If z = (2+ 3i) ( 1+ 2i) find z-¹

1.	If z = (2+ 3i) ( 1+ 2i) find z-¹
Answer» Z = (2 + 3i)(1 + 2i) = 2 + 4i + 3i + 6i² = 2 – 6 + 7i \((\because i^2 =-1)\) = –4 + 7i Now, Z^–1 \(=\frac{1}{z}=\frac{1}{-4+7i}\) \(=\frac{1}{-4+7i}\times \frac{4+7i}{4+7i}\) \(=\frac{4+7i}{49i^2-16}\) \(=\frac{4+7i}{-49-16}\) \((\because i^2=-1)\) \(=\frac{-1}{65}(4+7i)\)

Answer»

Z = (2 + 3i)(1 + 2i)

= 2 + 4i + 3i + 6i²

= 2 – 6 + 7i \((\because i^2 =-1)\)

= –4 + 7i

Now, Z^–1 \(=\frac{1}{z}=\frac{1}{-4+7i}\)

\(=\frac{1}{-4+7i}\times \frac{4+7i}{4+7i}\)

\(=\frac{4+7i}{49i^2-16}\)

\(=\frac{4+7i}{-49-16}\) \((\because i^2=-1)\)

\(=\frac{-1}{65}(4+7i)\)

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If z = (2+ 3i) ( 1+ 2i) find z-¹

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