What is the value of #(-sqrt(-1))^(4n+3)#? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer Shwetank Mauria Oct 22, 2016 #(-sqrt(-1))^(4n+3)=i# or #sqrt(-1)# Explanation: #(-sqrt(-1))^(4n+3)# = #(-i)^(4n+3)#, where #i^2=-1# = #(-1xxi)^(4n+3)# = #(-1)^(4n+3)xxi^(4n+3)# = #-1xxi^(4n)xxi^3#, as odd power of #-1# is #-1# = #-1xx1xxi^2xxi#, as #i^2=-1# #i^4=(i^2)^2=(-1)^2=1# = #-1xx1xx(-1)xxi# = #i# or #sqrt(-1)# Answer link Related questions How do I multiply complex numbers? How do I multiply complex numbers in polar form? What is the formula for multiplying complex numbers in trigonometric form? How do I use the modulus and argument to square #(1+i)#? What is the geometric interpretation of multiplying two complex numbers? What is the product of #3+2i# and #1+7i#? How do I use DeMoivre's theorem to solve #z^3-1=0#? How do I find the product of two imaginary numbers? How do you simplify #(2+4i)(2-4i)#? How do you multiply #(-2-8i)(6+7i)#? See all questions in Multiplication of Complex Numbers Impact of this question 18685 views around the world You can reuse this answer Creative Commons License