Division of Complex Numbers

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

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