How do I find the square roots of #i#? Precalculus Complex Numbers in Trigonometric Form Roots of Complex Numbers 1 Answer Wataru Oct 26, 2014 Let #z=re^{i theta}# be the square-roots of #i#. #z^2=i Rightarrow r^2e^{i(2theta)}=e^{i(pi/2+2npi)}# #Rightarrow {(r^2=1 Rightarrow r=1),(2theta=pi/2+2npi Rightarrow theta=pi/4+npi):}# #z={e^{i pi/4}, e^{i {5pi}/4}}# #={cos(pi/4)+isin(pi/4), cos({5pi}/4)+isin({5pi}/4)}# #={1/sqrt{2}+1/sqrt{2}i, -1/sqrt{2}-1/sqrt{2}i}# I hope that this was helpful. Answer link Related questions How do I find the cube root of a complex number? How do I find the fourth root of a complex number? How do I find the fifth root of a complex number? How do I find the nth root of a complex number? How do I find the square root of a complex number? What is the square root of #2i#? What is the cube root of #(sqrt3 -i)#? What are roots of unity? How do you solve #6x^2-5x+3=0#? If #z in CC# then what is #sqrt(z^2)#? See all questions in Roots of Complex Numbers Impact of this question 3252 views around the world You can reuse this answer Creative Commons License