locus/complex numbers (1 Viewer)

[M@C D@DDY]

can u guyz tell me when you get a qs such as

Arg(z-2i)-Arg(z+2i)=pi/2

how do u no which side the semi-circle is on
i no there's sumfing abt testin the points

fanx!!

 

[:: ck ::]

when its less than pi/2 its a major arc circle
greater than pi/2 its a minor arc circle
0 is two lines going outside like this <---. .---->
pi is a straight line between 2 pts


to tell which side the circle lies on.. take the pt represented by the numerator ... and create an angle with an line from the other point... in the anticlockwise direction

ur example : Arg(z-2i)-Arg(z+2i)=pi/2
simplify : Arg(z-2i)/Arg(z+2i)=pi/2

take the point 0,2 (pt representing z-2i) ... create an angle in anticlowise direction with the pt 0,-2 ... locus shud b a semicircle on the left side of the im axis, not including the points 0,2 and 0,-2

 

[M@C D@DDY]

i dun fink i get u sorri.....

 

[:: ck ::]

take the pt represented in the numerator... form an angle in the anticlockwise direction with the pt represented in the denominator.....

that is how i think of it anyway... =\

 

[Teoh]

form an angle in the anticlockwise direction with the pt represented in the denominator.....

so its always anti clock wise, from the deductee to the deductor (Making sense... )???

 

[:: ck ::]

yep.. if the angle is positive

if its negative.. go clockwise

thats just the way i do it.. i don't know if it will make sense to anyone though.. haha.. it gives the right answer thats all i care

 

[M@C D@DDY]

oh dat makes it so easi fanx m8

 

[Grey Council]

argh, mac daddy, can you please please please post in proper english! lol, it hurts my eyes to read all that, your breaking the arms and legs of our beloved language.

please?

 

[ND]

You should go through why it's a semicircle rather than just memorizing that something in that form is a semi-circle.

 

[:: ck ::]

hehe because angle formed by diameter of circle to circumference is a right angle?

 

[M@C D@DDY]

yeh that's right but aren't all of these types of questions a semi-circle??

 

[:: ck ::]

nop ... refer to my top post