COmplex Region (1 Viewer)

[RohitShubesh21]

|z^2 - zconjugate^2| >= 16

sorry I dont know how to make it look good, could i get sketch of locus please... my answer is -4i/x but not sure how to graph...

 

[ExtremelyBoredUser]

|z^2 - zconjugate^2| >= 16

sorry I dont know how to make it look good, could i get sketch of locus please... my answer is -4i/x but not sure how to graph...

Let z = x+iy

Lets just consider the modulus right now






Substituting back into the inequality





Edit: Mistake below, forgot case of modulus.

 

[RohitShubesh21]

Let z = x+iy

Lets just consider the modulus right now






Substituting back into the inequality




You should be good to go for graphing now!

ohokk thank u sir

 

[Trebla]

It's actually



so you'll need so consider the different cases for x and y.

 

[5uckerberg]

I would suggest you write the Q in latex, as it makes the question clear and that it resembles exactly the textbook. So instead of writing |z^2 - zconjugate^2| >= 16 you can write it as z^{2}-\bar{z}^{2} \geq{16} after you complete the left expression and note that the i gets removed in the process and is treated as one so that is why you have because the length of 4 is divided by both sides from

 

[icycledough]

|z^2 - zconjugate^2| >= 16

sorry I dont know how to make it look good, could i get sketch of locus please... my answer is -4i/x but not sure how to graph...

So as mentioned so far, the graph would be a hyperbole graph but on all 4 quadrants, because as Trebla said, there should be an absolute value. You then know that the shaded area will lie on the lines and outside, as with a hyperbole, anything above means greater.

 

[RohitShubesh21]

So as mentioned so far, the graph would be a hyperbole graph but on all 4 quadrants, because as Trebla said, there should be an absolute value. You then know that the shaded area will lie on the lines and outside, as with a hyperbole, anything above means greater.

yes thank you sirs