# Thread: Complex Numbers - Loci

1. ## Complex Numbers - Loci

Hi, I have two questions pertaining to loci of complex numbers and was hoping you guys could help me out here.

Essentially the first part of the problem calls for the sketching of $\displaystyle {|z-(1+i)|}^2 = 2$. I'm guessing it's a circle centred upon $\displaystyle 1+i$ with a radius of $\displaystyle \sqrt{2}$, but to do that would require taking roots on both sides of the equation: which would cause the right-hand-side to become $\displaystyle \pm \sqrt{2}$. Is it legitimate to just dispose of the negative root by reasoning that the length must be a non-negative value? Or is there something in the loci that I'm missing out?

The next part of the question involves the locus implied by the equation $\displaystyle {|z-(1-i)|}^2 = 2 {|z-1|}^2$. Provided taking roots on both sides as above is allowed, I understand from earlier questions that this defines a circle of Apollonius, with a fixed non-1 ratio between the points $\displaystyle 1-i$ and 1. The question then calls for the demonstration that this loci is the same as that for the first part ($\displaystyle {|z-(1+i)|}^2 = 2$), which is where I'm stuck. So the question is: given the locus of a circle of Apollonius in this format, is there a formula to find the centre of the circle described?

Essentially, I suppose the key thrust of the second part would be on how to convert $\displaystyle | \frac{z-(1-i)}{z-1}|$ into $\displaystyle |z-(1+i)|$. I have roughly managed to do it by using Cartesian coordinates (i.e. let $\displaystyle z=x+iy$) but I was wondering if there was a quicker or more elegant way to get to it.

Sorry for the very long question, I hope this is expressed clearly. >.< Thanks for all help in advance, it is greatly appreciated!

2. Originally Posted by synapse
Hi, I have two questions pertaining to loci of complex numbers and was hoping you guys could help me out here.

Essentially the first part of the problem calls for the sketching of $\displaystyle {|z-(1+i)|}^2 = 2$. I'm guessing it's a circle centred upon $\displaystyle 1+i$ with a radius of $\displaystyle \sqrt{2}$, but to do that would require taking roots on both sides of the equation:

Why? Says who? If you want to see what's going on, with this and other similar problems, put $\displaystyle z=x+yi$ and apply definitions:

$\displaystyle 2=|x+yi-1-i|^2=|(x-1)+(y-1)i|^2=(x-1)^2+(y-1)^2$ . and there we have the standard equation in the xy-plane $\displaystyle \sim \mathbb{C}$ of a nice circle of radius $\displaystyle \sqrt{2}$

and center $\displaystyle (1,1)$.

Tonio

which would cause the right-hand-side to become $\displaystyle \pm \sqrt{2}$. Is it legitimate to just dispose of the negative root by reasoning that the length must be a non-negative value? Or is there something in the loci that I'm missing out?

The next part of the question involves the locus implied by the equation $\displaystyle {|z-(1-i)|}^2 = 2 {|z-1|}^2$. Provided taking roots on both sides as above is allowed, I understand from earlier questions that this defines a circle of Apollonius, with a fixed non-1 ratio between the points $\displaystyle 1-i$ and 1. The question then calls for the demonstration that this loci is the same as that for the first part ($\displaystyle {|z-(1+i)|}^2 = 2$), which is where I'm stuck. So the question is: given the locus of a circle of Apollonius in this format, is there a formula to find the centre of the circle described?

Essentially, I suppose the key thrust of the second part would be on how to convert $\displaystyle | \frac{z-(1-i)}{z-1}|$ into $\displaystyle |z-(1+i)|$. I have roughly managed to do it by using Cartesian coordinates (i.e. let $\displaystyle z=x+iy$) but I was wondering if there was a quicker or more elegant way to get to it.

Sorry for the very long question, I hope this is expressed clearly. >.< Thanks for all help in advance, it is greatly appreciated!
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