z=1+i,那么z=1+iz=1+iz=1+i,那么
1.3 解方程组 ) 2(z)_(1)-(z)_(2)=i (1+i)(z)_(1)+i(z)_(2)=4-3i .
[单选题]复数z满足(1+i)z=1-i,则|Z|=()。A.√2/2B.√2C.2D.1
求下列函数的奇点:(1)dfrac (z+1)(z({z)^2+1)}; (2)dfrac (z+1)(z({z)^2+1)}求下列函数的奇点:(1)
直线-1=dfrac (y+3)(-2)=z-4与直线-1=dfrac (y+3)(-2)=z-4的夹角为( )-1=dfrac (y+3)(-2)=z-4-1
|z|lt dfrac (1)(4)-|||-(3) (z)=dfrac ({z)^-1-a}(1-a{z)^-1} , |z|gt a-|||-(4) (z)
考题9 设i是虚数单位,z表示复数z的共轭复数.若-|||-=1+i, 则 dfrac (z)(i)+icdot overline (z)= () .-|||-
[题目]如果复数z1,z2,z3满足等式 dfrac (({z)_(2)-(z)_(1))}(({z)_(3)-(z)_(1))}=dfrac (({z)_(1
(z)=(z)^2+dfrac (1)({z)^2-1},则其解析区域为()(z)=(z)^2+dfrac (1)({z)^2-1}(z)=(z)^2+dfra
(z)=(z)^2+dfrac (1)({z)^2+1},则其解析区域为( )(z)=(z)^2+dfrac (1)({z)^2+1}(z)=(z)^2+dfr