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The questions are to show $|z-w|=k$ defines a circle and find the centre and radius. $|z|=x+iy$ and $|w|=a+bi$.

So what I did was substitute the complex numbers in so $|x+iy-a-bi|=k$ eventually getting to $(x-a)^2+(y-b)^2=k$. Using the general equation for a circle, you can deduce that the center of a circle $(h,k)$ would in this case be $(a,b)$ and that $k=r^2$.

However, I'm confused as to how to state the radius because of the plus-minus signs and stuff. If I were to solve for r, it would be $r=\pm\sqrt{k}$ but I don't know if that's right because the radius is just the distance from the centre so it can't be negative. I'm really stuck on what I think is probably a really easy concept so I would appreciate some help.

anna
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C92wQQ
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    More context might help. If it is $z$ that is intended to vary, then $w$ is the center of the circle and its radius is $k$. But the assumed premises here are not clear from the stated information. – leslie townes Feb 14 '21 at 05:56
  • By arbitrary convention, $r$ is never negative. Further, this problem is best attacked with a preliminary intuition-only step. Construe $(z-w)$ to be a complex number $z_1$. Then, the constraint amounts to $|z_1| = k.$ Note, that as stated, the problem is ambiguous, because it is unclear which of $z,w$ is to be construed as a variable and which is to be construed as a fixed value. – user2661923 Feb 14 '21 at 05:59
  • @KaviRamaMurthy "...and $r^2 = k$..." No. The first paragraph of the query gives $|z-w|=k \implies r = k$, rather than $r = \sqrt{k}.$ The confusion is because the OP's subsequent analysis mistakenly concludes that $k = r^2.$ – user2661923 Feb 14 '21 at 07:25
  • @user2661923 wait why isn't k = r^2? If you put (x-a)^2+(y-b)^2=k equivalent to (x-h)^2+(y-k)^2=r^2 wouldn't k=r^2 thus r=√k ? – C92wQQ Feb 14 '21 at 08:02
  • The easiest way to see it is to pretend that $w$ is the fixed value $(0 + i[0])$ and consider the resultant meaning of $|z| = k$ (from the 1st line of your query). Normally, when $(x + iy) = z = re^{i\theta}$, then $\sqrt{x^2 + y^2} = |z| = r.$ Consider this against the backdrop of $|z| = k$, when $w = 0.$ Another way of saying the same thing is that $r$ is normally used to represent the distance of $z$ from the origin. – user2661923 Feb 14 '21 at 08:50
  • Still another way of saying the same thing is that when $z$ is represented by $re^{i\theta}$, then the polar equation $r = k$ would represent all complex numbers that are a distance of $k$ from the origin. That is, all complex numbers that are on the circle of radius $k$, that is centered at the origin. – user2661923 Feb 14 '21 at 08:53
  • As the point Z varies on the complex plane according to $$\left|z-w\right|=k,$$ it traces out a circle centred at the fixed point $W$ and of radius $k\in\mathbb{R}:;$

    if $z=(x+y,i)$ and $w=(a+b,i),$ then $$\left|z-w\right|=k\ \iff \left|(x-a)+(y-b)i\right|=k\ \iff \sqrt{(x-a)^2+(y-b)^2}=k\ \iff (x-a)^2+(y-b)^2=k^2.$$

    There are three careless/conceptual mistakes in the OP, all involving moduli of complex numbers; your confusion likely stems from those.

     – ryang Feb 14 '21 at 11:14

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