Find the locus of $w=1/z$ when $z$ lies on $y=2x+1$

Question

For the transformation $w=\frac{1}{z}$, find the locus of $w$ when $z$ lies on the line with equation $y=2x+1$

I'm not quite sure how or where to start on this question. I know for instance that the locus of $z$ such that

$$|z-2| = |z+2-2i|$$

is the line $y=2x+1$. I'm not sure where I could go from here though

Thanks in advance

Answer 1

If $z = t+i(2t+1)$ then $$ \frac{1}{z} = \frac{1}{t+i(2t+1)} = \frac{t-i(2t+1)}{5t^2+4t+1}. $$ So your locus is parametrized by $$ (u,v) = \left(\frac{t}{5t^2+4t+1},\frac{-2t-1}{5t^2+4t+1}\right) $$ where $w = u+iv$. Let's try to find an equation for the image of this parametrization. It looks like $$ u^2+v^2 = \frac{t^2+(-2t-1)^2}{(5t^2+4t+1)^2} = \frac{1}{5t^2+4t+1}, $$ and $$ 2u+v = \frac{-1}{5t^2+4t+1}. $$ So every point in the image of the parametrization satisfies $$ u^2+v^2+2u+v=0. $$ This can be rewritten as $(u+1)^2+(v+\frac{1}{2})^2=\frac{5}{4}$, a circle of radius $\sqrt{5}/2$.

The image of the parametrization isn't the whole circle; it's the whole circle minus one point. I will leave it to you to figure out which point is missing and why. (The point will be filled in if we let $t = \infty$, speaking informally.) (Which might be a bit of a hint for finding which point is missing, I guess.)

Answer 2

Suppose $z=x+iy$ then as $y=2x+1$ we have $w=\dfrac{1}{x +(2x+1)i}$

$\dfrac{1}{x +(2x+1)i}=\dfrac{x-(2x+1)i}{5 x^2+4 x+1}$

If $w=a+bi$ then we have $(a,b)=\left(\dfrac{x}{5 x^2+4 x+1},\dfrac{-2 x-1}{5 x^2+4 x+1}\right)$

So

$a=\dfrac{x}{5 x^2+4 x+1};\;b=\dfrac{-2 x-1}{5 x^2+4 x+1}$

$x=\dfrac{1-4a\pm\sqrt{-4 a^2-8 a+1}}{10 a}$

plugged in $b$ gives

$b=\dfrac{1}{2} \left(-1\pm\sqrt{1-4 a (a+2)}\right)$

$2b=-1\pm\sqrt{1-4 a (a+2)};\to (2b+1)^2=1-4 a (a+2)$

$a^2 + b^2+2a-b=0$

which is the equation of a circle. When $z$ varies on the blue line, $w$ varies on the orange circle

Hope it helps

remark

There are two points which are invariant

$\left(-\dfrac{4}{5}, -\dfrac{3}{5}\right),( 0,1)$

Answer 3

Let $\,z = x + i(2x + 1)=(1+2i)x + i \,\mid\, x \in \mathbb{R}\,$, then $\bar z = (1-2i)x - i\,$. Eliminating $\,x\,$ between the two gives the equation in complex numbers for the line $\,z\,$ lies on:

$$ (2+i)\,z \,+\, (2-i)\,\bar z + 2 \,=\, 0 $$

Dividing by $\,2z \bar z\,$ and substituting $\displaystyle\,\frac{1}{z}=w\,$:

$$ (1-i/2)\,w \,+\, (1+i/2)\,\bar w \,+\, w \,\bar w \color{red}{+(1+i/2)(1-i/2)-(1+i/2)(1-i/2)}\,=\, 0 \\[5px] \iff \quad (w+1+i/2)\,(\bar w + 1 - i/2) \,=\, |1+i/2|^2 \\[5px] \iff\quad \left|w+1+\frac{i}{2}\right|^2 = \frac{5}{4} $$

The latter is the equation of a circle of radius $\displaystyle\,\frac{\sqrt{5}}{2}\,$ centered at $\displaystyle\,-1-\frac{i}{2}\,$.