你如何以三角形式划分(-3-4i)/(5 + 2i)?

你如何以三角形式划分(-3-4i)/(5 + 2i)?
Anonim

回答:

#5 / SQRT(29)(COS(0.540)+ ISIN(0.540))~~ 0.79 + 0.48i#

说明:

#( - 3-4i)/(5 + 2I)= - (3 + 4I)/(5 + 2I)#

#Z = A + BI# 可写成 #Z = R(costheta + isintheta)#,哪里

  • #R = SQRT(A ^ 2 + B ^ 2)#
  • #THETA =黄褐色^ -1(B / A)#

对于 #Z_1 = 3 + 4I#:

#R = SQRT(3 ^ 2 + 4 ^ 2)= 5#

#THETA =黄褐色^ -1(4/3)= ~~ 0927#

对于 #Z_2 = 5 + 2I#:

#R = SQRT(5 ^ 2 + 2 ^ 2)= sqrt29#

#THETA =黄褐色^ -1(2/5)= 0.381 ~~#

对于 #Z_1 / Z_2#:

#Z_1 / Z_2 = R_1 / R_2(COS(theta_1-theta_2)+ ISIN(theta_1-theta_2))#

#Z_1 / Z_2 = 5 / SQRT(29)(COS(0.921-0.381)+ ISIN(0.921-0.381))#

#Z_1 / Z_2 = 5 / SQRT(29)(COS(0.540)+ ISIN(0.540))= 0.79 + 0.48i#

证明:

# - (3 + 4I)/(5 + 2I)*(5-2i)/(5-2i)= - (15 + 20 1-6I + 8)/(25 + 4)=(23 + 14 1)/ 29 = 0.79 + 0.48i#