For the circle $|z - 1|=1$ show that $z=1+\cos\theta-i\sin\theta$, where $-\pi<\theta\le\pi$

Question

The question is:

a). For the circle $|z - 1|=1$, show that $z=1+\cos\theta+i\sin\theta$, where $-\pi<\theta\le \pi$

b). Deduce that the point representing the complex number $\frac{1}{z}$ describes a straight line.

My attempts:

a). I simply thought that, $z-1=r\cos\theta + ir\sin\theta$, with $r=1$ rearranging for $z$ gives

$z = 1+\cos\theta+i\sin\theta$.

Not sure if this is the right approach?

b). I used the identity, $\frac{1}{z}=\frac{\overline{z}}{|z|^2}$,

with $z=1+\cos\theta+i\sin\theta$,

and $|z|^2=(1+\cos\theta)^2+\sin^2\theta$

I got, $\frac{1}{z}=\frac{(1+\cos\theta)}{2+2\cos\theta}-\frac{i\sin\theta}{2+2\cos\theta}$

which simplifies to,

$\frac{1}{z}=\frac{1}{2}-\frac{i\sin\theta}{2+2\cos\theta}$

In this form I can't see how to show this is a straight line?

But, I was able to show that $\frac{1}{z}$ forms a straight line using cartesian form of complex numbers:

let $w=\frac{1}{z} \implies z=\frac{1}{w}$

With $z=x+iy$ and $w=u+iv$, I got $x+iy=\frac{1}{u+iv}=\frac{u-iv}{u^2+v^2}$

Equating real and imaginary parts:

$x=\frac{u}{u^2+v^2}$ , and $y=\frac{-v}{u^2+v^2}$

Then for the circle,

$|z-1|=1 \implies |(x-1) +y|=1$

Squaring and rearranging to get,

$$x^2-2x+y^2=0 \tag{1}$$

I then substitute $x=\frac{u}{u^2+v^2}$ , and $y=\frac{-v}{u^2+v^2}$ into this equation (1).

After a bit of algebra I ended up with,

$(u^2+v^2)(1-2u)=0 \implies (1-2u)=0 \implies u=\frac{1}{2}$

Thus, $\frac{1}{z}$ transforms the circle $|z-1|=1$ into a straingth line $u=\frac{1}{2}$

Why or how is it that I can't get to this result using $z=1+\cos\theta + i\sin\theta$ ?

Answer 1

For part b. You can further write the expression you found in the the first way as

$\begin{align}w&=u+iv=\frac1z\\ &=\frac12-i\frac{\sin\theta}{2+2\cos\theta}\\ &=\frac12-i\frac{2\sin(\frac\theta 2)\cos(\frac\theta 2)}{2+2(2\cos^2(\frac\theta 2)-1)}\\ &=\frac12-\frac i2\tan(\frac\theta 2) \end{align}$

This is the line $u=\frac12$ in $w$-plane, because given $v\in\Bbb R$ we can find $\theta\in(-\pi,\pi)$ such that $v=-\frac12\tan(\frac\theta 2)$.