- #1
kishtik
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What is the argument of the complex number z which has the smallest argument in |z+8i|=4?
I solved the problem correctly but my answer is rather long
|z-(-8i)|=4 (drawing)
z_0=a+bi, z_1=-a+bi
|a+bi+8i|=4
sqrt(a^2+(b+8)^2)=4 (radius)
a^2+b^2+16b+64=16 (eq 1)
And from 4-8-sqrt(48) triangle
sqrt(a^2+b^2)=sqrt(48)
a^2+b^2=48 (eq 2)
Place into eq 1
48+16b+64=16
16b=-96
b=-6
Placing into eq 2
a^2+36=48
a^2=12
a=+-sqrt(12)
z_0=sqrt(12)-6i
z_1=-sqrt(12)-6i
From the drawing, the smallest argument is at the third zone.
tan theta=-6/-sqrt(12)
=sqrt(3)
so theta=240 degrees.
There must be a shorter solution, can you please help me?
I solved the problem correctly but my answer is rather long
|z-(-8i)|=4 (drawing)
z_0=a+bi, z_1=-a+bi
|a+bi+8i|=4
sqrt(a^2+(b+8)^2)=4 (radius)
a^2+b^2+16b+64=16 (eq 1)
And from 4-8-sqrt(48) triangle
sqrt(a^2+b^2)=sqrt(48)
a^2+b^2=48 (eq 2)
Place into eq 1
48+16b+64=16
16b=-96
b=-6
Placing into eq 2
a^2+36=48
a^2=12
a=+-sqrt(12)
z_0=sqrt(12)-6i
z_1=-sqrt(12)-6i
From the drawing, the smallest argument is at the third zone.
tan theta=-6/-sqrt(12)
=sqrt(3)
so theta=240 degrees.
There must be a shorter solution, can you please help me?