How do you simplify #i^803#? Precalculus Complex Numbers in Trigonometric Form Powers of Complex Numbers 1 Answer Alan P. · Nam D. Mar 9, 2018 #i^803=color(red)(-i)# Explanation: #i^1=i# #i^2=-1# (definition) #i^3=(-1)xxi=-i# #i^4=(-1)xx(-1)=1# #i^803=i^800 * i^3# #color(white)("XX")=(i^4)^200 * i^3# #color(white)("XX")=1^200 * (-i)# #color(white)("XX") =1xx(-i)# #color(white)("XX")=-i# 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 1446 views around the world You can reuse this answer Creative Commons License
