Question 394194
Note:
i = i
i^2 = -1
i^3 = -i
i^4 = 1
etc.
-------
1) 4i^5-6i^4+3i^3-9i^2
= 4i - 6*1+3(-i)-9(-1)
---
= -6+9+4i-3i
= 3+i
--------------------------
 
2) 3i^4-(2i-6)-5i^2+(2+12i)
= 3-2i+6-5(-1)+2+12i
---
= 3+6+4+2 -2i+12i
---
= 15+10i
----------------------------

3) (4i-5)*(5i-4) 
= 20i^2-16i-25i+20
= -20+20-41i
= -41i
--------------------------------- 
4) (5i+8)/(3i)
= [(-3i)(5i+8)]/[(-3i)(3i)]
---
= [15-24i]/9
---
= [5-8i]/3
--------------------------------- 
5) (11-3i)/(4+5i)
= [(11-3i)(4-5i)]/[(4+5i)(4-5i)]
---
= [44-55i-12i-15]/[16+25]
---
= [29-67i]/41
=================
Cheers,
Stan H.