Question

what is the answer of square root of iota?

Answer 1

can you write 'i' in polar (trigonometric ) form ?

Answer 2

\(\sqrt{\iota}\) ? or \\(\sqrt{i}\) ?

Answer 3

i think question is about sqare root of square root of -1

Answer 4

yes, i = 0+1i = 1 < 90
so its square root will be \(\sqrt 1 < (90/2)=1 <45 = (1/ \sqrt 2) (1+i)\)

Answer 5

how 1 < 90 ?

Answer 6

i = 0+1i =x+iy (x=0,y=1)
from rectangular to polar,
\(r=\sqrt{x^2+y^2}=1, \theta =arctan(y/x)=90 \\so, r<\theta =1<90\)

Answer 7

ok now i get it
thanks