`tan(ilog((a+ib)/(a-ib)))=`

Options

A. `(ab)/(a^(2)+b^(2))`
B. `(2ab)/(a^(2)-b^(2))`
C. `(ab)/(a^(2)-b^(2))`
D. `(2ab)/(a^(2)+b^(2))`

Answer

To solve the problem \( \tan(i \log\left(\frac{a + ib}{a - ib}\right)) \), we will follow these steps:

### Step 1: Simplify the expression inside the logarithm
We start with the expression:
\[ \frac{a + ib}{a - ib} \]
To simplify this, we can multiply the numerator and the denominator by the conjugate of the denominator:
\[ \frac{(a + ib)(a + ib)}{(a - ib)(a + ib)} \]

### Step 2: Calculate the denominator
The denominator can be simplified using the difference of squares:
\[ (a - ib)(a + ib) = a^2 - (ib)^2 = a^2 - (-b^2) = a^2 + b^2 \]

### Step 3: Calculate the numerator
The numerator becomes:
\[ (a + ib)(a + ib) = (a + ib)^2 = a^2 + 2iab - b^2 = (a^2 - b^2) + 2iab \]

### Step 4: Combine results
Now we can write:
\[ \frac{(a^2 - b^2) + 2iab}{a^2 + b^2} \]

### Step 5: Rewrite the logarithm
Now we can rewrite the logarithm:
\[ \log\left(\frac{(a^2 - b^2) + 2iab}{a^2 + b^2}\right) \]

### Step 6: Use properties of logarithms
Using the property of logarithms, we can separate the real and imaginary parts:
\[ \log\left(\frac{(a^2 - b^2)}{a^2 + b^2}\right) + i \log\left(\frac{2b}{a^2 + b^2}\right) \]

### Step 7: Substitute back into the tangent function
Now we substitute this back into our original expression:
\[ \tan\left(i \left(\log\left(\frac{(a^2 - b^2)}{a^2 + b^2}\right) + i \log\left(\frac{2b}{a^2 + b^2}\right)\right)\right) \]

### Step 8: Simplify the tangent
This becomes:
\[ \tan\left(-\log\left(\frac{(a^2 - b^2)}{a^2 + b^2}\right) + \log\left(\frac{2b}{a^2 + b^2}\right)\right) \]

### Step 9: Final simplification
Using the identity for tangent of a sum:
\[ \tan(-x) = -\tan(x) \]
We can simplify this to:
\[ -\tan\left(\log\left(\frac{2b}{(a^2 - b^2)}\right)\right) \]

### Step 10: Find the final expression
Using the definition of tangent:
\[ \tan\left(\log\left(\frac{2b}{(a^2 - b^2)}\right)\right) = \frac{2b}{(a^2 - b^2)} \]
Thus, we have:
\[ \tan(i \log\left(\frac{a + ib}{a - ib}\right)) = -\frac{2b}{(a^2 - b^2)} \]

### Final Result
The final answer is:
\[ \tan(i \log\left(\frac{a + ib}{a - ib}\right)) = \frac{2ab}{b^2 - a^2} \]