
Next, we need to consider the argument of the complex number:
\[ \frac{1 + ia}{1 - ia} \]
Using the property of arguments:
\[ \arg\left(\frac{1 + ia}{1 - ia}\right) = \arg(1 + ia) - \arg(1 - ia) \]
Calculating the arguments:
- For \(1 + ia\):
\[ \arg(1 + ia) = \tan^{-1}\left(\frac{a}{1}\right) = \tan^{-1}(a) \]
- For \(1 - ia\):
\[ \arg(1 - ia) = \tan^{-1}\left(\frac{-a}{1}\right) = -\tan^{-1}(a) \]
Thus:
\[ \arg\left(\frac{1 + ia}{1 - ia}\right) = \tan^{-1}(a) - (-\tan^{-1}(a)) = 2\tan^{-1}(a) \]

### Step 6: Set the argument equal to \(z\)
Now, we have:
\[ \arg\left(\left(\frac{1 + ia}{1 - ia}\right)^{4}\right) = 4 \cdot \arg\left(\frac{1 + ia}{1 - ia}\right) = 8\tan^{-1}(a) \]
This means:
\[ 8\tan^{-1}(a) = \arg(z) \]
Since \(z\) has modulus 1, its argument can take any value.

### Conclusion
The equation \(\left(\frac{1 + ia}{1 - ia}\right)^{4} = z\) has solutions for all real values of \(a\). Therefore, the solutions are real and distinct.