Question

两个复数商的模及辐角与被除数和除数的模和辐角关系

Answer 1

设复数z1=r1(cosa+isina),z2=r1(cosb+isinb)(|z1|=r1,|z2|=r2,z1辐角为a,z2辐角为b),

则z1/z2=r1(cosa+isina)/[r1(cosb+isinb)]=(r1/r2)(cosa+isina)/(cosb+isinb)

=(r1/r2)(cosa+isina)(cosb-isinb)/[(cosb+isinb)(cosb-isinb)]

=(r1/r2)[(cosacosb+sinasinb)+(sinacosb-cosasinb)i]/[(cosb)^2+(sinb)^2]

=(r1/r2)[cos(a-b)+isin(a-b)],z1/z2的辐角为a-b,

|z1/z2|=|(r1/r2)[cos(a-b)+isin(a-b)]|=|r1/r2|√{[cos(a-b)]^2+[sin(a-b)]^2},

=|r1/r2|=|z1|/|z2|,

两个复数商的模=模的商,两个复数商辐角=被除数的辐角-除数的辐角