Z1 + z2 = z1 + z2, če in samo če je arg (z1) = arg (z2), kjer sta z1 in z2 kompleksna števila. kako? prosim razloži!

Odgovor:

Prosimo, da se sklicujete na Diskusija v Razlaga.

Pojasnilo:

Pusti, # | z_j | = r_j; r_j gt 0 in arg (z_j) = theta_j v (-pi, pi; (j = 1,2).

#:. z_j = r_j (costheta_j + isintheta_j), j = 1,2.

Jasno je, # (z_1 + z_2) = r_1 (costheta_1 + isintheta_1) + r_2 (costheta_2 + isintheta_2), #

# = (r_1costheta_1 + r_2costheta_2) + i (r_1sintheta_1 + r_2sintheta_2).

Spomnimo se, da # z = x + iy rArr | z | ^ 2 = x ^ 2 + y ^ 2. #

#:. | (z_1 + z_2) | ^ 2 = (r_1costheta_1 + r_2costheta_2) ^ 2 + (r_1sintheta_1 + r_2sintheta_2) ^ 2, #

# = r_1 ^ 2 (cos ^ 2theta_1 + sin ^ 2theta_1) + r_2 ^ 2 (cos ^ 2theta_2 + sin ^ 2theta_2) + 2r_1r_2 (costheta_1costheta_2 + sintheta_1sintheta_2)

# = r_1 ^ 2 + r_2 ^ 2 + 2r_1r_2cos (theta_1-theta_2), #

#rArr | z_1 + z_2 | ^ 2 = r_1 ^ 2 + r_2 ^ 2 + 2r_1r_2cos (theta_1-theta_2) …. (zvezda ^ 1) #.

# "Zdaj glede na to," | z_1 + z_2 | = | z_1 | + | z_2 |, #

#iff | (z_1 + z_2) | ^ 2 = (| z_1 | + | z_2 |) ^ 2 = | z_1 | ^ 2 + | z_2 | ^ 2 + 2 | z_1 || z_2 |.

# | (z_1 + z_2) | ^ 2 = r_1 ^ 2 + r_2 ^ 2 + 2r_1r_2 ……. (zvezda ^ 2).

Od # (zvezda ^ 1) in (zvezda ^ 2) # dobimo, # 2r_1r_2cos (theta_1-theta_2) = r_1r_2.

# "Preklic" r_1r_2 gt 0, cos (theta_1-theta_2) = 1 = cos0. #

#:. (theta_1-theta_2) = 2kpi + -0, k v ZZ.

# "Ampak," theta_1, theta_2 v (pi, pi), theta_1-theta_2 = 0 ali, #

# theta_1 = theta_2, "dajanje," arg (z_1) = arg (z_2), # kot želeno!

Tako smo dokazali, da

# | z_1 + z_2 | = | z_1 | + | z_2 | rArr arg (z_1) = arg (z_2).

The pogovor mogoče dokazati na podobnih linijah.

Uživajte v matematiki!