{"id":6514,"date":"2021-10-17T19:05:42","date_gmt":"2021-10-17T13:35:42","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6514"},"modified":"2021-11-21T18:19:25","modified_gmt":"2021-11-21T12:49:25","slug":"division-of-complex-numbers","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/division-of-complex-numbers\/","title":{"rendered":"Division of Complex Numbers"},"content":{"rendered":"
Here you will learn what is the division of complex numbers with examples.<\/p>\n
Let’s begin –<\/p>\n
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\\).<\/p>\n
\nThus, \\(z_1\\over z_2\\) = \\(z_1.{z_2}^{-1}\\) = \\(z_1\\).(\\(1\\over z_2\\))<\/p>\n<\/blockquote>\n
How to Find Muliplicative Inverse :<\/strong><\/p>\n
Let z = a + ib be a non-zero complex number. Then,<\/p>\n
\\(1\\over z\\) = \\(1\\over a + ib\\)<\/p>\n
Multiply numerator and denominator by conjugate of denominator,<\/p>\n
\\(1\\over z\\) = \\(1\\over a + ib\\) \\(\\times\\) \\(a – ib\\over a – ib\\)<\/p>\n
\\(\\implies\\) \\(1\\over z\\) = \\(a – ib\\over a^2 – i^2b^2\\) = \\(a – ib\\over a^2 + b^2\\)<\/p>\n
\\(\\implies\\) \\(1\\over z\\) = \\(a\\over a^2 + b^2\\) + \\(i(-b)\\over a^2 + b^2\\)<\/p>\n
Clearly, \\(1\\over z\\) is equal to the multiplicatve inverse of z.<\/p>\n
How to Divide Two Complex Numbers :<\/strong><\/p>\n
Let \\(z_1\\) = \\(a_1 + ib_1\\) and \\(z_2\\) = \\(a_2 + ib_2\\). Then<\/p>\n
\\(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}\\)}<\/p>\n
[ \\(\\because\\) z = a + ib \\(\\implies\\) \\(1\\over z\\) = \\(a\\over a^2 + b^2\\) + \\(i(-b)\\over a^2 + b^2\\)<\/p>\n
By definition of multiplication,<\/p>\n
\n\\(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\\))<\/p>\n<\/blockquote>\n
Example<\/strong><\/span> : If \\(z_1\\) = 2 + 3i and \\(z_2\\) = 1 + 2i, then find \\(z_1\\over z_2\\).<\/p>\n
Solution<\/span><\/strong> : We have \\(z_1\\) = 2 + 3i and \\(z_2\\) = 1 + 2i,<\/p>\n
\\(\\implies\\) \\(1\\over z_2\\) = \\(1\\over 1 + 2i\\) = \\({1\\over 5} – {2\\over 5}i\\)<\/p>\n
Now, <\/p>\n
\\(z_1\\over z_2\\) = \\(z_1\\).\\(1\\over z_2\\) = (2 + 3i)( \\({1\\over 5} – {2\\over 5}i\\))<\/p>\n
= (\\({2\\over 5} + {6\\over 5}\\)) + i(\\({-4\\over 5} + {3\\over 5}\\)) = \\(8\\over 5\\) – \\(1\\over 5\\)i<\/p>\n\n\n