해법
을 위해 해결하다 x,fsif(x)=x2−3x+2
해법
x=2−81+1+16s2f2f(x)2sff(x)+3−81+1+16s2f2f(x)2+−81+1+16s2f2f(x)2i,x=−2−81+1+16s2f2f(x)2sff(x)−3−81+1+16s2f2f(x)2−−81+1+16s2f2f(x)2i,x=2−81−1+16s2f2f(x)2sff(x)+3−81−1+16s2f2f(x)2+−81−1+16s2f2f(x)2i,x=−2−81−1+16s2f2f(x)2sff(x)−3−81−1+16s2f2f(x)2−−81−1+16s2f2f(x)2i
솔루션 단계
fsif(x)=x2−3x+2
대체 x=a+bifsif(x)=(a+bi)2−3(a+bi)+2
(a+bi)2−3(a+bi)+2 확장 :(a2−b2−3a+2)+i(−3b+2ab)
fsif(x)=(a2−b2−3a+2)+i(−3b+2ab)
다시 쓰다 fsif(x) 표준복합형태로: 0+fsf(x)i0+fsf(x)i=(a2−b2−3a+2)+i(−3b+2ab)
복소수는 실수 부분과 허수 부분이 같을 때만 같을 수 있다방정식으로 다시 쓰시오:[0=a2−b2−3a+2fsf(x)=−3b+2ab]
[0=a2−b2−3a+2fsf(x)=−3b+2ab]:a=2−81+1+16s2f2f(x)2sff(x)+3−81+1+16s2f2f(x)2,a=−2−81+1+16s2f2f(x)2sff(x)−3−81+1+16s2f2f(x)2,a=2−81−1+16s2f2f(x)2sff(x)+3−81−1+16s2f2f(x)2,a=−2−81−1+16s2f2f(x)2sff(x)−3−81−1+16s2f2f(x)2,b=−81+1+16s2f2f(x)2b=−−81+1+16s2f2f(x)2b=−81−1+16s2f2f(x)2b=−−81−1+16s2f2f(x)2
뒤로 대체 x=a+bix=2−81+1+16s2f2f(x)2sff(x)+3−81+1+16s2f2f(x)2+−81+1+16s2f2f(x)2i,x=−2−81+1+16s2f2f(x)2sff(x)−3−81+1+16s2f2f(x)2−−81+1+16s2f2f(x)2i,x=2−81−1+16s2f2f(x)2sff(x)+3−81−1+16s2f2f(x)2+−81−1+16s2f2f(x)2i,x=−2−81−1+16s2f2f(x)2sff(x)−3−81−1+16s2f2f(x)2−−81−1+16s2f2f(x)2i