Finding the Multiplicative Inverse of a Complex Number

The question asks for the multiplicative inverse of the complex number \(2 - 3i\). The multiplicative inverse of a non-zero complex number \(z\) is the number \(z^{-1}\) such that \(z \times z^{-1} = 1\). For a complex number \(z = a + bi\), its multiplicative inverse can be found using the formula \(\frac{1}{a + bi}\).

To simplify \(\frac{1}{a + bi}\), we multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of \(a + bi\) is \(a - bi\). So, the multiplicative inverse is:

$$ \frac{1}{a + bi} = \frac{1}{a + bi} \times \frac{a - bi}{a - bi} $$

We know that \((a + bi)(a - bi) = a^2 - (bi)^2 = a^2 - b^2i^2\). Since \(i^2 = -1\), this simplifies to \(a^2 - b^2(-1) = a^2 + b^2\). So the formula becomes:

$$ \frac{1}{a + bi} = \frac{a - bi}{a^2 + b^2} = \frac{a}{a^2 + b^2} - \frac{b}{a^2 + b^2}i $$

Calculating the Multiplicative Inverse of 2 - 3i

Given the complex number \(z = 2 - 3i\), we can see that \(a = 2\) and \(b = -3\). We need to find its multiplicative inverse.

The conjugate of \(2 - 3i\) is \(2 - (-3)i = 2 + 3i\). Multiplying the complex number by its conjugate gives:

$$ (2 - 3i)(2 + 3i) = 2^2 + (-3)^2 $$

Here, we use the property that for a complex number \(a+bi\), \((a+bi)(a-bi) = a^2 + b^2\). In \(2 - 3i\), \(a=2\) and \(b=-3\). Using the property \((a-bi)(a+bi) = a^2+b^2\), we get:

$$ (2 - 3i)(2 + 3i) = (2)^2 + (3)^2 = 4 + 9 = 13 $$

Now, we can calculate the multiplicative inverse of \(2 - 3i\):

$$ \frac{1}{2 - 3i} = \frac{1}{2 - 3i} \times \frac{2 + 3i}{2 + 3i} $$

Multiplying the numerators and the denominators:

• Numerator: \(1 \times (2 + 3i) = 2 + 3i\)
• Denominator: \((2 - 3i)(2 + 3i) = 13\)

So, the multiplicative inverse is:

$$ \frac{2 + 3i}{13} $$

This can be written in the standard form \(a + bi\) as:

$$ \frac{2}{13} + \frac{3i}{13} $$

Comparing with Options

Let's look at the given options:

1. \(2 + 3i\)
2. \(3 - 2i\)
3. \(\frac{2}{13}+\frac{3 i}{13}\)
4. More than one of the above
5. None of the above

Our calculated multiplicative inverse is \(\frac{2}{13} + \frac{3i}{13}\). Comparing this with the options, we see that option 3 matches our result exactly.

Conclusion on Multiplicative Inverse

The multiplicative inverse of the complex number \(2 - 3i\) is \(\frac{2}{13} + \frac{3i}{13}\).