How do you find the square root of 8i?

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Answer 1 · 2018-03-25T17:19:35

#2sqrt(2i)#

#sqrt(8i)#

#sqrt(4*2*i) rarr# 4 is a perfect square; it can be taken out of the radical

#2sqrt(2i)#

Answer 2 · 2018-03-25T18:29:16

#sqrt(8i) = 2+2i#

Note that:

#8i = 8(cos (pi/2) + i sin (pi/2))#

Hence by de Moivre's theorem, one square root is:

#sqrt(8)(cos(pi/4)+isin(pi/4)) = 2sqrt(2)(1/sqrt(2)+1/sqrt(2)i) = 2+2i#

This is the principal square root. The other square root is #-2-2i#