Miten jaat (-3-4i) / (5 + 2i) trigonometrisessä muodossa?

Miten jaat (-3-4i) / (5 + 2i) trigonometrisessä muodossa?
Anonim

Vastaus:

# 5 / sqrt (29) (cos (0,540) + isin (0,540)) ~~ 0,79 + 0.48i #

Selitys:

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

# Z = a + bi # voidaan kirjoittaa # Z = r (costheta + isintheta) #, missä

  • # R = sqrt (a ^ 2 + b ^ 2) #
  • # Theta = tan ^ -1 (b / a) #

varten # Z_1 = 3 + 4i #:

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

# Theta = tan ^ -1 (4/3) = ~~ 0927 #

varten # Z_2 = 5 + 2i #:

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

# Theta = tan ^ -1 (2/5) = ~~ 0,381 #

varten # 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 #

Todiste:

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