Let,
#|z_j|=r_j; r_j gt 0 and arg(z_j)=theta_j in (-pi,pi]; (j=1,2).#
#:. z_j=r_j(costheta_j+isintheta_j), j=1,2.#
Clearly, #(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).#
Recall that, #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)....(star^1)#.
#"Now Given that, "|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|, i.e.,#.
# |(z_1+z_2)|^2=r_1^2+r_2^2+2r_1r_2.......(star^2).#
From #(star^1) and (star^2)# we get,
# 2r_1r_2cos(theta_1-theta_2)=r_1r_2.#
#"Cancelling "r_1r_2 gt 0, cos(theta_1-theta_2)=1=cos0.#
#:. (theta_1-theta_2)=2kpi+-0, k in ZZ.#
#"But, "theta_1, theta_2 in (pi,pi], theta_1-theta_2=0, or,#
#theta_1=theta_2," giving, "arg(z_1)=arg(z_2),# as desired!
Thus, we have shown that,
#|z_1+z_2|=|z_1|+|z_2| rArr arg(z_1)=arg(z_2).#
The converse can be proved on similar lines.
Enjoy Maths.!
