If `i^(2)=-1` and `((1+i)/(sqrt2))^(n)=((1-i)/(sqrt2))^(m)=1, AA n, m in N`, then the minimum value of `n+m` is equal to

Answer

To solve the problem, we start with the given equations:

\[ \left(\frac{1+i}{\sqrt{2}}\right)^n = \left(\frac{1-i}{\sqrt{2}}\right)^m = 1 \]

### Step 1: Convert to Exponential Form
We can express the complex numbers \(1+i\) and \(1-i\) in exponential form.

1. Calculate the modulus and argument of \(1+i\):
- Modulus:
\[ |1+i| = \sqrt{1^2 + 1^2} = \sqrt{2} \]
- Argument:
\[ \theta = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4} \]
- Therefore,
\[ 1+i = \sqrt{2} e^{i \frac{\pi}{4}} \]

2. Calculate the modulus and argument of \(1-i\):
- Modulus:
\[ |1-i| = \sqrt{1^2 + (-1)^2} = \sqrt{2} \]
- Argument:
\[ \theta = \tan^{-1}\left(\frac{-1}{1}\right) = -\frac{\pi}{4} \]
- Therefore,
\[ 1-i = \sqrt{2} e^{-i \frac{\pi}{4}} \]

### Step 2: Substitute into the Original Equations
Now we substitute these into the original equations:

1. For the first equation:
\[ \left(\frac{1+i}{\sqrt{2}}\right)^n = \left(\frac{\sqrt{2} e^{i \frac{\pi}{4}}}{\sqrt{2}}\right)^n = (e^{i \frac{\pi}{4}})^n = e^{i \frac{n \pi}{4}} = 1 \]

This implies:
\[ \frac{n \pi}{4} = 2k\pi \quad \text{for some integer } k \]
Thus,
\[ n = 8k \]

2. For the second equation:
\[ \left(\frac{1-i}{\sqrt{2}}\right)^m = \left(\frac{\sqrt{2} e^{-i \frac{\pi}{4}}}{\sqrt{2}}\right)^m = (e^{-i \frac{\pi}{4}})^m = e^{-i \frac{m \pi}{4}} = 1 \]

This implies:
\[ -\frac{m \pi}{4} = 2j\pi \quad \text{for some integer } j \]
Thus,
\[ m = -8j \]

### Step 3: Determine Natural Number Values
Since \(n\) and \(m\) must be natural numbers (positive integers), we can set \(k = 1\) and \(j = -1\) (the smallest positive values):

- For \(k = 1\):
\[ n = 8 \times 1 = 8 \]

- For \(j = -1\):
\[ m = -8 \times (-1) = 8 \]

### Step 4: Calculate \(n + m\)
Now, we find the minimum value of \(n + m\):
\[ n + m = 8 + 8 = 16 \]

### Final Answer
Thus, the minimum value of \(n + m\) is:
\[ \boxed{16} \]