Complex numbers help with a question.

z^2 = -5 + 12i
= +- (2 + 3i)

i dont get it !! can you recommend me some useful links

If you square a complex number a + bi, you get
a^2 + 2ab i + b^2 i^2
a^2 + 2ab i + b^2 (-1)
a^2 - b^2 + 2ab i

So you're looking for a and b where a^2 - b^2 = -5 and 2ab = 12.

Solve 2ab = 12 for b to get b = 6/a

a^2 - (6/a)^2 = -5
a^2 - 36 / a^2 = -5
(a^4 - 36) / a^2 = -5
a^4 - 36 = -5 a^2
a^4 + 5 a^2 - 36 = 0
(a^2 - 9)(a^2 - 4) = 0
a = +/- 3, +/- 2

However, by multiplying by a^2, we introduced two extraneous solutions. Go back to the original equation a^2 - (6/a)^2 = -5 and you will see that only a = +/- 2 actually works.

b = 6/a = +/- 3

So a + bi = +/- (2 + 3i)

z = a + bi
z^2 = (a+ bi)^2 = a^2 + 2ab i + b^2 .i^2= a^2 - b^2 + 2abi = -5 + 12 i
--> a^2 - b^2 = -5 and 2ab i = 12i----> a = (6/b)
a^2 - b^2 = -5 = (6/b)^2 - b^2 = 12 ---> b^4 -5b^2 - 36 = 0; take x = b^2
then x^2 - 5x - 36 = 0; x = ( 5 + - sqrt(25 + 4.36))/2 = (5 +- 13)/2 = -4 or 9
for x = 9 = b^2 , so b = + - 3
for b = 3 ; a = 6/3 = 2 ( z1 = 2 + 3i )
for b = -3; a = (-6/3) = -2 ( z2 = -2 - 3i )

for x = -4 = b^2 we obtain the same values ​​of z
solution z = + or - ( 2 + 3i)