The value of `sqrti+sqrt(-i)` is :

Options

A. 1
B. `sqrt2`
C. -i
D. none of these

Answer

To solve the expression \( \sqrt{i} + \sqrt{-i} \), we will follow these steps:

### Step 1: Understand the values of \( i \) and \( -i \)
We know that:
- \( i = \sqrt{-1} \)
- Therefore, \( -i = -\sqrt{-1} \)

### Step 2: Find \( \sqrt{i} \)
To find \( \sqrt{i} \), we can express \( i \) in polar form. The polar form of \( i \) is:
- \( i = \cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right) \)

Using the polar form, we can find the square root:
\[ \sqrt{i} = \sqrt{r} \left( \cos\left(\frac{\theta}{2}\right) + i\sin\left(\frac{\theta}{2}\right) \right) \]
Where \( r = 1 \) (the modulus of \( i \)) and \( \theta = \frac{\pi}{2} \):
\[ \sqrt{i} = \sqrt{1} \left( \cos\left(\frac{\pi/2}{2}\right) + i\sin\left(\frac{\pi/2}{2}\right) \right) = \cos\left(\frac{\pi}{4}\right) + i\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2} \]

### Step 3: Find \( \sqrt{-i} \)
Similarly, for \( -i \):
\[ -i = \cos\left(-\frac{\pi}{2}\right) + i\sin\left(-\frac{\pi}{2}\right) \]
Using the same method:
\[ \sqrt{-i} = \sqrt{1} \left( \cos\left(-\frac{\pi/2}{2}\right) + i\sin\left(-\frac{\pi/2}{2}\right) \right) = \cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2} \]

### Step 4: Combine \( \sqrt{i} \) and \( \sqrt{-i} \)
Now we add \( \sqrt{i} \) and \( \sqrt{-i} \):
\[ \sqrt{i} + \sqrt{-i} = \left(\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}\right) + \left(\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}\right) \]
Combining the real and imaginary parts:
\[ = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2} = \sqrt{2} + 0i = \sqrt{2} \]

### Final Answer
Thus, the value of \( \sqrt{i} + \sqrt{-i} \) is \( \sqrt{2} \).

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