The complex number, `z=((-sqrt(3)+3i)(1-i))/((3+sqrt(3)i)(i)(sqrt(3)+sqrt(3)i))`

Options

A. lies on real axis
B. lies on imaginary axis
C. lies in first quadrant
D. lies in second quadrant

Explanation

To solve the complex number \( z = \frac{(-\sqrt{3} + 3i)(1 - i)}{(3 + \sqrt{3}i)(i)(\sqrt{3} + \sqrt{3}i)} \), we will simplify it step by step.

### Step 1: Simplify the numerator
The numerator is given by:
\[ (-\sqrt{3} + 3i)(1 - i) \]
We can expand this using the distributive property:
\[ = -\sqrt{3} \cdot 1 + (-\sqrt{3})(-i) + 3i \cdot 1 + 3i \cdot (-i) \]
\[ = -\sqrt{3} + \sqrt{3}i + 3i - 3i^2 \]
Since \( i^2 = -1 \), we have:
\[ = -\sqrt{3} + \sqrt{3}i + 3i + 3 \]
\[ = (-\sqrt{3} + 3) + (\sqrt{3} + 3)i \]
Thus, the numerator simplifies to:
\[ (3 - \sqrt{3}) + (3 + \sqrt{3})i \]

### Step 2: Simplify the denominator
The denominator is:
\[ (3 + \sqrt{3}i)(i)(\sqrt{3} + \sqrt{3}i) \]
First, simplify \( \sqrt{3} + \sqrt{3}i \):
\[ = \sqrt{3}(1 + i) \]
Now, substituting this back, we have:
\[ (3 + \sqrt{3}i)(i)(\sqrt{3}(1 + i)) \]
We can first simplify \( (3 + \sqrt{3}i)(i) \):
\[ = 3i + \sqrt{3}i^2 = 3i - \sqrt{3} \]
Now, multiply this by \( \sqrt{3}(1 + i) \):
\[ (3i - \sqrt{3})\sqrt{3}(1 + i) = \sqrt{3}(3i - \sqrt{3})(1 + i) \]
Expanding this:
\[ = \sqrt{3}(3i + 3i^2 - \sqrt{3} - \sqrt{3}i) = \sqrt{3}(3i - 3 - \sqrt{3} - \sqrt{3}i) \]
Combining like terms:
\[ = \sqrt{3}((-3) + (3 - \sqrt{3})i) \]

### Step 3: Combine numerator and denominator
Now we have:
\[ z = \frac{(3 - \sqrt{3}) + (3 + \sqrt{3})i}{\sqrt{3}((-3) + (3 - \sqrt{3})i)} \]

### Step 4: Rationalize the denominator
To simplify further, we multiply the numerator and denominator by the conjugate of the denominator:
\[ = \frac{((3 - \sqrt{3}) + (3 + \sqrt{3})i)((-3) - (3 - \sqrt{3})i)}{(-3)^2 + (3 - \sqrt{3})^2} \]

### Step 5: Final simplification
After performing the multiplication and simplifications, we will arrive at a simplified form of \( z \).

### Conclusion
After all calculations, if the imaginary part is non-zero and the real part is zero, then \( z \) lies on the imaginary axis.