Author Topic: Graphs- Complex number help (Read 762 times) Tweet Share

Damo17 · « **on:** January 06, 2009, 09:20:05 pm »

I can do part a) i) and ii) but part b and c i don't know how to do. This is an analysis question from the chapter review of the second chapter of complex numbers. Any help would be much appreciated.

the answer to a- i) and ii) are:
i) $(x+1)^2+y^2=4$
ii) $(x-2)^2+(y+2)^2=72$

Mao · « **Reply #1 on:** January 06, 2009, 09:55:51 pm »

$(x+1)^2+y^2=4\implies 18(x+1)^2+18y^2=72=(x-2)^2+(y+2)^2$

$\implies 18x^2+18x+18+18y^2=x^2-4x+4+y^2+4y+4$

$\implies 17x^2+22x+10+17y^2-4y=0$

$\implies x^2+\frac{22}{17}x+\frac{10}{17}+y^2-\frac{4}{17}y=0$

$\implies \left(x+\frac{11}{17}\right)^2 + \frac{170}{289}-\frac{121}{289}+\left( y-\frac{2}{17}\right)^2 -\frac{4}{289}=0$

$\implies \left(x+\frac{11}{17}\right)^2 + \left( y-\frac{2}{17}\right)^2 =-\frac{45}{289}$

since there are two squares on the LHS (positive), the RHS cannot be negative. Hence, this relationship is false over $(x,y)\in \mathbb{R}^2$ , the graphs do not intersect.

TrueTears · « **Reply #2 on:** January 06, 2009, 10:14:28 pm »

hehe Mao i was gonna do exactly what u were doing but i stuffed my working in the middle and i thought i was wrong LOL

/0 · « **Reply #3 on:** January 06, 2009, 10:43:41 pm »

I was going for the discriminant being negative, but it seems Mao's method is more elegant.

Damo17 · « **Reply #4 on:** January 07, 2009, 08:29:09 am »

Thanks Mao, that is a very smart solution.

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Author Topic: Graphs- Complex number help (Read 762 times) Tweet Share

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