什么是-8-i的三角形式?

什么是-8-i的三角形式?
Anonim

回答:

# - (8 + I)~~ -sqrt58(COS(0.12)+ ISIN(0.12))#

说明:

#-8-1 = - (8 + I)#

对于给定的复数, #Z = A + BI#, #Z = R(costheta + isintheta)#

#R = SQRT(A ^ 2 + B ^ 2)#

#THETA =黄褐色^ -1(B / A)#

我们来处理吧 #8 + I#

#Z = 8 + 1 = R(costheta + isintheta)#

#R = SQRT(8 ^ 2 + 1 ^ 2)= sqrt65#

#THETA =黄褐色^ -1(1/8)0.12 ~~ ^ C#

# - (8 + I)~~ -sqrt58(COS(0.12)+ ISIN(0.12))#