Complex numbers - $z^{2} = 4i$

Question

$z^{2} = 4i$

$z'=4i=4(\cos{\pi/2}+i\sin{\pi/2})$

$\sqrt{z'}=2\left[\cos\left(\dfrac{\pi/2+2k\pi}{2}\right)+i\sin\left(\dfrac{\pi/2+2k\pi}{2}\right)\right]$

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I got from it

$z_{0} = \sqrt{2}+i\sqrt{2}$

$z_{0} = -\sqrt{2}+i\sqrt{2}$

And I don't know why in the 2nd one there should 2 minuses.

Answer 1

Because $$\cos(180^{\circ}+\alpha)=-\cos\alpha$$ and $$\sin(180^{\circ}+\alpha)=-\sin\alpha$$ and in $z_0$ you got two pluses.

Answer 2

with $$z=x+iy$$ we get $$x^2-y^2+2xyi=4i$$ so we get $$x^2-y^2=0$$ and $$2xy=4$$