How do I find the square roots of ii? Precalculus Complex Numbers in Trigonometric Form Roots of Complex Numbers 1 Answer Wataru Oct 26, 2014 Let z=re^{i theta}z=reiθ be the square-roots of ii. z^2=i Rightarrow r^2e^{i(2theta)}=e^{i(pi/2+2npi)}z2=i⇒r2ei(2θ)=ei(π2+2nπ) 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 3718 views around the world You can reuse this answer Creative Commons License