Z1 + z2 = z1 + z2 kung at kung lamang kung arg (z1) = arg (z2), kung saan ang z1 at z2 ay mga kumplikadong numero. paano? pakipaliwanag!

Sagot:

Mabait sumangguni sa Usapan nasa Paliwanag.

Paliwanag:

Hayaan, # | z_j | = r_j; r_j gt 0 at arg (z_j) = theta_j sa (-pi, pi); (j = 1,2). #

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

Malinaw, # (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). #

Tandaan na, # 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) …. (bituin ^ 1) #.

# "Ngayon Given that," | z_1 + z_2 | = | z_1 | + | z_2 |, #

#api | (z_1 + z_2) | ^ 2 = (| z_1 | + | z_2 |) ^ 2 = | z_1 | ^ 2 + | z_2 | ^ 2 + 2 | z_1 || z_2 |, i.e., #.

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

Mula sa # (bituin ^ 1) at (bituin ^ 2) # makukuha natin, # 2r_1r_2cos (theta_1-theta_2) = r_1r_2. #

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

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

# "Ngunit," theta_1, theta_2 in (pi, pi), theta_1-theta_2 = 0, o, #

# theta_1 = theta_2, "giving," arg (z_1) = arg (z_2), # bilang ninanais!

Kaya, ipinakita namin na, # | z_1 + z_2 | = | z_1 | + | z_2 | rArr arg (z_1) = arg (z_2). #

Ang usap ay maaaring patunayan sa mga katulad na linya.

Tangkilikin ang Matematika.!