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Could anyone help me with this question, I get a different answer to the textbook.

Question The unit circle $\left|z\right|=1$ in the z-plane is transformed to the w-plane by the transformation $w=\frac{z}{z-2}$. Determine the locus of w.

My Answer

Let $z = x + iy$ and $w = u+iv$ then we have:

$u+iv=\frac{x+iy}{x+iy-2}=\frac{x+iy}{x+iy-2}*\frac{x-iy-2}{x-iy-2}=\frac{x^2+y^2-2x-2yi}{x^2+y^2-4x+4}$

But from $\left|z\right|=1$ we have that $x^2+y^2 = 1$

and so we have:

$u+iv=\frac{1-2x-2yi}{5-4x}$

Therefore $u=\frac{1-2x}{5-4x}$ and $v=\frac{-2y}{5-4x}$

And so, from the first of these, $x=\frac{1-5u}{2-4u}$

And substituting this value for x into the second and rearranging gives, $y=\frac{-3v}{2-4u}$

Now, using $x^2+y^2=1$ we obtain:

$\frac{(1-5u)^2}{(2-4u)^2} + \frac{9v^2}{(2-4u)^2} = 1$

So already I can see that as there is no term in v, other than the $v^2$ term, that this is going to be a circle with centre of y-ordinate 0, which is different from the books answer.

Pushing on I rearrange this to give:

$(u+\frac{1}{3})^2+v^2=\frac{10}{9}$

and so I have a circle centre $(-\frac{1}{3},0)$ radius $\frac{\sqrt{10}}{3}$

Book Answer

Circle centre $(-\frac{1}{3},1)$ radius $\frac{2}{3}$

Thanks for any help, Mitch.

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2 Answers 2

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You might find it easier to rearrange the equation as $$z=\frac{2w}{w-1}$$ and then apply the condition $$|z|=1\Rightarrow 2|w|=|w-1|$$.

This leads to the equation of the circle $$3u^2+3v^2+2u-1=0$$

This has centre $(-\frac 13,0)$ and radius $\frac 23$, so there appears to be a misprint in the book.

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  • $\begingroup$ Thanks for that. I agree the book is wrong and also see my mistake. Solved. $\endgroup$
    – MAry
    Commented Jun 15, 2016 at 13:45
  • $\begingroup$ Your comment implies that you have found David's answer useful. If that's the case you should upvote and check it. You should also upvote other answers according to their helpfulness so that we can discern whether you've got it. $\endgroup$
    – user127032
    Commented Jun 15, 2016 at 15:23
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If you do it the way you started, you have to push all the way for both $x$ and $y$:

$$ \begin{align} u&=\Re\left(\frac{x+yi}{x+yi-2}\right)\\ v&=\Im\left(\frac{x+yi}{x+yi-2}\right) \end{align} $$

getting:

$$ \begin{align} u&=\frac{x(x-2)+y^2}{(x-2)^2+y^2}\text{ (1)}\\ v&=\frac{y(x-2)-xy}{(x-2)^2+y^2} \end{align} $$

System (1) has the solution:

$$ \begin{align} x&=\frac{2(u^2+v^2-u)}{u^2+v^2-2u+1}\text{ (2)}\\ y&=\frac{-2v}{u^2+v^2-2u+1} \end{align} $$

Now push $x$ and $y$ from (2) into equation $x^2+y^2=1$ to get:

$$\frac{4(u^2+v^2)}{u^2+v^2-2u+1}=1$$

which if you simplify, will give you the equation:

$$3u^2+2u+3v^2=1$$

which is the same as that in the answer of David.

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