Find the set of points belonging to the coordinate plane

Sovardipk

Sovardipk

Answered question

2022-07-04

Find the set of points belonging to the coordinate plane xy, for which the real part of the complex number ( 1 + i ) z 2 is positive.

Answer & Explanation

eurgylchnj

eurgylchnj

Beginner2022-07-05Added 14 answers

Step 1
After having cos ( 2 θ + π 4 ) > 0 , you have
π 2 < ( 2 θ + π 4 ) < π 2
which is incorrect.
To make it easy to understand why this is incorrect, let α := 2 θ + π 4 .
Then, we want to solve
cos α > 0 and π θ < π , ,
i.e.
cos α > 0 and 7 4 π α < 9 4 π
which sould be easier to solve, to have
7 4 π α < 3 2 π or π 2 < α < π 2 or 3 2 π < α < 9 4 π , ,
i.e.
π θ < 7 8 π or 3 8 π < θ < π 8 or 5 8 π < θ < π
as always more elegant solutions are welcome
(not sure if this is more elegant, but) another solution :
Let z = x + i y where x , y R . Then,
(1) ( ( 1 + i ) z 2 ) = ( ( 1 + i ) ( x + i y ) 2 ) = x 2 y 2 2 x y
When we solve y 2 + 2 x y x 2 = 0 for y, we get
y = x ± x 2 + x 2 = x ± 2   x = ( ± 2 1 ) x
so from (1),
( ( 1 + i ) z 2 ) > 0 ( ( 2 1 ) x y ) ( ( 2 + 1 ) x + y ) > 0 ( 2 + 1 ) x < y < ( 2 1 ) x or ( 2 1 ) x < y < ( 2 + 1 ) x
Ximena Skinner

Ximena Skinner

Beginner2022-07-06Added 7 answers

Step 1
I would find the general solutions of the inequation first:
cos ( 2 θ + π 4 ) > 0 π 2 < 2 θ + π 4 < π 2 3 π 4 < 2 θ < π 4 mod 2 π 3 π 8 < θ < π 8 mod π
Now that if conventionally, we choose π < θ π , we obtain
3 π 8 < θ < π 8 , π < θ < 7 π 8 , 5 π 8 < θ < π .

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