复数指数形式
回答
爱扬教育
2022-06-16
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扩展资料
e^(iθ)=1+iθ+(iθ)^2/2!+(iθ)^3/3!+........(iθ)^k/k!+..........
sinθ=θ-θ^3/3!+θ^5/5!+..............+(-1)^(k-1) [θ^(2k-1)/(2k-1)!]+.........
cosθ=1-θ^2/2!+θ^4/4!+...........+(-1)^(k-1) [θ^(2k)/(2k)!]+.........
三角函数
sin(a+bi)=sin(a)cos(bi)+sin(bi)cos(a)
=sin(a)cosh(b)+isinh(b)cos(a)
cos(a-bi)=cos(a)cos(bi)+sin(bi)sin(a)
=cos(a)cosh(b)+isinh(b)sin(a)
tan(a+bi)=sin(a+bi)/cos(a+bi)
cot(a+bi)=cos(a+bi)/sin(a+bi)
sec(a+bi)=1/cos(a+bi)
csc(a+bi)=1/sin(a+bi)
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)/(c+d)+(bc-ad)i/(c+d)
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)]