How do you simplify #3i^2 - 4i^4 + 5i^8 + 3# and write in a+bi form? Precalculus Complex Numbers in Trigonometric Form Powers of Complex Numbers 1 Answer Vikki Nov 27, 2015 Remember #color(red)(i^2 = -1)# ; #color(blue)(i^4 = 1)# #3i^2 -4i^4 +5i^8 +3# Can be rewrite as #3i^2 -4i^4 _5(i^4)^2 +3# #=> =3*color(red)(-1)-4*(color(blue)1)+5*color(blue)(1)^2+3# #=>= -3 -4 +5 +3# #=> =1# In #a+-bi# form #1 +- 0i# Answer link Related questions How do I use DeMoivre's theorem to find #(1+i)^5#? How do I use DeMoivre's theorem to find #(1-i)^10#? How do I use DeMoivre's theorem to find #(2+2i)^6#? What is #i^2#? What is #i^3#? What is #i^4#? How do I find the value of a given power of #i#? How do I find the #n#th power of a complex number? How do I find the negative power of a complex number? Write the complex number #i^17# in standard form? See all questions in Powers of Complex Numbers Impact of this question 1738 views around the world You can reuse this answer Creative Commons License