Find the multiplicative inverse of `(3+4i)/(4-5i)`

Answer

To find the multiplicative inverse of the complex number \(\frac{3+4i}{4-5i}\), we follow these steps:

### Step 1: Write the given complex number
We start with the given complex number:
\[ z = \frac{3 + 4i}{4 - 5i} \]

### Step 2: Find the multiplicative inverse
The multiplicative inverse of a complex number \(z\) is given by \(\frac{1}{z}\). This can be expressed as:
\[ \frac{1}{z} = \frac{4 - 5i}{3 + 4i} \]

### Step 3: Multiply numerator and denominator by the conjugate
To simplify \(\frac{4 - 5i}{3 + 4i}\), we multiply both the numerator and the denominator by the conjugate of the denominator, which is \(3 - 4i\):
\[ \frac{4 - 5i}{3 + 4i} \cdot \frac{3 - 4i}{3 - 4i} = \frac{(4 - 5i)(3 - 4i)}{(3 + 4i)(3 - 4i)} \]

### Step 4: Simplify the numerator
Now we calculate the numerator:
\[ (4 - 5i)(3 - 4i) = 4 \cdot 3 + 4 \cdot (-4i) - 5i \cdot 3 - 5i \cdot (-4i) \]
Calculating each term:
- \(4 \cdot 3 = 12\)
- \(4 \cdot (-4i) = -16i\)
- \(-5i \cdot 3 = -15i\)
- \(-5i \cdot (-4i) = 20\) (since \(i^2 = -1\))

Combining these:
\[ 12 - 16i - 15i + 20 = (12 + 20) + (-16i - 15i) = 32 - 31i \]

### Step 5: Simplify the denominator
Now we simplify the denominator:
\[ (3 + 4i)(3 - 4i) = 3^2 - (4i)^2 = 9 - 16(-1) = 9 + 16 = 25 \]

### Step 6: Combine results
Now we have:
\[ \frac{32 - 31i}{25} \]
This can be separated into real and imaginary parts:
\[ \frac{32}{25} - \frac{31}{25}i \]

### Final Result
Thus, the multiplicative inverse of \(\frac{3 + 4i}{4 - 5i}\) is:
\[ \frac{32}{25} - \frac{31}{25}i \]

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