Question

Комплексные числа:

Решить уравнение
$z^2=i$

Answer 1

$z^{2}=i\\(a+bi)^{2}=i\\a^{2}+2abi+(bi)^{2}=i\\a^{2}+2abi+b^{2}i^{2}=i\\a^{2}+2abi+b^{2}(-1)=i\\a^{2}+2abi-b^{2}=i\\(a^{2}-b^{2})+2abi-i=0\\(a^{2}-b^{2})+(2ab-1)i=\\(a^{2}-b^{2})+(2ab-1)i=0+0i \\ \\ a^{2}-b^{2}=0\\a^{2}=b^{2}\\a=\pm\sqrt{b^{2}}\\a=\pm b \\ \\ 2ab-1=0\\2b*b-1=0\\2b^{2}-1=0\\2b^{2}=1\\b^{2}=1/2\\b=\pm\sqrt{1/2}\\b=\pm\sqrt{2}/2 \\ \\a=\pm\sqrt{2}/2 \\ z=(\sqrt{2}/2)+(\sqrt{2}/2)i \\ z=-(\sqrt{2}/2)-(\sqrt{2}/2)i$