If z_{1} = 2 + 5i, z_{2} = – 3 – 4i, and z_{3} = 1 + i, find the additive and multiplicative inverse of z_{1}, z_{2} and z_{3}

#### Solution

**(i)** z_{1} = 2 + 5i

Additive inverse z_{1 }is – z_{1}

⇒ – (2 + 5i) = – 2 – 5i

Multiplicative inverse z_{1 }is (z_{1})^{–1}

We know

z_{1}z_{1}^{–1} = 1

⇒ (2 + 5i)(u + iv) = 1 ......[∵ z^{–1} = u + iv]

2u + 2iv + 5iu – 5v = 1

(2u – 5v) + i(5u + 2v) = 1 + i0

Equating real and imaginary parts

2u – 5v = 1

5u + 2v = 0

Solving them, we get

u = `2/29`

v = `- 5/29`

∴ (z_{1})^{–1 }= `2/29 + (- 5/29)"i"`

(z_{1})^{–1 }= `1/29(2 - 5"i")`

**(ii)** z_{2} = – 3 – 4i

Additive inverse z_{2 }is – z_{2}

⇒ – (3 – 4i) = 3 + 4i

Multiplicative inverse z_{2 }is (z_{2})^{–1}

We know

z_{2} z_{2}^{–1} = 1

⇒ (– 3 – 4i)(u + iv) = 1 ......[∵ z_{2}^{–1} = u + iv]

– 3u – 3iv – 4iu + 4v = 1

(– 3u + 4v) + i(– 4u – 3v) = 1 + i 0

We get – 3u + 4v = 1

– 4u – 3v = 0

Solving them, u = `-3/25`

v = `4/25`

∴ (z_{2})^{–1 }= `1/25(- 3 + 4"i")`

**(iii)** z_{3} = 1 + i

**(а)** Additive inverse of z_{3}

= – z_{3}

= – (1 + i)

= – 1 – i

**(b)** Multiplicative inverse of z_{3}

= `1/"z"_3 1/(1 + "i") xx (1 - "i")/(1 - "i")`

= `(1 - "i")/2`