一.在复数集C中分解因式:
1) x^2+5=x^2-(-5)=(x+√5i)(x-√5i)
2) 2x^2-6x+5=2(x^2-3x+5/2)=2(x^2-3x+9/4+1/4)=2(x-3/2+i/2)(x-3/2-i/2)
3) x^2-2xcosα+1=x^2-2xcosα+cos^2α+sin^2α=(x-cosα+isinα)(x-cosα-isinα)
4) x^6-1=(x^3+1)(x^3-1)=(x+1)(x-1)(x^2+x+1)(x^2-x+1)=(x+1)(x-1)(x-(1+√3i)/2)(x-(1-√3i)/2)(x-(-1+√3i)/2)(x-(-1-√3i)/2)
二.解下列方程:
1) x^4+24i=0
x^4-12(1-i)^2=0
(x^2+2√3(1-i))(x^2-2√3(1-i))=0
(x+r(icosθ+sinθ))(x-r(icosθ+sinθ))(x+r(cosθ-isinθ))(x-r(cosθ-isinθ))=0
于是得到四个解.其中,r=4√24（四次根号24）,θ=π/8,cosθ=√(2+√2)/2,sinθ=√(2-√2)/2.
2) (x+1)^9=(1+i)^9
先分解因式x^9-y^9=(x-y)(x^2+xy+y^2)(x^6+x^3y^3+y^6),然后用前两个式子可以得到三个根
x=i
x=(-3-i+√3(-1+i))/2
x=(-3-i-√3(-1+i))/2
至于剩下六个根繁得出奇,略.
三.巳知复数Z=x+yi(x,y全属于R),求下列各式的实部与虚部:
1) Z^2=(x^2-y^2)+2xyi
2) Z^3=(x^3-3xy^2)+(3x^2y-y^3)i
3) 1/Z=1/(x+yi)=(x-yi)/(x^2+y^2)=x/(x^2+y^2)-yi/(x^2+y^2)
4) V0Z+(M/2π)*1/Z=V0x+V0yi+Mx/2π(x^2+y^2)-Myi/2π(x^2+y^2)
=[V0x+Mx/2π(x^2+y^2)]+[V0y-Myi/2π(x^2+y^2)]i
从以上结果很容易得出实部和虚部.
四.巳知(x+yi)^3=a+bi,这里a,b,x,y全属于R,求证
(a/x)+(b/y)=4(x^-y^2)
证明：由上题可知,(x+yi)^3=(x^3-3xy^2)+(3x^2y-y^3)i,根据复数相等的定义,a=(x^3-3xy^2),b=(3x^2y-y^3).
所以(a/x)+(b/y)
=(x^2-3y^2)+(3x^2-y^2)
=4(x^-y^2).