Question

复数的指数形式是

Answer 1

复数的指数形式是：

证明方法就是把e^(iθ)和sinθ，cosθ展开成无穷级数，

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)!]+.........

扩展资料

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)/(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)]