How do you find all solutions to #x^4-i=0#?

1 Answer
Salvatore I.
Nov 8, 2016

4 solutions
#x_k=cos(pi/8+kpi/2)+isin(pi/8+kpi/2)# with #k=0,1,2,3#

Explanation:

#x^4=i#
it is better to express the immaginary unit in exponential form
#i=e^(ipi/2)# in such a way we can rewrite the equation like this
#x^4=e^(i(pi/2+2kpi)# and as a consequence it must be
#x=root4(e^(i(pi/2+2kpi)))=[e^(i(pi/2+2kpi))]^(1/4)=e^(i(pi/8+kpi/2)# with #k=0,1,2,3#

Related questions
See all questions in Roots of Complex Numbers
Impact of this question
9803 views around the world
You can reuse this answer
Creative Commons License