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

Answer 2

круто , я понял )