Question
Hi, there is a problem on my math book that has been giving me headaches. That's why I'm looking for help. The problem is:
Given z and w, two complex numbers such as |z+w|=|z-w|, prove that
Arg z - arg w = plusminus (pi/2)
Given z and w, two complex numbers such as |z+w|=|z-w|, prove that
Arg z - arg w = plusminus (pi/2)
Answers
Steve
In other words, the two vectors are perpendicular
If the two magnitudes are equal, then assuming that |z| >= |w|
|z| = √(a^2+b^2)
|w| = √(c^2+d^2)
z+w = (a+c)+(b+d)i
z-w = (a-c)+(b-d)i
√((a+c)^2+(b+d)^2) = √((a-c)^2+(b-d)^2)
(a+c)^2+(b+d)^2 = (a-c)^2+(b-d)^2
a^2+2ac+c^2 + b^2+2bd+d^2 = a^2-2ac+c^2 + b^2-2bd+d^2
2ac+2bd = -2ac-2bd
ac = -bd
a/b = -d/c
In other words, the slopes of the two vectors are negative reciprocals. The vectors are perpendicular.
If the two magnitudes are equal, then assuming that |z| >= |w|
|z| = √(a^2+b^2)
|w| = √(c^2+d^2)
z+w = (a+c)+(b+d)i
z-w = (a-c)+(b-d)i
√((a+c)^2+(b+d)^2) = √((a-c)^2+(b-d)^2)
(a+c)^2+(b+d)^2 = (a-c)^2+(b-d)^2
a^2+2ac+c^2 + b^2+2bd+d^2 = a^2-2ac+c^2 + b^2-2bd+d^2
2ac+2bd = -2ac-2bd
ac = -bd
a/b = -d/c
In other words, the slopes of the two vectors are negative reciprocals. The vectors are perpendicular.
Steve
sorry - forgot to eliminate my first line of thought; skip that stuff about |z| and |w|.
There's probably a nice complex-number way to do this, but this was my first approach.
There's probably a nice complex-number way to do this, but this was my first approach.