Bagaimanakah anda membahagikan (i + 2) / (9i + 14) dalam bentuk trigonometri?

Bagaimanakah anda membahagikan (i + 2) / (9i + 14) dalam bentuk trigonometri?
Anonim

Jawapan:

# 0.134-0.015i #

Penjelasan:

Untuk nombor kompleks # z = a + bi # ia boleh diwakili sebagai # z = r (costheta + isintheta) # di mana # r = sqrt (a ^ 2 + b ^ 2) # dan # theta = tan ^ -1 (b / a) #

(2 + i) / (14 + 9i) = (sqrt (2 ^ 2 + 1 ^ 2) (cos (tan ^ -1 (1/2)) (/ cos (tan ^ -1 (9/14)) + isin (tan ^ -1 (9/14)))) ~~ (sqrt5 (cos (0.46) + isin (0.46))) / (sqrt277 (cos (0.57) + isin (0.57))) #

Diberikan # z_1 = r_1 (costheta_1 + isintheta_1) # dan # 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 #

Bukti:

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