locus of complex numbers

IntegralAnomaly

can some1 just verify this for me,is the locus of arg(z+i/z+2)=pi/3 this:

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Reply 1

elpaw

15

IntegralAnomaly

can some1 just verify this for me,is the locus of arg(z+i/z+2)=pi/3 this:

shouldnt it be (0,-2) and (0,-1) ?

Reply 2

IntegralAnomaly

OP

oops the question is arg(z-i/z+2)=pi/3.

Reply 3

IntegralAnomaly

OP

anyone?

Reply 4

Willa

11

if that's supposed to be a circle, then the angle certainly isnt pi/3

and i think the two points of interest are (-2,0) and (0,-1) because you have a -i involved in there....i cant remember anything else, sorry

Reply 5

IntegralAnomaly

OP

Willa

if that's supposed to be a circle, then the angle certainly isnt pi/3

and i think the two points of interest are (-2,0) and (0,-1) because you have a -i involved in there....i cant remember anything else, sorry

no its ment to be an arc of a circle.
If arg(z-i/z+2)=pi/3 then arg(z-i)-arg(z+2)=pi/3.So the arguement of z-i minus the arguement of z+2 is always pi/3.

Reply 6

Squishy

11

Um, I get a Cartesian equation of the locus as:

x^2 + 2x + x/sqrt(3) + y^2 - y - 2y/sqrt(3) + 2/sqrt(3) = 0

Now, this can be rearranged to show it's a circle, and it does go through the point (0, 1), but from what I can tell, the centre and radius of the circle involves horrible multiples of sqrt(3) and in particular, it cuts the x-axis at some peculiar number.

What I've said is probably bollocks, but my maths isn't too hot right now. :smile:

locus of complex numbers

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