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Homework Help: Smallest Argument

  1. Jan 30, 2004 #1
    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
    sqrt(a^2+(b+8)^2)=4 (radius)
    a^2+b^2+16b+64=16 (eq 1)
    And from 4-8-sqrt(48) triangle
    a^2+b^2=48 (eq 2)
    Place into eq 1
    Placing into eq 2
    From the drawing, the smallest argument is at the third zone.
    tan theta=-6/-sqrt(12)
    so theta=240 degrees.

    There must be a shorter solution, can you please help me?
  2. jcsd
  3. Jan 30, 2004 #2


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    Saying that |z+8i|= 4 is the same as saying (geometrically, in the complex plane) that the distance from z to -8i is 4. That is all z satisfying that are on a circle with center at -8i and radius 4.

    The argument of a complex number is the angle the line from 0 to the number makes with the real axis. In this case it is geometrically clear that that will happen when the line from 0 to the circle is tangent to the circle (in the first quadrant). That tangent line, the line from 0 to -8i and the line from -8i to the point on the circle make a right triangle with hypotenuse of length 8 and one leg of length 4. (edited: I just realized that the problem only asks for the argument, not for the actual number itself.)

    The angle from the (negative) imaginary axis is given by sin(φ)= 4/8= 1/2 and so φ= 30 degrees. The argument is 270- 30= 240 degrees just as you got.
    Last edited by a moderator: Jan 30, 2004
  4. Jan 31, 2004 #3
    Knew these.
    BLAAAAH!!! 4*2=8 I shoulda seen that!!!
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