Question

Select the correct answer.

Answer 1

Answer:

$\dfrac{31}{2}+\dfrac{27}{2}i$

Step-by-step explanation:

The given complex number is:

$z=\dfrac{(2+3i)(4-7i)(1+i)}{(1^2-i^2)}$

Firstly we can solve (2+3i)(4-7i)(1+i)

= 2(8)+2(-7i)+4(3i)+3i(-7i²)× (1+i)

= (16-14i+12i-21i²i)× (1+i)

Since, i² = -1

= (16-14i+12i-21(-1)i)× (1+i)

= (16-14i+12i+21)× (1+i)

= (29-2i)(1+i)

= 31+27i

Denominator : (1-i²)

= (1-(-1))

= 2

$\dfrac{(2+3i)(4-7i)(1+i)}{(1^2-i^2)}=\dfrac{31+27i}{2}\\\\=\dfrac{31}{2}+\dfrac{27}{2}i$

Hence, the correct answer is $\dfrac{31}{2}+\dfrac{27}{2}i$.