How do you find the fourth roots of #i#?

1 Answer
Jan 24, 2017

Fourth roots of #i# are #0.9239+0.3827i#,
#-0.3827+0.9239i#,
#-0.9239-0.3827i# and
#+0.3827-0.9239i#

Explanation:

We can use the De Moivre's Theorem, according to which

If #z=re^(itheta)=r(costheta+isintheta)#

then #z^n=r^n e^(ixxntheta)=r(cosntheta+isinntheta)#

As #i=cos(pi/2)+isin(pi/2)# can be written as #i=cos(2npi+pi/2)+isin(2npi+pi/2)#

#root(4)i=i^(1/4)=cos((2npi)/4+pi/8)+isin((2npi)/4+pi/8)# or

= #cos((npi)/2+pi/8)+isin((npi)/2+pi/8)#

Note that #n=0,1,2,3# and after #n=3#, it will repeat.

This gives us four roots of #i#, which are

#cos(pi/8)+isin(pi/8)#,

#cos(pi/2+pi/8)+isin(pi/2+pi/8)=-sin(pi/8)+icos(pi/8)#

#cos(pi+pi/8)+isin(pi+pi/8)=-cos(pi/8)-isin(pi/8)#

and #cos((3pi)/2+pi/8)+isin((3pi)/2+pi/8)=sin(pi/8)-icos(pi/8)#

and to get exact values we can use #sin(pi/8)=sqrt(2-sqrt2)/2=0.3827# and #cos(pi/8)=sqrt(2+sqrt2)/2=0.9239#

Hence, fourth roots of #i# are #0.9239+0.3827i#, #-0.3827+0.9239i#, #-0.9239-0.3827i# and #+0.3827-0.9239i#