Complex locus. (1 Viewer)

[tommykins]

|z^2 - z-^2| < 4
where z- is conjugate z.

An actual graph + shaded area would be sexy.

 

[namburger]

lool, i got ixy < 1 ...

 

[tommykins]

回复: Re: Complex locus.

yeah. i got

|ixy| < 1 as well, the issue here is what next?

If we square both sides, -x^2y^2 < 1
x^2y^2 > 1?

But my friend says you can't square the i as the | | is not abs value, its a modulus.

 

[namburger]

Re: 回复: Re: Complex locus.

yeah. i got

|ixy| < 1 as well, the issue here is what next?

If we square both sides, -x^2y^2 < 1
x^2y^2 > 1?

But my friend says you can't square the i as the | | is not abs value, its a modulus.

you dont square the i. Remember | x + iy | = sqrt (x^2 +y^2)

So in this case, sqrt (x^2y^2) < 1

| xy | < 1 ( note: | | is absolute value)
not to good on the abs value let me think about it haha

 

[lolokay]

Re: 回复: Re: Complex locus.

it would just be the area in between the hyerbolas for xy = 1 and xy = -1 wouldn't it?

 

[tommykins]

回复: Re: 回复: Re: Complex locus.

lolokay- if you don't square the i, then yes.

 

[namburger]

yep lolokay is correct

EDIT: from | xy | < 1
split the cases
xy < 1 or -xy < 1

Draw both xy = 1 and xy = -1 and test point to satisfy inequality?
Sorry if i get it wrong, im not good with abs values lol

 

[tacogym27101990]

ok this is my solution:
|4ixy|<4
|i|.|xy|<1
|xy|<1
so the locus is the region between the graphs xy=1 and xy=-1

EDIT: i see now theres been about 4 solutions

 

[munch0r]

Re: 回复: Re: Complex locus.

heres my graph

 

[tommykins]

回复: Re: 回复: Re: Complex locus.

thanks for repliess x]