Question

Solve:
(1 +3i) (1-3i) - (-i-2) (i-2)​

Answer 1

Answer:

$(1 + 3i)(1 - 3i) - ( - i - 2)(i - 2)$

$= > {1}^{2} - {(3i)}^{2} + (i + 2)(i - 2)$

$= > 1 - \: ( - 9) + {i}^{2} - {2}^{2}$

$= > 1 + 9 - 1 - 4$

$= > 5$

Because value of

${i}^{2} = - 1$

Hence 5 is the required answer. I hope it will help you. Thank you

Answer 2

Step-by-step explanation:

[(1 + 3i) × (1 - 3i)] - [(-i -2) × (i - 2)]

= [1^2 - (3i)^2] - [(-2 - i) × (-2 + i)]

[(a + b)×(a - b) = a^2 -b^2]

= [1 - 9i^2] - [(-2)^2 - i^2]

= [1 - 9i^2] - [4 - i^2]

= 1 - 9i^2 - 4 + i^2

= 1 - 4 - 9i^2 + i^2

= -3 - 8i^2

Hope it helps.