0
$\begingroup$

The transformation at $T$ given by $w=kz/(i+z)$ where $z\neq -i$, $k$ a real number, maps the complex number $2+i$ in the $z$-plane to its image $\frac{(3-i)}{2}$ in the $w$-plane.

a) Show that $k=2$

Point $P$ represents the complex number $z$ where $|z|=\sqrt{3}$. $T$ maps the point $P$ to point $Q$ in the w-plane.

b) Show that the locus of $Q$ is a circle with the cartesian equation given by: $(u-3)^2+v^2=3$ for $u, v\in\mathbb{R}$.

$T$ maps the point $z_0$ in the locus of $P$ to the point $w_0$ in the locus of $Q$, where the acute angle $\arg w_0$ is as large as possible.

c) Find the exact value of $|i+z_0|$

I did part (a) easily, but I couldn't do part (b). So far I did this: $$ iw=z(2-w)\Rightarrow z=iw/(2-w)\Rightarrow |z|=\left|\frac{iw}{2-w}\right|\Rightarrow \sqrt{3}=\frac{|iw|}{|(2-w)|}$$

$\endgroup$
3
  • $\begingroup$ This is rather obnoxious to read. You don't need to put every instance of math in double dollar signs. You should only put major equations in double dollars. Small equations should be enclosed in single dollar signs. $\endgroup$ Commented Dec 28, 2014 at 11:42
  • $\begingroup$ Sorry, I'm new to this. I'll edit it. $\endgroup$ Commented Dec 28, 2014 at 11:43
  • $\begingroup$ I've already edited it. Please read meta.math.stackexchange.com/questions/5020/… $\endgroup$
    – hjhjhj57
    Commented Dec 28, 2014 at 11:45

1 Answer 1

1
$\begingroup$

Assume $z$=$x+iy$ then $|z|$=$x^2+y^2$=$3$

Now you apply the T on z.

Which gives $$w=\frac{2(x+iy)}{i+x+iy}$$ then make the denominator real by multiplying it with the complex conjugate. Then once you sorted the resulting equation into real and imaginary parts, equate u to the Real part and v to the imaginary part. Then substitute in the given equation $$(u-3)^2+v^2=3$$

Finally, evaluate the equation it should give you $x^2+y^2=3$ which is true. Hence QED.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .