#i ^i = # ? Precalculus Complex Numbers in Trigonometric Form Multiplication of Complex Numbers 1 Answer Cesareo R. Jun 25, 2016 #i^i=0.20788# Explanation: Any complex number can be written as #x +i y = (sqrt(x^2+y^2))e^{i phi}# with #{x,y}in RR^2# where #phi = arctan(y/x)# then #(x +i y)^{x+iy} equiv ((sqrt(x^2+y^2))e^{i phi})^{(sqrt(x^2+y^2))e^{i phi}}# making #x = 0, y = 1# #i^i = (e^{i pi/2})^{e^{i pi/2}} = (e^{i pi/2})^i = e^{-pi/2} = 0.20788# 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 2066 views around the world You can reuse this answer Creative Commons License