求解复数方程z∧2-4iz-4+9i=0 求大神帮忙

2025-06-21 15:02:44
推荐回答(1个)
回答1:

z=-3√2/2+(2+3√2/2)i 或3√2/2+(2-3√2/2)i。

设 z=a+bi

(z-2i)^2+9i=0

(a+(b-2)i)^2+9i=0

a^2-(b-2)^2+【2a(b-2)+9】i=0

a^2-(b-2)^2=0 a=b-2(舍) 或 a=2-b

2a(b-2)+9=0

(b-2)^2=9/2 b=2+3√2/2 或2-3√2/2

所以z=-3√2/2+(2+3√2/2)i 或3√2/2+(2-3√2/2)i

扩展资料

复数的四则运算

(a+bi)±(c+di)=(a±c)+(b±d)i

(a+bi)(c+di)=(ac-bd)+(ad+bc)i

(a+bi)/(c+di)=(ac+bd)/(c²+d²)+(bc-ad)i/(c²+d²)

r1(isina+cosa)r2(isinb+cosb)=r1r2[cos(a+b)+isin(a+b)]

r1(isina+cosa)/r2(isinb+cosb)=r1/r2[cos(a-b)+isin(a-b)]