Hvordan multipliserer du e ^ ((2 pi) / 3 i) * e ^ (pi / 2 i) i trigonometrisk form?

Svar:

#cos ((7pi) / 6) + ISIN ((7pi) / 6) = e ^ ((7pi) / 6i) #

Forklaring:

# E ^ (itheta) = cos (theta) + isin (theta) #

# E ^ (itheta_1) * e ^ (itheta_2) == cos (theta_1 + theta_2) + isin (theta_1 + theta_2) #

# Theta_1 + theta_2 = (2 pi) / 3 + pi / 2 = (7pi) / 6 #

#cos ((7pi) / 6) + ISIN ((7pi) / 6) = e ^ ((7pi) / 6i) #

Svar:

Svaret er # == - sqrt3 / 2 + 1 / 2i #

Forklaring:

En annen metode.

# I ^ 2 = -1 #

Eulers forhold

# E ^ (itheta) = costheta + isintheta #

Derfor, # E ^ (2 / 3pii) * e ^ (pi / 2i) = (cos (2 / 3n) + isin (2 / 3n)) (cos (pi / 2) + isin (pi / 2)) #

# = (1/2 + isqrt3 / 2) (0 + i) #

# = 1 / 2i-sqrt3 / 2 #

# = - sqrt3 / 2 + 1 / 2i #