Jak se dělí (-3-4i) / (5 + 2i) v trigonometrickém tvaru?

Odpovědět:

# 5 / sqrt (29) (cos (0.540) + isin (0.540)) ~ ~ 0.79 + 0.48i #

Vysvětlení:

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

# z = a + bi # lze psát jako # z = r (costheta + isintheta) #, kde

• # r = sqrt (a ^ 2 + b ^ 2) #
• # theta = tan ^ -1 (b / a) #

Pro # z_1 = 3 + 4i #:

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

# theta = tan ^ -1 (4/3) = ~ ~ 0,927 #

Pro # z_2 = 5 + 2i #:

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

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

Pro # 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 #

Důkaz:

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