Question

Find the multiplicative inverse of Z is equal to 2 + 3 iota

Answer 1

To find the multiplicative inverse of a complex number \( z = 2 + 3i \), we need to find a complex number \( w = a + bi \) such that:

\[
z \cdot w = 1
\]

This can be expressed as:

\[
(2 + 3i)(a + bi) = 1
\]

We can multiply this out:

\[
2a + 2bi + 3ai + 3bi^2 = 1
\]

Recalling that \( i^2 = -1 \), we rewrite \( 3bi^2 \) as \( -3b \):

\[
(2a - 3b) + (2b + 3a)i = 1 + 0i
\]

This gives us two equations by separating the real and the imaginary parts:

1. \( 2a - 3b = 1 \) (real part)
2. \( 2b + 3a = 0 \) (imaginary part)

Now, we can solve this system of equations. From the second equation, \( 2b + 3a = 0 \), we can express \( b \) in terms of \( a \):

\[
2b = -3a \quad \Rightarrow \quad b = -\frac{3}{2}a
\]

Substituting \( b \) into the first equation:

\[
2a - 3\left(-\frac{3}{2}a\right) = 1
\]

This simplifies to:

\[
2a + \frac{9}{2}a = 1
\]

Combining the terms, we have:

\[
\frac{4}{2}a + \frac{9}{2}a = 1 \quad \Rightarrow \quad \frac{13}{2}a = 1
\]

Now, solve for \( a \):

\[
a = \frac{2}{13}
\]

Next, substitute \( a \) back to find \( b \):

\[
b = -\frac{3}{2}\left(\frac{2}{13}\right) = -\frac{3}{13}
\]

Thus, the multiplicative inverse \( w \) is:

\[
w = a + bi = \frac{2}{13} - \frac{3}{13}i
\]

To express this in standard form, we write:

\[
w = \frac{2}{13} - \frac{3}{13}i
\]

Thus, the multiplicative inverse of \( z = 2 + 3i \) is

\[
\frac{2}{13} - \frac{3}{13}i
\]