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 $-2-2i$