Question

find the multiplicative inverse of z=(2+√3i)²​

Answer 1

Step-by-step explanation:

Given,

$z = {(2 + \sqrt{3}i) }^{2}$

$= > z = {(2)}^{2} + {( \sqrt{3}i) }^{2} + 2.2. (\sqrt{3}) i$

$z = 4 - 3 + (4 \sqrt{3}) i$

$= > z = 1 + (4 \sqrt{3}) i$

So,

${z}^{ - 1} = \frac{1}{1 + 4 \sqrt{3} i }$

$= > {z}^{ - 1} = \frac{1 - 4 \sqrt{3}i }{(1 + 4 \sqrt{3}i)(1 - 4 \sqrt{3} i) }$

$= > {z}^{ - 1} = \frac{1 - 4 \sqrt{3} i}{1 + 48}$

$= > {z}^{ - 1} = \frac{1}{49} - \frac{4 \sqrt{3} }{49} i$