complex numbers (1 Viewer)

samuelclarke

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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

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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?
 
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seanieg89

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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

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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

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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.
 

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