Simple De Moivre's Q. (1 Viewer)

[Utility]

Use De Moivre's theorem to find all z where
(i) z^2 = i

 

[nightweaver066]

Let z = cis@

(cis@)^2 = i
cis2@ = i (By De Moivre's theorem)
cos2@ + isin2@ = i

Equating real and imaginary terms,
cos2@ = 0, sin2@ = 1

Therefore 2@ = pi/2, 5pi/2
@ = pi/4, 5pi/4

 

[mathsbrain]

z^2=i

let z=rcis(x), polar form for i=cis(pi/2)
this means r^2 cis(2x) = 1cis (pi/2 +2kpi)

therefore equating moduli and arguments gives:
r=1, x= (pi/2 +2k pi)/2 where k=0,1
so your answers are z1=cis(pi/4) and z2 = cis(3 pi/4)

 

[Utility]

so your answers are z1=cis(pi/4) and z2 = cis(3 pi/4)

Nice, that is the method I used (except 3pi / 4 should be 5pi / 4).