Operations on complex numbers | Part-2

Equality and Inverses of Complex Numbers

Equality of Two Complex Numbers:
Two complex numbers \( z_1 = a + ib \) and \( z_2 = c + id \) are equal if and only if their real and imaginary parts are equal, that is: \( a = c \) and \( b = d \).

Additive Inverse of a Complex Number:
The additive inverse of a complex number \( z = a + ib \) is \( -z = -a - ib \). Adding these gives:
\( z + (-z) = (a + ib) + (-a - ib) = 0 \)

Multiplicative Inverse of a Complex Number:
For a complex number \( z = a + ib \), where z ≠ 0 , the multiplicative inverse is another complex number \( z^{-1} \) such that:
\( z \cdot z^{-1} = 1 \)
It is given by:
\( z^{-1} = \frac{a}{a^2 + b^2} - i \cdot \frac{b}{a^2 + b^2} \)

Example 1: Multiplicative Inverse of \( 5 - 3i \)

Let \( z = 5 - 3i \). Here, \( a = 5 \), \( b = -3 \). Then:

\( z^{-1} = \frac{5}{5^2 + (-3)^2} + i \cdot \frac{3}{5^2 + 9} = \frac{5}{34} + i \cdot \frac{3}{34} \)

Example 2: Finding the Multiplicative Inverse of the Complex Expression \( \frac{5 - 3i}{3 + 2i} \)

First simplify the expression:

\( z = \frac{5 - 3i}{3 + 2i} \cdot \frac{3 - 2i}{3 - 2i} = \frac{(5 - 3i)(3 - 2i)}{(3 + 2i)(3 - 2i)} \)

\( = \frac{15 - 10i - 9i + 6i^2}{9 + 4} = \frac{15 - 19i - 6}{13} = \frac{9 - 19i}{13} \)

\( = \frac{9}{13} - i \cdot \frac{19}{13} \)

Now, find the inverse of \( \frac{9}{13} - i \cdot \frac{19}{13} \). Set ( a = frac{9}{13} ) and ( b = -frac{19}{13} ):

\( a^2 + b^2 = \left(\frac{9}{13}\right)^2 + \left(\frac{19}{13}\right)^2 = \frac{81 + 361}{169} = \frac{442}{169} \)

\( z^{-1} = \frac{a}{a^2 + b^2} - i \cdot \frac{b}{a^2 + b^2} = \frac{9}{13} \cdot \frac{169}{442} + i \cdot \frac{19}{13} \cdot \frac{169}{442} \)

\( = \frac{1521}{5746} + i \cdot \frac{3211}{5746} \)

Final Answer:
The reciprocal (or multiplicative inverse) of \( \frac{5 - 3i}{3 + 2i} \) is given by: \( z^{-1} = \frac{1521}{5746} + i \cdot \frac{3211}{5746} \)