Re: HSC 2016 Complex Numbers Marathon
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Consider (1 + i(x - y)) (1+ i(x + y)) (x2 - y2 - 1 + 2ix) and expanding
= (1 + 2ix - (x2) - y2 - 1)) ((x2 - y2 - 1 + 2ix))
= -4x2 - (x2 - y2 - 1)^2
= some negative real.
Now we use our first hypothesis:
x > √(y2 + 1 (thus x >= 1) . . . (*)
x2 > y2 + 1
x2 -y2 > 1 . . . (**)
(x+y)(x-y) > 1
Consider the graph for (**), then for all x > 1, there is both +y and -y corresponding to each x and |y| does not exceed x otherwise hypothesis (*) is violated. That means we have (x + |y|) > 0.
That means the complex numbers (1 + i(x + y)) and (1+ i(x - y)) are either both in the 1st quadrant, by (***). Hence their arguments are arctan(x + y) and arctan(x - y) respectively.
By (*) and (**), we have that the complex number (x2 - y2 - 1 + 2ix) must lie in the first quadrant. Thus, we can use the arctan(b/a) formula to compute the argument.
It's time to run home. Take argument of both sides of the very first equation
arg((1 + i(x - y)) (1+ i(x + y)) (x2 - y2 - 1 + 2ix)) = arg(some negative real)
arg((1 + i(x - y)) + arg(1+ i(x + y)) + arg(x2 - y2 - 1 + 2ix)) = pi
arctan(x + y) + arctan(x - y) + arctan( 2x / (x2 - y2 - 1)) = pi
Lol, I was to slow...
