The modulus of `sqrt(2i)-sqrt(-2i)` is

Options

A. 2 `sqrt( 2)`
B. 0
C. `sqrt(2)`
D. 2

Answer

To find the modulus of \( \sqrt{2i} - \sqrt{-2i} \), we will follow these steps:

### Step 1: Rewrite the square roots
We start with the expression:
\[ \sqrt{2i} - \sqrt{-2i} \]
We can express \( \sqrt{-2i} \) as \( \sqrt{2} \cdot \sqrt{-i} \). Since \( -i = e^{-i\pi/2} \), we can write:
\[ \sqrt{-2i} = \sqrt{2} \cdot \sqrt{-i} = \sqrt{2} \cdot \sqrt{e^{-i\pi/2}} = \sqrt{2} \cdot e^{-i\pi/4} = \sqrt{2} \left( \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}} \right) = 1 - i \]

### Step 2: Find \( \sqrt{2i} \)
Similarly, for \( \sqrt{2i} \):
\[ \sqrt{2i} = \sqrt{2} \cdot \sqrt{i} = \sqrt{2} \cdot e^{i\pi/4} = \sqrt{2} \left( \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}} \right) = 1 + i \]

### Step 3: Substitute back into the expression
Now substituting back into our original expression:
\[ \sqrt{2i} - \sqrt{-2i} = (1 + i) - (1 - i) = 1 + i - 1 + i = 2i \]

### Step 4: Find the modulus
The modulus of a complex number \( z = a + bi \) is given by \( |z| = \sqrt{a^2 + b^2} \). Here, \( z = 2i \) which can be expressed as \( 0 + 2i \):
\[ |2i| = \sqrt{0^2 + 2^2} = \sqrt{4} = 2 \]

### Final Answer
Thus, the modulus of \( \sqrt{2i} - \sqrt{-2i} \) is:
\[ \boxed{2} \]