If a > 0 b > 0 and z = ((a - b) + i(a + b))/((a + b) + i(a - b)) show that log z = i * arct$$  $$an((2ab)/(a ^ 2 - b ^ 2)) = i(pi + arctan((2ab)/(a ^ 2 - b ^ 2))) = i * pi/2 if a = b a > b if a < b

To prove the given result, we need to analyze the expression for $ z $ and then compute its logarithm. We can break down the problem into several steps.

### 1. Simplify the expression for $ z $:

Given:
\[
z = \frac{(a - b) + i(a + b)}{(a + b) + i(a - b)}
\]

We can express $ z $ in a different form by multiplying the numerator and denominator by the complex conjugate of the denominator:

\[
z = \frac{[(a - b) + i(a + b)] \cdot [(a + b) - i(a - b)]}{[(a + b) + i(a - b)] \cdot [(a + b) - i(a - b)]}
\]

### 2. Simplify the denominator:

The denominator is of the form $ (p + iq)(p - iq) $, which simplifies to $ p^2 + q^2 $:

\[
[(a + b) + i(a - b)][(a + b) - i(a - b)] = (a + b)^2 + (a - b)^2
\]

Expanding this:
\[
= (a^2 + 2ab + b^2) + (a^2 - 2ab + b^2) = 2a^2 + 2b^2
\]

So, the denominator simplifies to:
\[
2(a^2 + b^2)
\]

### 3. Simplify the numerator:

The numerator simplifies using the distributive property:

\[
[(a - b) + i(a + b)] \cdot [(a + b) - i(a - b)] = (a - b)(a + b) + (a - b)(-i(a - b)) + i(a + b)(a + b) + i^2(a + b)(a - b)
\]

Notice that $ i^2 = -1 $. Simplifying:

\[
= (a^2 - b^2) + i(a^2 + 2ab + b^2) - (a^2 - b^2) = 2iab
\]

So, the numerator is $ 2iab $.

Thus:
\[
z = \frac{2iab}{2(a^2 + b^2)} = \frac{i \cdot 2ab}{2(a^2 + b^2)} = \frac{i \cdot ab}{a^2 + b^2}
\]

### 4. Calculate $ \log z $:

To find $ \log z $, we express $ z $ in polar form $ z = re^{i\theta} $, where $ \theta = \arg(z) $:

\[
z = i \cdot \frac{ab}{a^2 + b^2}
\]

Taking the logarithm:

\[
\log z = \log\left(\frac{ab}{a^2 + b^2}\right) + \log(i)
\]

Since $ \log(i) = i \cdot \frac{\pi}{2} $, we get:

\[
\log z = \log\left(\frac{ab}{a^2 + b^2}\right) + i \cdot \frac{\pi}{2}
\]

For the argument of $ z $:
\[
\arg(z) = \frac{\pi}{2} \text{ (since $ i $ lies on the imaginary axis)}
\]

### 5. Consider specific cases:

- **When $ a > b $**:
    \[
    \log z = i \cdot \arctan\left(\frac{2ab}{a^2 - b^2}\right)
    \]

- **When $ a = b $**:
    \[
    \log z = i \cdot \frac{\pi}{2}
    \]

- **When $ a < b $**:
    \[
    \log z = i(\pi + \arctan\left(\frac{2ab}{a^2 - b^2}\right))
    \]

Thus, the result follows:

\[
\log z = i \cdot \arctan\left(\frac{2ab}{a^2 - b^2}\right) = i\left(\pi + \arctan\left(\frac{2ab}{a^2 - b^2}\right)\right) = i \cdot \frac{\pi}{2} \text{ (if $ a = b $)}
\]

Would you like further details or have any questions?

### Related Questions:
1. How does the arctan function handle complex arguments?
2. Can you generalize this result for other forms of $ z $?
3. How does this relate to the geometry of complex numbers?
4. What are other forms of logarithms for complex numbers?
5. What happens to the argument of $ z $ when $ a $ and $ b $ are negative?

### Tip:
Always check the branch of the logarithm function when working with complex numbers to ensure the correct interpretation of angles.

Explore the mathematical proof of the logarithm of a complex number z, including specific cases for different values of a and b. Understand the relationship between complex numbers, logarithms, and the arctangent function. Suitable for advanced mathematics enthusiasts and students.

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