Question

if z1= -2+i and z2=1-3i then find Re (1/z1z2)​

Answer 1

Given :-

$\leadsto z_1 = - 2 + i$

$\leadsto z_2 = 1 - 3i$

To find :-

$\leadsto Re\left(\dfrac{1}{z_1.z_2}\right)$

Solution :-

$\implies Re\left(\dfrac{1}{z_1.z_2}\right)$

$\implies Re\left(\dfrac{1}{( - 2 + i)(1 - 3i)}\right)$

$\implies Re\left(\dfrac{1}{ - 2(1 - 3i) + i(1 - 3i)}\right)$

$\implies Re\left(\dfrac{1}{ - 2 + 6i + i - 3i^{2} }\right)$

$\implies Re\left(\dfrac{1}{ - 2 + 7i- 3i^{2} }\right)$

$\implies Re\left(\dfrac{1}{ - 2 + 7i- 3( - 1) }\right)$

$\implies Re\left(\dfrac{1}{ - 2 + 7i + 3}\right)$

$\implies Re\left(\dfrac{1}{ 1 + 7i }\right)$

$\implies Re\left(\dfrac{1}{ 1 + 7i } \times \dfrac{1 - 7i}{1 - 7i} \right)$

$\implies Re\left( \dfrac{1 - 7i}{(1)^{2} - (7i)^{2} } \right)$

$\implies Re\left( \dfrac{1 - 7i}{1 - 49 {i}^{2} } \right)$

$\implies Re\left( \dfrac{1 - 7i}{1 - 49 ( - 1)} \right)$

$\implies Re\left( \dfrac{1 - 7i}{1 + 49} \right)$

$\implies Re\left( \dfrac{1 - 7i}{50} \right)$

$\implies Re\left( \dfrac{1 }{50} + \dfrac{ - 7i}{50} \right)$

$\boxed{ \implies \dfrac{1 }{50}}$

Hence the required answer is 1/50.

Note : i = √-1 , i² = 1