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

Question

35069 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-24T06:04:19

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

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 $+0.3827-0.9239i$

How do you find the fourth roots of ii?