complex numbers (1 Viewer)

[samuelclarke]

describe the locus of the point representing z on the argand diagram if lz²-(12i-5)l=lz-(2+3i)l given that root 12i-5= +- (2+3i) and solutions to z²+(2+i)z+2-2i=0 are i and -2-2i.

(this is from the fort street ext 2 2001 paper btw)

 

[barbernator]

circle with centre (-2,-3) and radius 1. post working in a sec

<a href="http://www.codecogs.com/eqnedit.php?latex=|z^2-(12i-5)|=|z-(2@plus;3i)|\\ |(z@plus;(2@plus;3i))(z-(2@plus;3i))|=|z-(2@plus;3i)|\\ |z@plus;(2@plus;3i)||z-(2@plus;3i)|=|z-(2@plus;3i)|\\ |z@plus;(2@plus;3i)|=1\\ |z-(-2-3i)|=1\\ therefore~we~have~a~circle~with~centre~(-2,-3)~with~radius=1" target="_blank"><img src="http://latex.codecogs.com/gif.latex?|z^2-(12i-5)|=|z-(2+3i)|\\ |(z+(2+3i))(z-(2+3i))|=|z-(2+3i)|\\ |z+(2+3i)||z-(2+3i)|=|z-(2+3i)|\\ |z+(2+3i)|=1\\ |z-(-2-3i)|=1\\ therefore~we~have~a~circle~with~centre~(-2,-3)~with~radius=1" title="|z^2-(12i-5)|=|z-(2+3i)|\\ |(z+(2+3i))(z-(2+3i))|=|z-(2+3i)|\\ |z+(2+3i)||z-(2+3i)|=|z-(2+3i)|\\ |z+(2+3i)|=1\\ |z-(-2-3i)|=1\\ therefore~we~have~a~circle~with~centre~(-2,-3)~with~radius=1" /></a>

I feel like I have made a mistake though because I didnt use the second piece of given info

can someone please confirm whether this is correct or not?

 

[samuelclarke]

ohh didn't see that thanks man!

 

[seanieg89]

The point (2,3) also satisfies the given conditions. So the correct solution is the circle of radius 1 about (-2,-3) and the point (2,3).

 

[barbernator]

The point (2,3) also satisfies the given conditions. So the correct solution is the circle of radius 1 about (-2,-3) and the point (2,3).

ahh yes good point missed that, also, is there an alternative solution to this that uses the other info, or is that just superfluous?

 

[bleakarcher]

he just used inspection.

 

[barbernator]

he just used inspection.

i mean this info and solutions to z²+(2+i)z+2-2i=0 are i and -2-2i.

 

[samuelclarke]

could i just ask why (2,3)m is also a considered centre?

 

[bleakarcher]

You mean why it is also considered a solution?
Because on the complex plane the point (2,3) represents the complex number 2+3i. Sub that into the equation and you have a valid statement meaning it is a solution.