Here you will learn what is the division of complex numbers with examples.
Let’s begin –
Division of Complex Numbers
The division of a complex number \(z_1\) by a non-zero complex number \(z_2\) is defined as the multiplication of \(z_1\) by the multiplicative inverse of \(z_2\) and is denoted by \(z_1\over z_2\).
Thus, \(z_1\over z_2\) = \(z_1.{z_2}^{-1}\) = \(z_1\).(\(1\over z_2\))
How to Find Muliplicative Inverse :
Let z = a + ib be a non-zero complex number. Then,
\(1\over z\) = \(1\over a + ib\)
Multiply numerator and denominator by conjugate of denominator,
\(1\over z\) = \(1\over a + ib\) \(\times\) \(a – ib\over a – ib\)
\(\implies\) \(1\over z\) = \(a – ib\over a^2 – i^2b^2\) = \(a – ib\over a^2 + b^2\)
\(\implies\) \(1\over z\) = \(a\over a^2 + b^2\) + \(i(-b)\over a^2 + b^2\)
Clearly, \(1\over z\) is equal to the multiplicatve inverse of z.
How to Divide Two Complex Numbers :
Let \(z_1\) = \(a_1 + ib_1\) and \(z_2\) = \(a_2 + ib_2\). Then
\(z_1\over z_2\) = (\(a_1 + ib_1\)){\(a_2\over {a_2}^2 + {b_2}^2\) + \(i{(-b_2)\over {a_2}^2 + {b_2}^2}\)}
[ \(\because\) z = a + ib \(\implies\) \(1\over z\) = \(a\over a^2 + b^2\) + \(i(-b)\over a^2 + b^2\)
By definition of multiplication,
\(z_1\over z_2\) = (\(a_1a_2 + b_1b_2\over {a_2}^2 + {b_2}^2\)) + i(\(a_2b_1 – a_1b_2\over {a_2}^2 + {b_2}^2\))
Example : If \(z_1\) = 2 + 3i and \(z_2\) = 1 + 2i, then find \(z_1\over z_2\).
Solution : We have \(z_1\) = 2 + 3i and \(z_2\) = 1 + 2i,
\(\implies\) \(1\over z_2\) = \(1\over 1 + 2i\) = \({1\over 5} – {2\over 5}i\)
Now,
\(z_1\over z_2\) = \(z_1\).\(1\over z_2\) = (2 + 3i)( \({1\over 5} – {2\over 5}i\))
= (\({2\over 5} + {6\over 5}\)) + i(\({-4\over 5} + {3\over 5}\)) = \(8\over 5\) – \(1\over 5\)i
