Answer:
[tex]\dfrac{31}{2}+\dfrac{27}{2}i[/tex]
Step-by-step explanation:
The given complex number is:
[tex]z=\dfrac{(2+3i)(4-7i)(1+i)}{(1^2-i^2)}[/tex]
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
[tex]\dfrac{(2+3i)(4-7i)(1+i)}{(1^2-i^2)}=\dfrac{31+27i}{2}\\\\=\dfrac{31}{2}+\dfrac{27}{2}i[/tex]
Hence, the correct answer is [tex]\dfrac{31}{2}+\dfrac{27}{2}i[/tex].
