Miten jaat (i + 2) / (9i + 14) trigonometrisessä muodossa?

Miten jaat (i + 2) / (9i + 14) trigonometrisessä muodossa?
Anonim

Vastaus:

# 0.134-0.015i #

Selitys:

Monimutkainen numero # Z = a + bi # se voidaan esittää # Z = r (costheta + isintheta) # missä # R = sqrt (a ^ 2 + b ^ 2) # ja # Theta = tan ^ -1 (b / a) #

# (2 + i) / (14 + 9 i) = (sqrt (2 ^ 2 + 1 ^ 2) (cos (tan ^ -1 (1/2)) + isin (tan ^ -1 (1/2)))) / (sqrt (14 ^ 2 + 9 ^ 2) (cos (tan ^ -1 (9/14)) + isin (tan ^ -1 (9/14)))) ~~ (sqrt5 (cos (0,46) + isin (0,46))) / (sqrt277 (cos (0,57) + isin (0,57))) #

tietty # Z_1 = r_1 (costheta_1 + isintheta_1) # ja # Z_2 = R_2 (costheta_2 + isintheta_2) #, # Z_1 / z_2 = r_1 / R_2 (cos (theta_1-theta_2) + isin (theta_1-theta_2)) #

# Z_1 / z_2 = sqrt5 / sqrt277 (cos (0,46-0,57) + isin (0,46-0,57)) = sqrt1385 / 277 (cos (-0,11) + isin (-0,11)) ~~ sqrt1385 / 277 (0.99-0.11i) ~~ 0.134-0.015i #

Todiste:

# (2 + i) / (14 + 9 i) * (14-9i) / (14-9i) = (28-4i + 9) / (14 ^ 2 + 9 ^ 2) = (37-4i) / 277 ~~ 0.134-0.014i #